\frontmatter \title \mbox\chancery F A N C y [2mm] \chancery\Large an online documentation project on [3mm] \chancery \underlineFunctional \underlineAnalysis and \underlineNon-\underlineCommutative Geometr\underliney \addcontentslinetocchapterPreliminaries \addcontentslinetocsectionTitlepage \authorFounded by \scshape Markus J.~Pflaum \maketitle \phantomsection \mainmatter \partFundamentals \setcounterchapter0
In this section we collect several inequalities from real analysis which will
be of use later in this monograph.
Theorem 1.1.1[Young's inequality]
Let a,b \geq 0, and assume that p,q > 1 satisfy the relation \frac 1p + \frac 1q =1. Then
ab \leq \frac 1p a^p + \frac 1q b^q \: .
Equality holds if and only if a^p = b^q.
Proof.
Since the second derivative \exp'' of the exponential function attains
only positive values, the function \exp is strictly convex that means satisfies
\exp \big( \lambda x + (1-\lambda) y \big) \leq
\lambda \exp ( x ) + (1-\lambda) \exp ( y )
for all x,y\in \mathbb{R} and \lambda \in [0,1] with equality holding true if and only if x = y or \lambda \in \{ 1,0 \}.
Putting x = p \ln a, y = q \ln b, and \lambda = \frac 1p one obtains
ab = \exp\big( \lambda x + (1-\lambda) y \big) \leq
\lambda \exp ( x ) + (1-\lambda) \exp ( y ) = \frac 1p a^p + \frac 1q b^q \: .
Equality holds if and only if x = y which is equivalent to a^p = b^q.
Theorem 1.1.2[Cauchy--Schwarz inequality for sums]
Let v,w \in \mathbb{C}^n. Then
\Big| \sum_{i=1}^n v_i \overline{w_i} \: \Big|^2 \leq \Big( \sum_{i1}^n |v_i|^2 \Big) \Big( \sum_{i=1}^n |w_i|^2 \Big).
Equality holds true if and only if v and w are linearly dependant.
Proof.
Let us use the inner product notation
\langle v,w\rangle := \sum_{i=1}^n v_i \overline{w_i} \quad \text{for } v,w\in \mathbb{C}^n.
Then the \ell^2-norm
\| v \| := \left( \sum_{i=1}^n |v_i|^2\right)^{1/2} = \langle v , v \rangle^{1/2}
is well-defined and non-negative for any v\in \mathbb{C}^n. If \|v \| =0 or \| w \|= 0, then v=0 or w=0, and
the claim is trivial. So we assume \|v \|, \| w \| > 0 and compute
\tag{1.1.1}
0 \leq \, \big\langle \|w\| v - \|v\| w , \|w\| v - \|v\| w \big\rangle =
\sum_{i=1}^n \big( \|w\| v_i - \|v\| w_i \big)\big( \|w\| \overline{v_i} - \|v\| \overline{w_i} \big) =
= \, \sum_{i=1}^n \|w\|^2 v_i\overline{v_i} - \|w\| \|v\| v_i \overline{w_i} - \|w\| \|v\| w_i \overline{v_i}
+ \|v\|^2 w_i\overline{w_i} =
= \, 2 \|v\|\|w\| \Big( \|v\|\|w\| - \Re \langle v,w \rangle \Big) .
Now choose c \in \mathbb{C} with |c|=1 such that c \langle v,w \rangle = |\langle v,w \rangle|. Replacing v by cv
in inequality \eqrefeq:inequality-chain and observing that \|cv \| and \| w \| are positive then entails
0 \leq \|c v\|\|w\| - \Re \langle c v,w \rangle = \|v\|\|w\| - \Re (c \langle v,w \rangle) =
\|v\|\|w\| - |\langle v,w \rangle|,
which is the claimed Cauchy--Schwartz inequality for sums in abbreviated form.
Equality holds true if and only if \|w\| c v - \|v\| w =0. So if \|v\|\|w\| = |\langle v,w \rangle| ,
then v and w are linearly dependant. To show the converse, assume that av =bw for some a,b\in \mathbb{C}
with (a,b) \neq (0,0). Because we consider the nontrivial case where both v and w are nonzero, we can
assume without loss of generality that b=1. But then
|\langle v,w \rangle |= |\langle v, av \rangle | = |a| \|v\|^2 = \|v\| \, \|w\| ,
hence equality holds in this case. The proof is finished.
1.1.3
Besides the \ell^2-norm on \mathbb{C}^n one has the so-called \ell^p-norms
\| \cdot \| : \mathbb{C}^n \to \mathbb{R}_{\geq 0} for p \geq 1.
They are defined by
\| v \|_p = \left( \sum_{k=1}^n |v_k|^p \right)^{1/p} \quad \text{for } v \in \mathbb{C}^n .
The maximum norm or \ell^\infty-norm \| \cdot \|_\infty is given by
\| v \|_\infty = \sup \big\{ |v_k| \bigm\vert k = 1,\ldots , n \big\} .
The \ell^p-norms are all norms indeed as we will later see.
Theorem 1.1.4[H\"older's inequality for sums]
Let p,q \in [1, \infty ) such that \frac 1p + \frac 1q =1. Then
\sum_{k=1}^n | v_k w_k | \leq \| v \|_p \cdot \| w\|_q \quad \text{for all } v,w\in \mathbb{C}^n .
Proof.
If p=1 or q=1 the claim is immediate, because then q=\infty or p=\infty, respectively,
and the two estimates
\sum_{k=1}^n | v_k w_k | \leq \left( \sum_{k=1}^n | v_k | \right) \cdot
\sup\big\{ |w_k| \bigm\vert k = 1,\ldots , n \big\}
and
\sum_{k=1}^n | v_k w_k | \leq \left( \sum_{k=1}^n | w_k | \right) \cdot
\sup \big\{ |v_k| \bigm\vert k = 1,\ldots , n \big\}
obviously hold. So we can assume 1 < p,q < \infty. Moreover we can assume that both v and w
are nonzero because otherwise the claim is trivial. Now observe that by Young's inequality
\frac{|v_k|}{\| v \|_p} \cdot \frac{|w_k|}{\| w \|_q} =
\left(\frac{|v_k|^p}{\| v \|_p^p} \right)^{1/p} \cdot \left(\frac{|w_k|^q}{\| w \|_q^q} \right)^{1/q}
\leq \frac 1p \frac{|v_k|^p}{\| v \|_p^p} + \frac 1q \frac{|w_k|^q}{\| w \|_q^q} \quad
\text{for } k=1,\ldots , n.
Summing over all k gives
\sum_{k=1}^n \frac{|v_k|}{\| v \|_p} \cdot \frac{|w_k|}{\| w \|_q} \leq
\frac 1p \frac{\|v\|_p^p}{\| v \|_p^p} + \frac 1q \frac{\|w\|_q^q}{\| w \|_q^q} =
\frac 1p + \frac 1q = 1 .
Multiplication of both sides by \| v \|_p \cdot \| w\|_q entails H\"older's inequality.
Theorem 1.1.5[Minkowski's inequality for sums]
Let p \in [1, \infty ). Then
\| v + w \|_p \leq \| v\|_q + \| w \|_p \quad \text{for all } v,w\in \mathbb{C}^n .
Proof.
For p=1 the claim is trivial, likewise for p=\infty. So assume 1 < p < \infty
and put q := \frac{p}{p-1}. Then \frac 1p + \frac 1q =1, and we can apply
H\"older's inequality to compute
\| v + w \|_p^p\, = \sum_{k=1}^n | v_k + w_k|^p \leq
\sum_{k=1}^n |v_k| \, | v_k + w_k|^{p-1} + |v_k| \, | v_k + w_k|^{p-1}
\leq
\leq \| v \|_p \cdot \left( | v_k + w_k|^{(p-1)q} \right)^{1/q} +
\| w \|_p \cdot \left( | v_k + w_k|^{(p-1)q} \right)^{1/q} =
= \left( \| v \|_p + \| w \|_p \right) \, \| v + w \|_p^{p/q} .
Minkowski's inequality follows.
\addcontentslinetocsubsectionDefinitions and first examples
Definition 3.1.1
Let \Omega be a set. By an algebra on \Omega one understands a collection \mathscr{A} of subsets of \Omega
or in other words an element \mathscr{A}\in \mathscr{P}(\Omega) such that
If in addition
then the algebra \mathscr{A} is called a \sigma-algebra.
A set \Omega equipped with a \sigma-algebra \mathscr A is called a
measurable space. The elements of \mathscr A are termed the
measurable subsets of \Omega.
Proposition 3.1.2
If \mathscr{A} is an algebra on a set \Omega, the empty set and the intersection of finitely many
elements of \mathscr{A} lies in \mathscr{A}.
If \mathscr{A} is a \sigma-algebra on \Omega, then the intersection of countably many
measurable sets is also measurable.
Proof.
These facts follow immediately from the axioms and the set-theoretic de Morgan's laws.
Examples 3.1.3
Remark 3.1.4
Obviously, the set of algebras on a set \Omega and the set of \sigma-algebras on \Omega are both ordered
by set-theoretic inclusion. When talking about a ``smaller`` \sigma-algebra or a ``largest'' one we always
implicitely mean in regard to set-theoretic inclusion as underlying order relation.
The following two results are extremely useful when constructing examples.
Proposition 3.1.5
Let (\mathscr{A}_i)_{i\in I} a family of algebras on a set \Omega.
Then the intersection
\mathscr{A} = \bigcap\limits_{i\in I} \mathscr{A}_i is an algebra on \Omega.
If each of the \mathscr{A}_i is a \sigma-algebra, then \mathscr{A} is so, too.
Proof.
Assume first that each \mathscr{A}_i is an algebra on \Omega.
Obviously, \Omega \in \mathscr{A} because \Omega \in \mathscr{A}_i for all i\in I.
Similarly, if A \in \mathscr{A}, then A \in \mathscr{A}_i, hence \complement A \in \mathscr{A}_i
for all i\in I. Therefore \complement A is in the intersection
\mathscr{A} = \bigcap\limits_{i\in I} \mathscr{A}_i. Now assume that (A_k)_{k=1}^n is a finite
sequence of sets belonging to \mathscr{A}. Then A_k \in \mathscr{A}_i for k=1,\ldots,n and all
i\in I which entails that \bigcup\limits_{k=1}^n A_k is in each of the \mathscr{A}_i, hence in the
intersection \mathscr{A}. The latter argument also works under the condition that each \mathscr{A}_i is a \sigma-algebra
to verify that for a sequence (A_k)_{k\in\mathbb{N}} in \mathscr{A} the union \bigcup\limits_{k\in\mathbb{N}} A_k
is in \mathscr{A}. So the proposition is proved.
Corollary 3.1.6
Let \mathscr F be a collection of subsets of a set \Omega. Then there is a unique smallest \sigma-algebra,
\sigma (\mathscr F), on \Omega containing \mathcal F. It is called the \sigma-algebra generated by
\mathscr F.
Proof.
Let {\mathcal M} be the family of all \sigma-algebras which contain \mathscr F. The set of all subsets of \Omega is
certainly a \sigma-algebra, so {\mathcal M} \neq \emptyset. Let \sigma (\mathscr F) be the intersection of all
\sigma-algebras in the family {\mathcal M}. By the preceding proposition \sigma (\mathscr F) is a \sigma-algebra.
Since every element of {\mathcal M} contains \mathscr F, the intersection
\sigma (\mathscr F) = \bigcap\limits_{\mathscr{A}\in \mathcal{M}}\mathscr{A} contains \mathscr F as well. By construction,
\sigma (\mathscr F) is minimal with that property.
Example 3.1.7
Let X be a topological space. The \sigma-algebra, \mathscr B (X), generated by all open subsets of X is called the
\em Borel \sigma-algebra on X. Its elements are the \em Borel measurable sets or simply the \em Borel sets of X.
Obviously all open and all closed sets of X are Borel measurable, as are all countable unions of closed sets and countable
intersections of open sets.
Example 3.1.8
Definition 3.1.9
Let \Omega be a measurable space, and Y a topological space. A map f\colon \Omega \rightarrow Y is termed \em measurable if the set f^{-1}[U] is measurable for every open subset U\subseteq Y.
Proposition 3.1.10
Let f\colon \Omega \rightarrow Y be a measurable function. Then the inverse image f^{-1}[B] is measurable whenever B\subseteq Y is a Borel set.
Proof.
Let \mathcal M be the collection of all subsets E\subseteq Y such that the inverse image f^{-1}[E] \subseteq \Omega is measurable. It is easy to check the axioms required to show that \mathcal M is a \sigma-algebra.
Since the function f is measurable, the \sigma-algebra \mathcal M contains all open sets of Y, and therefore all Borel sets. Thus, by definition of the \sigma-algebra \mathcal M, the set f^{-1}[B] is measurable whenever B is a Borel set.
Definition 3.1.11
Let f\colon X\rightarrow Y be a mapping between topological spaces. If f is measurable with respect to the \sigma-algebra of all Borel sets in X, then we call f a Borel function.
Thus a function is a Borel function if the inverse image of any open set is a Borel set. In particular, any continuous function is a Borel function.
Proposition 3.1.12
Let f\colon \Omega \rightarrow X be a measurable function, and let g\colon X\rightarrow Y be a Borel function. Then the composite g\circ f \colon \Omega \rightarrow Y is measurable.
Proof.
Let U\subseteq Y be an open set. Then the inverse image g^{-1}[U] is a Borel set, so by proposition 3.1.10 the inverse image (g\circ f)^{-1}[U] is measurable.
As a corollary, the composite of a measurable and a continuous function is measurable.
Example 3.1.13
Let X be a measure space, and let E\subseteq X be a measurable set. Then the function
\chi_E \colon X\rightarrow {\mathbb C} given by the formula
\chi_E (x) = \left\{ \begin{array}{ll} 1 & x\in E \\
0 & x\not\in E \\
\end{array} \right.
is measurable.
The function \chi_E is called the \em characteristic function of E.
\addcontentslinetocsubsectionAlgebras of real and complex valued Borel measurable functions
Proposition 3.1.14
Let u,v\colon X\rightarrow {\mathbb R} be measurable functions, and let \Phi \colon {\mathbb R}^2 \rightarrow Y be continuous. Define a function h\colon X\rightarrow Y by the formula
h(x) = \Phi (u(x),v(x))
Then the function h is measurable.
Proof.
Define f\colon X\rightarrow {\mathbb R}^2 by the formula f(x) = (u(x),v(x)). In view of proposition 3.1.12 it suffices to prove that the function f is measurable. Observe that:
f^{-1}((a,b)\times (c,d)) = u^{-1}(a,b)\cap v^{-1}(c,d)
so the inverse image f^{-1}((a,b)\times (c,d)) is measurable since u and v are measurable functions.
But every open set U\subseteq {\mathbb R}^2 is a countable union of rectangles of the form (a,b)\times (c,d). The \sigma-algebra axioms thus ensure that the inverse image f^{-1}[U] is measurable whenever U\subseteq {\mathbb R}^2 is an open set.
The above proposition and proof still function if the function \Phi is a Borel function rather than a continuous function.
Corollary 3.1.15
Let f\colon X\rightarrow {\mathbb C} be a function on a measurable space X. Then the function f is measurable if and only if the functions \Re (f) and \Im (f) are measurable.
Proof.
Let u,v\colon X\rightarrow {\mathbb R} be measurable functions. Define \Phi \colon {\mathbb R}^2\rightarrow {\mathbb C} by the formula \Phi (x,y) = x+iy. Then the function u+iv is measurable by proposition 3.1.14.
The converse follows immediately from proposition 3.1.12 since the functions \Re and \Im are continuous.
Corollary 3.1.16
Let f,g\colon X\rightarrow {\mathbb C} be measurable functions. Then the functions f+g and fg are measurable.
Proof.
In view of corollary 3.1.15 it suffices to prove this result for real-valued measurable functions. If we define continuous functions \Phi_1,\Phi_2 \colon {\mathbb R}^2\rightarrow {\mathbb R} by the formulae \Phi_1 (s,t)=s+t and \Phi_2 (s,t)=st, then the result follows immediately from proposition 3.1.14
Proposition 3.1.17
Let f\colon X\rightarrow {\mathbb C} be a measurable function. Then the function |f| is measurable, and there is a measurable function \alpha \colon X\rightarrow {\mathbb C} such that |\alpha (x)| =1 for all x\in X, and f=\alpha |f|.
Proof.
Let E = \{ x\in X |f(x)=0 \}. Then the set E is the inverse image of a closed subset, and so measurable. We can define a continuous function \varphi \colon {\mathbb C}\backslash \{ 0\} \rightarrow {\mathbb C} by the formula \varphi (z) = z/|z|. It follows from example 3.1.13, corollary 3.1.16, and proposition 3.1.12 that the function \alpha \colon X\rightarrow {\mathbb C} defined by the formula
\alpha (x) = \varphi (f(x) + \chi_E (x))
is measurable. The formulae |\alpha (x)| =1 and f=\alpha |f| are easy to check.
\addcontentslinetocsubsectionMeasurable functions to the extended real line
Definition 3.1.18
Let (a_n) be a sequence of real numbers. Then we define
{\lim \sup}_{n\rightarrow \infty} a_n = {\lim \sup}_{n\rightarrow \infty} \{ a_n , a_{n+1} , a_{n+2} , \ldots \}
and
{\lim \inf}_{n\rightarrow \infty} a_n = {\lim \inf}_{n\rightarrow \infty} \{ a_n , a_{n+1} , a_{n+2} , \ldots \}
We can pass from results about \lim \sup to results about \lim \inf, or conversely, by the observation
{\lim \sup}_{n\rightarrow \infty} a_n = - {\lim \inf}_{n\rightarrow \infty} (-a_n)
It will occasionally be convenient to us to allow \infty and -\infty as values of limits and functions. This is a safe enough option provided we do not attempt to do arithmetic with these symbols; for example, expressions such as `\infty - \infty' are completely meaningless.
However, we can form `intervals'
[a,\infty ] = [a ,\infty )\cup \{ \infty \} \qquad [\infty ,b] = (\infty ,b]\cup \{ \infty \}
and so on. These intervals are topological spaces. We can also allow ourselves the inequality
-\infty < a < \infty
for all a\in {\mathbb R}. The standard result about \lim \sup and \lim \inf can now be expressed quite simply; although a number of special cases need to be examined in the proof.
Theorem 3.1.19
Let (a_n) be a real-valued sequence. Then the limits
{\lim \inf}_{n\rightarrow \infty} a_n \in [-\infty , \infty ) \qquad {\lim \sup}_{n\rightarrow \infty} a_n \in (-\infty , \infty ]
exist and satisfy the inequality
{\lim \inf}_{n\rightarrow \infty} a_n \leq {\lim \sup}_{n\rightarrow \infty} a_n
Further, the equality
{\lim \inf}_{n\rightarrow \infty} a_n = a = {\lim \sup}_{n\rightarrow \infty} a_n
holds precisely when the sequence (a_n) converges to the real number a.
proof to be filled in!
Note that the number a in the above result must be finite.
Proposition 3.1.20
Let \Omega be a measurable space, and let f\colon \Omega \rightarrow [\infty , \infty] be any map. Suppose that the inverse image f^{-1}((\alpha , \infty ]) is measurable for every point \alpha \in {\mathbb R}. Then the function f is measurable.
Proof.
Let
{\mathcal M} = \{ E\subseteq [-\infty ,\infty]|f^{-1}[E] \textrm{ is measurable } \}
By proposition 3.1.10 the set \mathcal M is a \sigma-algebra. Choose points \alpha \in {\mathbb R} and \alpha_n < \alpha such that \lim_{n\rightarrow \infty} \alpha_n = \alpha. Since the set (\alpha_n , \infty ] is measurable by hypothesis, and
[-\infty , \alpha ) = \bigcup_{n=1}^\infty [-\infty , \alpha_n ] = \bigcup_{n=1}^\infty [-\infty , \infty ]\backslash (\alpha_n , \infty]
it follows that [-\infty , \alpha )\in \Omega. Hence
(\alpha , \beta ) = [-\infty , \beta ) \cap (\alpha , \infty ] \in \Omega
for every point \alpha , \beta \in {\mathbb R}. Since every open set in [-\infty , \infty ] is a countable union of such open intervals, the collection \mathcal M contains every open set. Thus the map f is measurable.
Corollary 3.1.21
Let f_n\colon X\rightarrow [-\infty , \infty] be measurable functions for n\in {\mathbb N}. Then the functions
\sup \{ f_n \} \quad {\lim \sup}_{n\rightarrow \infty} f_n \quad \inf \{ f_n \} \quad {\lim \inf}_{n\rightarrow \infty} f_n
are measurable.
Proof.
Let a\in {\mathbb R}. Observe that the set
(\sup \{ f_n \})^{-1} (a,\infty ] = \bigcup_{n=1}^\infty f_n^{-1}(a,\infty ]
is measurable. Hence by the above proposition, the function \sup \{ f_n \} is measurable. The formula \inf \{ f_n \} = - \sup \{ -f_n \} tells us that the function \inf \{ f_n \} is also measurable.
Now, for each point x\in \Omega, the sequence of numbers
g_n (x) = \sup \{ f_n (x) , f_{n+1} (x) , f_{n+2}(x) , \ldots \}
is monotonic increasing. It follows that
{\lim \sup }_{n\rightarrow \infty} f_n (x) = \inf \{ g_n (x) \}
We know that each function f_n is measurable. The above argument tells us that each function g_n is measurable, and that the function {\lim \sup}_{n\rightarrow \infty}f_n is measurable. A similar argument tells us that the function {\lim \inf}_{n\rightarrow \infty}g_n is measurable.
Corollary 3.1.22
If f,g\colon X\rightarrow [-\infty, \infty ] are measurable functions, then so are the functions \max \{ f,g \} and \min \{ f,g \}.
proof to be filled in!
Corollary 3.1.23
The limit of a pointwise-convergent sequence of meaurable functions is measurable.
proof to be filled in!
Definition 3.2.1
Let \Omega be a measurable space, equipped with a \sigma-algebra \mathscr A.
A measure on \Omega is a function \mu \colon {\mathscr A}\rightarrow [0,\infty ] such that:
The number \mu (A) is called the measure of a set A. A measurable space equipped with some measure is called a
measure space.
For the above definition to make sense, we need to make a convention concerning our `number' \infty, namely that a + \infty = \infty whenever a\in [0,\infty ].
Example 3.2.2
Let \Omega be a measurable space. For any measurable set E\subseteq \Omega, let us define
\mu (E) = |E|, where |E| denotes the number of elements of E. Then \mu is a measure on \Omega, called the \em counting measure.
Example 3.2.3
Let \Omega be a measurable space, and let x_0 \in \Omega. For any measurable set E\subseteq \Omega, let us define
\mu (E) = \left\{ \begin{array}{ll}
1 & x_0 \in E \\
0 & x_0 \not\in E \\
\end{array} \right.
Then \mu is a measure on \Omega, called the \em Dirac measure.
Proposition 3.2.4
Let \Omega be a measure space, with measure \mu. Then \mu (\emptyset ) =0.
Proof.
Choose a measurable set A such that \mu (A) < \infty. Then
\mu (A) = \mu (A) + \mu (\emptyset ) + \mu (\emptyset ) +\cdots
Hence \mu (\emptyset ) =0.
Corollary 3.2.5
Let A_1 , \ldots , A_n be disjoint measurable sets. Then
\mu (A_1 \cup \cdots \cup A_n ) = \mu (A_1 ) + \cdots + \mu (A_n)
proof to be filled in!
Corollary 3.2.6
Let A and B be measurable set where A\subseteq B. Then \mu (A) \leq \mu (B).
Proof.
The set B\backslash A = B\cap (\Omega \backslash A ) is measurable, the sets A and B\backslash A are disjoint, and B = A\cup B\backslash A. By the above corollary
\mu (B) = \mu (A) + \mu (B\backslash A)
The inequality \mu (A)\leq \mu (B) follows since \mu (B\backslash A)\geq 0.
Proposition 3.2.7
Let (A_n ) be a sequence of measurable sets such that A_n \subseteq A_{n+1} for all n. Let A = \bigcup_{n=1}^\infty A_n. Then \lim_{n\rightarrow \infty} \mu (A_n ) = \mu (A).
Proof.
Let B_1 =A_1, and B_n = A_n \backslash A_{n-1} when n\geq 2. Then the sets B_n are measurable and disjoint. Further
A_n = B_1 \cup \cdots \cup A_n \qquad A= \bigcup_{n=1}^\infty B_n
Hence
\mu (A) = \sum_{n=1}^\infty \mu (B_n) = \lim_{N\rightarrow} \sum_{n=1}^N \mu (B_n) = \lim_{N\rightarrow \infty}\mu (A_N)
Corollary 3.2.8
Let (A_n) be a sequence of measurable sets such that \mu (A_1) < \infty and A_{n+1}\subseteq A_n for all n. Let A = \bigcap_{n=1}^\infty A_n. Then \lim_{n\rightarrow \infty} \mu (A_n ) = \mu (A).
Proof.
Let C_n =A_1\backslash A_n. Then the set C_n is measurable, C_n\subseteq C_{n+1} for all n, and \bigcup_{n=1}^\infty C_n = A_1 \backslash A. Hence, by the above proposition
\lim_{n\rightarrow \infty}\mu (C_n) = \mu (A_1 \backslash A)
We know that the measure \mu (A_1) is finite, and that we have disjoint unions
A_1 = A_n \cup C_n \qquad A_1 = A_1\backslash A \cup A
Hence
\mu (A_1 )- \lim_{n\rightarrow \infty} \mu (A_n) = \mu (A_1 ) - \mu (A)
and
\lim_{n\rightarrow \infty} \mu (A_n ) = \mu (A)
The above corollary is false if we omit the assumption that \mu (A_1) < \infty.
\addcontentslinetocsubsectionSimple Functions
Definition 3.3.1
A function s\colon \Omega \rightarrow {\mathbb C} on a measurable space \Omega is called \em simple if the range of s is a finite set of points.
Let s\colon \Omega \rightarrow {\mathbb C} be a simple function, with image s[X] = \{ 0 \} \cup \{ \alpha_1 , \ldots , \alpha_n \}. Write A_i = s^{-1}(\alpha_i ). Then clearly
s = \sum_{i=1}^n \alpha_i \chi_{A_i}
and the function s is measurable if and only if each set A_i is measurable.
Proposition 3.3.2
Let f\colon \Omega \rightarrow [0,\infty ] be a measurable function. Then there are simple measurable functions s_n \colon X\rightarrow [0,\infty ) such that the sequence (s_n (x)) is monotonically increasing, with limit f(x) for each point x\in X.
Proof.
Let n\in {\mathbb N}, and t\in [0,\infty ]. Then there is a unique integer k_n (t) such that
k_n (T) 2^{-n} \leq t \leq (k_n (t) +1)2^{-n}
Define
\varphi_n (t) = \left\{ \begin{array}{ll}
k_n (t)2^{-n} & 0\leq t < n \\
n & n\leq t\leq \infty \\
\end{array} \right.
The function \varphi_n \colon [0,\infty ]\rightarrow [0,\infty ] is a Borel function, and
t-2^{-n} \leq \varphi_n (t) \leq t
if 0\leq t\leq n. Thus we have a monotonically increasing sequence (\varphi_n (t)) with limit t. If we write s_n = \varphi_n \circ f, then (s_n) is a monotonically increasing sequence of simple measurable functions, with pointwise limit f as required.
We now come to the first of our definitions of the integral.
Definition 3.3.3
Let \Omega be a measure space, with measure \mu. Let s\colon \Omega \rightarrow {\mathbb C} be a measurable simple function, with set of non-zero values \{ \alpha_1 ,\ldots , \alpha_n \}. Write
s = \sum_{k=1}^n \alpha_k \chi_{A_k}
Let E\subseteq \Omega be a measurable subset of \Omega. Then we define the \em integral of s over E to be the complex number
\int_E sd\mu = \sum_{k=1}^n \alpha_k \mu (A_k \cap E )
There are several simple computations we can do immediately with integrals. For example, with s as above:
\int_\Omega s\chi_E d\mu = \sum_{k=1}^\infty \alpha_k \mu (A_k \cap E) = \int_E sd\mu
Lemma 3.3.4
Let \Omega be a measure space, with measure \mu. Let s\colon \Omega \rightarrow [0,\infty ) be a measurable simple function. Then we can define a new measure \varphi on \Omega by the formula
\varphi (E) = \int_E sd\mu
Proof.
To begin with, observe that \varphi (E)\geq 0 for every measurable set E, and that if \mu (E) < \infty, then \varphi (E) < \infty, so there is at least one measurable set with finite measure. We need to test \sigma-additivity.
Let (E_n) be a sequence of disjoint measurable sets. We know that
\mu (\bigcup_{i=1}^\infty E_i ) = \sum_{i=1}^\infty \mu (E_i)
Let \{ \alpha_1 , \ldots , \alpha_k \} be the set of non-zero values of the simple function s. Then
\varphi (\bigcup_{i=1}^\infty E_i ) = \sum_{k=1}^n \sum_{i=1}^\infty \alpha_k \mu (A_k \cap E_i )
Exchanging the summation signs is possible since all of the numbers involved in the above equation are positive. Therefore
\varphi (\bigcup_{i=1}^\infty E_i ) = \sum_{i=1}^\infty \sum_{k=1}^n \alpha_k \mu (A_k \cap E_i ) = \sum_{i=1}^\infty \varphi (E_i)
and we are done.
Proposition 3.3.5
Let s,t\colon \Omega \rightarrow [0,\infty ] be simple functions. Then
\int_\Omega s+td\mu = \int_\Omega sd\mu + \int_\Omega td\mu
Proof.
Write as usual
s = \sum_{i=1}^m \alpha_k \chi_{A_i} \qquad t = \sum_{j=1}^n \beta_j \chi_{B_j}
Let E_{ij} = A_i \cap B_j. Then certainly
int_{E_{ij}} (s+t)d\mu = (\alpha_i +\beta_j)\mu (E_{ij}) = \int_{E_{ij}}sd\mu + \int_{E_{ij}}td\mu
Now the sets \{ 0, \alpha_1 , \ldots , \alpha_m \} and \{ 0, \beta_1 , \ldots ,\beta_n \} are the ranges of the functions s and t respectively. Let A_0 = s^{-1}[0] and B_0 = t^{-1}[0]. Then
\Omega = \bigcup_{i=0}^m A_i = \bigcup_{j=0}^n B_j
Hence
\Omega = \bigcup_{i,j=0}^{m,n}E_{ij}
The sets E_{ij} are certainly disjoint. Hence by the above lemma, we know that
\int_\Omega s+t d\mu = int_\Omega sd\mu + \int_\Omega td\mu
and we are done.
If s is a step function, and \alpha \in {\mathbb C}, then clearly
\int_\Omega \alpha sd\mu = \alpha \int_\Omega sd\mu
Hence we have proven linearity for integrals of positive-valued step functions.
Definition 3.4.1
Let \Omega be a measure space, with measure \mu. Let f\colon \Omega \rightarrow [0,\infty ] be a measurable function, and let E\subseteq \Omega be a measurable set. Let S be the set of simple functions, s\colon \Omega \rightarrow [0,\infty ), such that s(x) \leq f(x) for all x\in \Omega. Then we define the \em integral of f over E:
\int_E fd\mu = \sup \{ \int_E sd\mu |s\in S \}
A few properties of the integral are easy to prove. For example:
\beginitemize
\item Let f\colon \Omega \rightarrow [0,\infty ] and E\subseteq \Omega be measurable. Then
\int_\Omega fd\mu = \int_\Omega f\chi_Ed\mu
\item Let f,g\colon \Omega \rightarrow [0,\infty ] be measurable functions such that f\leq g, that is to say f(x)\leq g(x) for all x\in \Omega. Then
\int_E f \leq \int_E g
whenever E\subseteq \Omega is a measurable subset.
\enditemize
Theorem 3.4.2[The Monotone Convergence Theorem]
Let f_n\colon \Omega \rightarrow [0,\infty ] be a sequence of measurable functions, such that for each point x\in \Omega the sequence (f_n (x)) is monotonically increasing, with limit f(x). Then the function f\colon \Omega \rightarrow [0,\infty ] is measurable, and
\int_\Omega fd\mu = \lim_{n\rightarrow \infty} \int_\Omega f_n d\mu
Proof.
As the limit of a sequence of measurable functions, the function f is measurable. Since the seqyebce (f_n (x)) is monotonic increasing, with limit f(x), we know that F_n \leq f_{n+1} \leq f for all n. Therefore the sequence of integrals \left( \int_\Omega f_n \right) is monotonic increasing, and
\int_\Omega f_n d\mu \leq \int_\Omega f d\mu
for all n.
Choose a simple function s such that 0\leq s\leq f. Let 0 < \alpha < 1, and write
E_n = \{ x\in \Omega |f_n (x) \geq \alpha s(x) \}
Each set E_n is measurable, and E_n \subseteq E_{n+1} for all n since the sequence (f_n) is monotonic increasing. Since the sequence (f_n) has pointwise limit f, we see that
\Omega = \bigcup_{n=1}^\infty E_n
Further
\int_\Omega f_nd\mu \geq \int_{E_n} f_n d\mu \geq \alpha \int_{E_n}sd\mu \qquad (\ast )
By lemma 3.3.4 we can define a measure on the set \Omega by the formula
\varphi (E) = \int_E sd\mu
Hence
\int_\Omega sd\mu = \varphi (\Omega ) = \lim_{n\rightarrow \infty} \varphi (E_n ) = \lim_{n\rightarrow \infty} \int_{E_n}sd\mu
by proposition 3.2.7.
Taking limits in inequality (\ast ), we see that
\lim_{n\rightarrow \infty } \int_\Omega f_n d\mu \geq \alpha \int_\Omega sd\mu
In particular, this inequality holds whenever 0 < \alpha < 1 and s\leq f. By the definition of the integral, it follows that
\lim_{n\rightarrow \infty}\int_\Omega f_n d\mu \geq \int_\Omega f d\mu
and we are done.
Let f\colon \Omega \rightarrow [0,\infty ] be a measurable function. By proposition 3.3.2, there is a monotonically increasing sequence of simple functions s\colon \Omega \rightarrow [0,\infty ) with pointwise limit f.
The monotone convergence theorem tells us that
\int_\Omega f = \lim_{n\rightarrow \infty} \int_\Omega s_n
and so gives us a new way of viewing the definition of the integral. Using this viewpoint, the following result follows immediately from proposition 3.3.5
Corollary 3.4.3
Let f,g\colon \Omega \rightarrow [0,\infty ] be measurable functions, and let \alpha, \beta \in [0, \infty ). Then
\int_\Omega (\alpha f+ \beta g)d\mu = \alpha \int_\Omega fd\mu + \beta \int_\Omega gd\mu
proof to be filled in!
We can also immediately deduce the following result from the monotone convergence theorem.
Corollary 3.4.4
Consider a sequence of measurable functions f_n \colon \Omega \rightarrow [0,\infty ]. Then for any measurable subset E\subseteq \Omega we have the formula
\sum_{n=1}^\infty \int_E f_nd\mu = \int_E \left( \sum_{n=1}^\infty f_n \right) d\mu
proof to be filled in!
Theorem 3.4.5[Fatou's lemma]
Let f_n \colon \Omega \rightarrow [0,\infty ] be a sequence of measurable functions. Then
\int_\Omega {\lim \inf}_{n\rightarrow \infty} f_n \leq {\lim \inf}_{n\rightarrow \infty} \int_\Omega f_n
Proof.
Let
g_n (x) = \inf \{ f_n (x) , f_{n+1} (x) , f_{n+2} (x) , \ldots \}
Then the function g_n is measurable, the sequence (g_n) is monotonic increasing, and the inequality g_n \leq f_n holds for all n.
We know that
\lim_{n\rightarrow \infty} g_n (x) = {\lim \inf}_{n\rightarrow \infty} f_n (x)
Hence, by the monotone convergence theorem
\int_\Omega {\lim \inf}_{n\rightarrow \infty} f_n = \lim_{n\rightarrow \infty} \int_\Omega g_n \leq {\lim \inf}_{n\rightarrow \infty} \int_\Omega f_n
and we are done.
The inequality
\int_\Omega {\lim \sup}_{n\rightarrow \infty} f_n \geq {\lim \sup}_{n\rightarrow \infty} \int_\Omega f_n
is easily deduced from Fatou's lemma.
Definition 3.5.1
Let \Omega be a measure space, with measure \mu. We call a measurable function f\colon \Omega \rightarrow {\mathbb C} \em integrable if
\int_\Omega |f|d\mu < \infty
We write L^1 (\Omega ) to denote the set of all integrable functions.
Suppose we have a measurable function f and a positive-valued integrable function g such that |f|\leq g. Then it follows by the above definition that the function f is integrable. This integrability criterion is often used.
Definition 3.5.2
Let f\colon \Omega \rightarrow {\mathbb R} be any real-valued function. Then we define functions f^+ , f^- \colon \Omega \rightarrow [0, \infty ) by the formulae
f^+ (x) = \max (f(x), 0)) \qquad f^- (x) = \max (-f(x) , 0)
respectively.
Observe that f = f^+ - f^-. If the function f is measurable, then so are the functions f^+ and f^-.
Proposition 3.5.3
Let f\colon \Omega \rightarrow {\mathbb R} be an integrable function. Then the functions f^+ and f^- are also integrable.
Proof.
The functions f^+ and |f| are positive-valued, and f^+ \leq |f|. We know that \int_\Omega |f| < \infty, so \int_\Omega f^+ < \infty.
The proof that the function f^- is integrable is identical to the above.
Definition 3.5.4
Let f\colon \Omega \rightarrow {\mathbb R} be an integrable function. Then we define we define the integral
\int_\Omega fd\mu := \int_\Omega f^+d\mu - \int_\Omega f^-d\mu
It is easy to see that definition agrees with the previous definition when the function f is positive-valued. Further, the equation
\int_\Omega (\alpha f + \beta g)d\mu = \alpha \int_\Omega f d\mu + \beta \int_\Omega gd\mu
holds for all real numbers \alpha , \beta \in {\mathbb R} and integrable functions f,g\colon \Omega \rightarrow {\mathbb R}.
Definition 3.5.5
Let f,g\colon \Omega \rightarrow {\mathbb C} be integrable functions. Then we define the integral
\int_\Omega fd\mu := \int_\Omega \Re (f)d\mu +i \int_\Omega \Im (f)d\mu
An argument similar to that made above tells us that this integral is well-defined, agrees with the previous definition for real-valued functions, and is linear.
Proposition 3.5.6
Let f\colon \Omega \rightarrow {\mathbb C} be an integrable function. Then
\left| \inf_\Omega fd\mu \right| \leq \int_\Omega |f|d\mu
Proof.
Choose \alpha \in {\mathbb C} such that |\alpha |=1 and
\left| \inf_\Omega fd\mu \right| = \alpha \int_\Omega fd\mu = \int_\Omega \alpha fd\mu
Let g = \Re (\alpha f ) and h= \Im (\alpha f). Then
\left| \inf_\Omega fd\mu \right| = \int_\Omega gd\mu + i \int_\Omega hd\mu
Certainly, \left| \inf_\Omega fd\mu \right| \in {\mathbb R}, so
\int_\Omega hd\mu
and
\left| \inf_\Omega fd\mu \right| = \int_\Omega gd\mu
However
g\leq |g| \leq |\alpha f| = |f|
It follows that
\left| \inf_\Omega fd\mu \right| \leq \int_\Omega |f|d\mu
and we are done.
Observe that the proof of the above result uses only positivity and linearity of the integral.
Theorem 3.5.7[The Dominated Convergence Theorem]
Let (f_n) be a sequence of measurable functions f_n \colon \Omega \rightarrow {\mathbb C} such that:
\beginitemize
\item The limit
f(x) = \lim_{n\rightarrow \infty} f_n (x)
exists for all x\in \Omega.
\item There is an integrable function g\in L^1 (\Omega ) such that |f_n (x)|\leq g(x) for all x\in \Omega and n\in {\mathbb N}.
\enditemize
Then f\in L^1 (\Omega ), and
\lim_{n\rightarrow \infty} \int_\Omega |f_n - f|d\mu =0
Proof.
Since each fucntion f_n is measurable, the limit function f is also measurable. We know that |f_n|\leq g for all n. Therefore |f|\leq g. It follows that f\in L^1 (\Omega ).
Now, let
h_n = 2g - |f_n -f |
Observe that h_n \geq 0 for all n. Hence by Fatou's lemma
\int_\Omega {\lim \inf}_{n\rightarrow \infty} h_n \leq {\lim \inf}_{n\rightarrow \infty} \int_\Omega h_n
that is
\int_\Omega 2gd\mu \leq \int_\Omega 2gd\mu - {\lim \inf}_{n\rightarrow \infty} \int_\Omega |f_n -f| d\mu
and so
{\lim \inf}_{n\rightarrow \infty} \int_\Omega |f_n -f| d\mu \leq 0
Since |f_n -f | \geq 0 for all n, we deduce that
\lim_{n\rightarrow \infty} \int_\Omega |f_n - f|d\mu =0
as required.
Combining the dominated convergence theorem with proposition 3.5.6 we obtain the following corollary, also referred to as the dominated convergence theorem.
Corollary 3.5.8[The Dominated Convergence Theorem]
Let (f_n) be a sequence of measurable functions f_n \colon \Omega \rightarrow {\mathbb C} such that:
\beginitemize
\item The limit
f(x) = \lim_{n\rightarrow \infty} f_n (x)
exists for all x\in \Omega.
\item There is an integrable function g\in L^1 (\Omega ) such that |f_n (x)|\leq g(x) for all x\in \Omega and n\in {\mathbb N}.
\enditemize
Then f\in L^1 (\Omega ), and
\lim_{n\rightarrow \infty} \int_\Omega f_n d\mu = \int_\Omega fd\mu
proof to be filled in!
Definition 3.6.1
Let \Omega be a measure space, with measure \mu. Then a set E\subseteq \Omega is called a \em null set if E is measurable, and \mu (E) =0.
The measure space \Omega is called \em complete if every subspace of a null set is measurable.
The usual manipulations of the axioms tell us that every measure space is contained in a unique smallest complete measure space. To be more precise, we have the following result.
Proposition 3.6.2
Let \Omega be a measure space, equipped with \sigma-algbebra \mathcal M, and measure \mu. Let us define
{\mathcal M}^\star := \{ E\subseteq \Omega |A\subseteq E\subseteq B,\, A,B\in \Omega ,\, \mu (B\backslash A) = 0 \}
Then the set {\mathcal M}^\star is a \sigma-algebra. We can define a measure \mu^\star on the set {\mathcal M}^\star by writing
\mu^\star (E) = \mu (A) \qquad A\subseteq E\subseteq B,A,B\in \Omega ,\mu (B\backslash A) =0
proof to be filled in!
As we might expect from the terminology, null sets are irrelevant from the point of view of integration theory.
Theorem 3.6.3
Let f\colon \Omega \rightarrow [0,\infty ] be a measurable function. Then the integral of f is zero if and only if the function f is equal to zero except on a null set.
Proof.
Suppose that the set
N = \{ x\in \Omega |f(x)\neq 0 \}
is a null set. Let s\colon \Omega \rightarrow [0,\infty ] be a simple function such that s\leq f. Then s(x)=0 when x\not\in N. The definition of the integral of a simple function tells us that
\int_\Omega s d\mu =0
The definition of the integral of a non-negative function now implies that
\int_\Omega fd\mu =0
Conversely, suppose that the integral of the function f is zero. Let
A_n = \{ x\in \Omega |f(x) > 1/n \}
Then clearly
\frac{1}{n} \mu (A_n ) \leq \int_{A_n} fd\mu \leq \int_\Omega f d\mu =0
so \mu (A_n ) =0. But
\{ x\in \Omega |f(x) > 0 \} = \bigcup_{n=1}^\infty A_n
Thus \sigma-additivity implies that the set of all points x\in \Omega such that f(x)\neq 0 has measure zero.
Given two functions f,g\colon \Omega \rightarrow {\mathbb C}, let us say that f and g are equal \em almost everywhere if they are equal outside of some set of measure zero.
Corollary 3.6.4
Let f,g\colon \Omega \rightarrow {\mathbb C} be integrable functions that are equal almost everywhere.
Corollary 3.6.5
Let f\colon \Omega \rightarrow {\mathbb C} be an integrable function. Suppose that
\int_E f =0
whenever the subset E\subseteq \Omega is measurable. Then the function f is equal to zero almost everywhere.
Proof.
Let us write
f(x) = u(x) + iv(x) = (u^+(x)- u^- (x)) + i(v^+(x) - v^-(x))
where the functions u and v are real and integrable, and the functions u^\pm and v^\pm are integrable and non-negative.
Let
E = \{ x\in \Omega |u(x)\geq 0 \}
Then
\Re \left( \int_E f \right) = \int_E u^+ =0
By the above theorem, it follows that u^+ =0 except on a null set. Similarly, it follows that u^- =0 except on a null set. Since the union of two null sets is also a null set, we have shown that u=0 almost everywhere.
A similar argument tells us that v=0 almost everywhere. We conclude that f=0 almost everywhere.
Before we are ready to state the Riesz representation theorem, we need some terminology from point-set topology.
Definition 3.7.1
Let X be a topological space. Then we define the \em support of a continuous function f\colon X\rightarrow {\mathbb C} to be the closure
\operatorname{supp} (f):= \overline{ \{ x\in X |f(x)\neq 0 \} }
We write C_c (X) to denote the set of all continuous compactly supported functions f\colon X\rightarrow {\mathbb C}. The set C_c (X) is a vector space under the operations of pointwise addition and scalar multiplication.
Definition 3.7.2
A linear map \Lambda \colon C_c (X) \rightarrow {\mathbb C} is said to be a \em positive functional if \Lambda (f) \geq 0 whenever f\geq 0.
 Then the integration map
f\mapsto \int_X f
defines a positive linear functional.
The Riesz representation theorem is essentially a converse of the above observation.
Theorem 3.7.3
Let X be a locally compact Hausdorff space, and let \Lambda \colon C_c (X) \rightarrow {\mathbb C} be a positive linear functional.
Then the set X has a \sigma-algebra \Omega containing all Borel sets, and a unique measure \mu on \Omega such that
\Lambda (f) = \int_X fd\mu
whenever f\in C_c (X).
The proof of this theorem is in a series of lemmas; the proof is quite long. Before we begin the proof, let us note a theorem from general topology which we shall need.
Theorem 3.7.4
Let X be a locally compact Hausdorff space, and let {\mathcal U} = \{ U_\alpha |\alpha \in A \} be an open cover of the space X. Then there is a \em partition of unity subordinate to the cover \mathcal U, that is to say a set of continuous functions u_\alpha \colon X\rightarrow [0,1] such that \operatorname{supp} u_\alpha \subseteq U_\alpha and
\sum_{\alpha \in A} u_\alpha (x) =1
whenever x\in X.
proof to be filled in!
The following corollary is known as \em Urysohn's lemma.
 Then there is a continuous function f\colon X\rightarrow [0,1]
such that
\chi_K(x) \leq f(x) \leq \chi_U (x)
Proof.
The collection \{ U, X\backslash K \} is an open cover of the space X. There is therefore a partition of unity \{ f,g\} subordinate to this open cover.
The definition of a partition of unity gives us the required inequality for the function f.
We now begin our proof of the Riesz representation theorem with the definition of the measure we are looking for.
 Let U\subseteq X be open. Then we define
\mu (U):= \sup \{ \Lambda f |f\leq \chi_U \}
In general, for a subset, E\subset X, we define
\mu (E) = \inf \{ \{ \mu (U) |U \textrm{ open }, E\subseteq U \}
Proposition 3.7.7
Let f,g\in C_c (X), and let f\leq g. Then \Lambda f \leq \Lambda g.
Proof.
Observe g-f \geq 0. The result follows from positivity and linearity of the function \Lambda.
Corollary 3.7.8
Let A and B be subsets of the space X where A\subseteq B. Then \mu (A)\leq \mu (B).
proof to be filled in!
Although we have defined a function \mu for every subset of E, the definition is only sensible for a certain \sigma-algebra.
Definition 3.7.9
We define \Omega_F to be the sollection of all subsets E\subseteq X such that \mu (E) < \infty and
\mu (E) = \sup \{ \mu (K) |K\subseteq E, K \textrm{ compact} \}
We define \Omega to be the collection of all subsets E\subseteq X such that E\cap K \in \Omega_F whenever K is compact.
We need to prove that the set \Omega is a \sigma-algebra which contains all Borel sets; this statement is not obvious.
 Then V\in \Omega_F.
Proof.
Choose a number a < \mu (V). By the definition of \mu, there is a function f\in C_c (X) such that f\leq \chi_V and a < \Lambda f. Write K=\operatorname{supp} (f), and let W be an open set that contains K. Then \Lambda f \leq \mu (W), so \Lambda f \leq \mu (K), using the above proposition and corollary, and the definition of the function \mu.
Thus K\subseteq V and \mu (K) > a. It follows that
\mu (V) = \sup \{ \mu (K) |K\subseteq E, K \textrm{ compact} \}
and we are done.
Proposition 3.7.11
Let U_1 , \ldots , U_N\subseteq X be open sets. Then
\mu (U_1\cup \cdots \cup U_N) \leq \mu (U_1)+ \cdots +\mu (U_N).
Proof.
Let N=2. Choose a function g\in C_c (X) such that g\leq \chi_{U_1\cup U_2}. By theorem 3.7.4 there are functions u_1 , u_2\in C_c (X) such that u_1\leq \chi_{U_1}, u_2\leq \chi_{u_2}, and u_1(x) + u_2 (x) =1 whenever x\in U_1\cup U_2. It follows that
u_1 g\leq \chi_{U_1},u_2g \leq \chi_{U_2} \qquad g=u_1 g + u_2 g
and therefore
\Lambda g = \Lambda (u_1 g) +\Lambda (u_2 g) \leq \mu (U_1) + \mu (U_2)
Since the above inequality holds for every function g\in C_c (X) such that g\leq \chi_{U_1\cup U_2}, the result follows from the definition of \mu when N=2. The general result follows by induction.
Lemma 3.7.12
Let E_1,E_2,E_3,\ldots be subsets of the space X. Write
E = \bigcup_{n=1}^\infty E_n
Then
\mu (E) \leq \sum_{n=1}^\infty \mu (E_n)
Proof.
If \mu (E_n )=\infty for some n, then the result is obviously true. Thus, let us suppose that \mu (E_n) < \infty for all n. Choose \varepsilon > 0. By definition of the function \mu, there are open sets U_n \supseteq V_n such that
\mu (V_n) < \mu (E_n) +2^{-n} \varepsilon
for all n.
Let U = \bigcup_{n=1}^\infty U, and choose f\in C_c (X) such that f\leq \chi_U. The support of the function f is covered by the collection of sets \{ U_n |n=1,2,3,\ldots \}. Since the function f has compact support, it follows that it has a finite subcovering, and so
f\leq \chi_{U_1 \cup \cdots \cup U_N }
for some N. By the above proposition, we see that
\Lambda f \leq \mu (U_1 \cup \cdots \cup U_N) \leq \mu(V_1 ) + \cdots + \mu (V_N) \leq \sum_{n=1}^\infty \mu (E_n) + \varepsilon
Since the above inueqality holds for every funtion f\subseteq \chi_U, and E\subseteq U, we see that
\mu (E) \leq \sum_{n=1}^\infty \mu (E_n) +\varepsilon
But this inequality holds whenever \varepsilon > 0, so the result follows.
Proposition 3.7.13
Let K\subseteq X be compact. Then \mu (K)\leq \Lambda f whenever f\geq \chi_K, and K\in \Omega_F.
Proof.
Let 0 < a < 1, and choose f\in C_c (X) such that f\geq \chi_K. Write
V_a = \{ x\in X |f(x) > a \}
Then K\subseteq V_a, and ag\leq f whenever f\leq \chi_{V_a}. Therefore
\mu (K) \leq \mu (V_a ) = \sup \{ \Lambda g | g\leq \chi_{V_a} \} \leq a^{-1}\Lambda f
Since this inequaltity holds whenever 0 < a < 1, it follows that \mu (K)\leq \Lambda f. It follows that \mu (K) < \infty, and so K\in \Omega_F.
Lemma 3.7.14
Let K\subseteq X be compact. Then
\mu (K) =\inf \{ \Lambda f |\chi_K \leq f \}
Proof.
Let \varepsilon > 0. Then there is an open set U\supseteq K such that \mu (U) < \mu (K) +\varepsilon. By Urysohn's lemma there is a continuous function f\colon [0,1]\rightarrow X such that \chi_K \leq f \leq \chi_U. It follows that
\Lambda f\leq \mu(U) < \mu (K) +\varepsilon
The result follows from the above inequality combined with the previous proposition.
Proposition 3.7.15
Let K_1,\ldots K_N be disjoint compact sets. Then
\mu (K_1 \cup \cdots \cup K_N) \leq \mu (K_1 ) + \cdots + \mu (K_N)
Proof.
Let N=2. We can find an open set U such that U\supseteq K_1 and U\cap K_2 = \emptyset. It follows by Urysohn's lemma that we can find a compactly supported function u\colon X\rightarrow [0,1] such that u(x)=1 whenever x\in K_1, and u(x) =0 whenever x\in K_2.
Let \varepsilon > 0. By lemma 3.7.14 there is a function g\in C_c (X) such that
\chi_{K_1 \cup K_2}\leq g \qquad \Lambda g \leq \mu (K_1 + K_2) +\varepsilon
Observe that
\chi_{K_1} \leq fg \qquad \chi_{K_2} \leq (1-f)g
Hence
\mu (K_1 ) + \mu (K_2) \leq \Lambda (fg) + \Lambda (g-fg) \leq \mu (K_1 \cup K_2 ) + \varepsilon
Since the above inequality holds whenever \varepsilon > 0, the desired result follows when N=2. The general result follows by induction.
Lemma 3.7.16
Let E_1,E_2,E_3,\ldots be pairwise disjoint members of the collection \Omega_F. Write
E = \bigcup_{n=1}^\infty E_n
Then
\mu (E) = \sum_{n=1}^\infty \mu (E_n)
Further, if \mu (E) < \infty, then E\in \Omega_F.
Proof.
Observe that the result follows from lemma 3.7.12 when \mu (E)=\infty. Let us therefore assume that \mu (E) < \infty. Choose \varepsilon > 0. Since E_n\in \Omega_F, we can find a compact set K_n \subseteq E_n such that
\mu (K_n ) > \mu (E_n ) - 2^{-n} \varepsilon
for each n. Let H_N = K_1 \cup \cdots \cup K_N. Then by the above propositiion:
\mu E) \geq \mu (H_N) = \sum_{n=1}^N \mu (K_n) > \sum_{n=1}^N \mu (E_n) - \varepsilon
Since the above inequality holds whenever \varepsilon > 0, combining it with the inequality in lemma 3.7.12, we see that
\mu (E) = \sum_{n=1}^\infty \mu (E_n)
Now, if \mu (E) < \infty, and \varepsilon > 0, then we can find N such that
\mu (E) \leq \sum_{n=1}^N \mu (E_n) + \varepsilon
It follows that \mu (E)\leq \mu (H_N) +2\varepsilon, and so E\in \Omega_F.
Proposition 3.7.17
Let E\subseteq \Omega_F, and let \varepsilon > 0. Then there is a compact set K and an open set V such that K\subseteq E\subseteq V, and \mu (V\backslash K) < \varepsilon.
Proof.
by definition of the collection \Omega_F, we can find a compact set K\subseteq E and an open set U\supseteq E such that
\mu (V) - \frac{\varepsilon}{2} < \mu (E) < \mu(K) + \frac{\varepsilon}{2}
By lemma 3.7.10, we see that V\backslash K \in \Omega_F. By lemma 3.7.16, we see
\mu (K) + \mu (U\backslash K) = \mu (U) < \mu (K) +\varepsilon
and we are done.
Proposition 3.7.18
Let A,B\in \Omega_F. Then the sets A\backslash B, A\cup B, and A\cap B belong to the collection \Omega_F.
Proof.
By the above proposition, there are compact sets K and K', and open sets U and U' such that
K\subseteq A\subseteq U \qquad K'\subseteq B\subseteq U'
and
\mu (U\backslash K) < \varepsilon \qquad \mu (U'\backslash K') < \varepsilon
Observe
A\backslash B \subseteq U\backslash K' \subseteq U\backslash K \cup K\backslash U'\cup U'\backslash K'
Hence by lemme 3.7.12:
\mu (A\backslash B) \subseteq \mu (K\backslash V') +2\varepsilon
Further, the set K\backslash V' is compact, so the above inequality tells us that A\backslash B \in \Omega_F.
But A\cup B = (A\backslash B)\cup B, so A\cup B \in \Omega_F by lemma 3.7.16. Finally, A\cap B = A\backslash (A\backslash B), so A\cap B\in \Omega_F by the above calculation.
We are now nearly done, and can prove a slightly less technical result.
Theorem 3.7.19
The set \Omega is a \sigma-algebra containing all Borel sets.
Proof.
Let K\subseteq X be compact. If A\in \Omega, then X\backslash A \cap K = K \backslash (A\cap K), so X\backslash A \cap K\in \Omega_F by the above proposition, and X\backslash A\in \Omega.
Suppose that
A = \bigcup_{n=1}^\infty A_n \qquad A_n \in \Omega
Let B_1 = A_1 \cap K, and
B_n = (A_n \cap K) \backslash (B_1 \cup \cdots \cup B_n) \qquad n\geq 2
Then the collection \{ B_n |n=1,2,\ldots \} is a pairwise disjoint, and B_n \in \Omega_F for all n by the above lemma. But A\cap K =\bigcup_{n=1}^\infty B_n, so A\cap K \in \Omega_F by lemma 3.7.16. It follows that A\in \Omega.
We have proved that the collection \Omega is a \sigma-algebra. If C\subseteq X is a closed subset, then the intersection K\cap C is compact. Thus C\cap K \in \Omega_F, and so C\in \Omega. Thus every closed set belongs to the collection \Omega. It follows that the \sigma-algebra \Omega contains all Borel sets.
Lemma 3.7.20
\Omega_F = \{ E\in \Omega |\mu (E) < \infty \}
Proof.
Let E\in \Omega_F. Then by lemmas 3.7.14 and 3.7.16, we see E\cap K \in \Omega_F whenever K\subseteq X is compact. Then E\in \Omega. By definition of the set \Omega_F, \mu (E) < \infty.
Conversely, suppose that E\in \Omega and \mu (E) < \infty. Let \varepsilon > 0. We can certainly find an open set U\supseteq E such that \mu (E) < \infty. By propositions 3.7.10 and 3.7.17, there is a compact set K\subseteq U such that \mu (U\backslash K) < \varepsilon.
We know that E\cap K\in \Omega_F. There is therefore a compact set H\subseteq E\cap K such that
\mu (E\cap K) < \mu (H) +\varepsilon
But E\subseteq (E\cap K) \cup (U\backslash K). Therefore
\mu (E)\subseteq \mu (E\cap K) + \mu (V\backslash K) < \mu (H)+\varepsilon
and we see that E\in \Omega_F.
We can now prove our main result.
Theorem 3.7.21
The function \mu is a measure on the \sigma-algebra \Omega. It is the unique measure with the property
\Lambda f = \int_X f(x)d\mu (x)
for all f\in C_c (X).
Proof.
It follows immediately that \mu is a measure from lemmas 3.7.16 and 3.7.20. Our next step is to prove the inequality
\Lambda f \leq \int_X f(x)d\mu (x)
for every real-valued compactly supported function f. To do this, let K = \operatorname{supp} (f), and choose a,b\in {\mathbb R} such that f[K]\subseteq [a,b]. Let \varepsilon > 0, and choose y_0 ,\ldots ,y_N such that
a=y_0 < \cdots < y_N \qquad y_n - y_{n-1} < \varepsilon \textrm{ for all }n
We can form Borel sets
E_n := \{ x\in X | y_{n-1} < f(x) \leq y_n \}
The sets E_n are pairwise disjoint with union K. We can find open sets U_n \supseteq E_n such that
\mu (U_k ) < \mu (E_k ) + \frac{\varepsilon}{n} \qquad f(x) < y_n + \varepsilon
whenever x\in U_n.
By theorem 3.7.4, we can choose a partition of unity \{ u_1 , \ldots , u_N \} subordinate to the open cover \{ U_1 ,\ldots , U_N \}. It follows that
f = \sum_{n=1}^N u_n f
and by lemma 3.7.14
\mu (K) \leq \Lambda \left( \sum_{n=1}^N u_n \right) = \sum_{n=1}^N \Lambda (u_n)
But by construction u_nf \leq (y+n + \varepsilon )u_n, and y_n - \varepsilon < f(x) for all x\in E_n, so
\Lambda f \leq \sum_{n=1}^N (y_k + \varepsilon) \Lambda (u_n) = \sum_{n=1}^N (|a|+ y_k + \varepsilon) \Lambda (u_n) - |a| \sum_{n=1}^N \Lambda (u_n)
and
\Lambda f \leq \sum_{n=1}^N (|a|+ y_k + \varepsilon) (\mu (E_n) + \varepsilon /n ) - |a| \mu (K)
Multiplying out, we see that
\Lambda f \leq \sum_{n=1}^N (y_n - \varepsilon )\mu (E_n) + 2\varepsilon \mu (K) \frac{\varepsilon}{n} \sum_{n=1}^N (|a| + y_n + \varepsilon )
so by construction of the integral
\Lambda f \leq \int_X fd\mu + \varepsilon (2\mu (K) + |a| +b +\varepsilon )
Since the above inequality must hold for every choice of \varepsilon > 0, we see that
\Lambda f \leq \int_X f(x)d\mu (x)
as required.
Now, if we replace the function f by the function -f, we see that
-\Lambda f \leq -\int_X f(x)d\mu (x)
Combining the above two inequalities, we have the equation
\Lambda f = \int_X f(x)d\mu (x)
for every real-valued compactly supported function f. The proof of the above equation for complex-valued functions follows by splitting such a function into real and imaginary parts, and using linearity.
All that remains is to show uniqueness. Let \mu' be a measure such that the eqation
\Lambda f = \int_X f(x)d\mu' (x)
holds for every compactly supported function f. Let K be a compact set. By theorem 3.7.4, given an open set U\supseteq K, there is a compactly supported function g such that \chi_K \leq g\leq \chi_U. Hence
\mu' (K) \leq \int_X f d\mu' \leq \mu' (U)
and
\mu'(U) = \sup \{ \Lambda f |f\leq \chi_U \} = \mu (U)
It follows that \mu (B)= \mu' (B) whenever B is a Borel set, and we are done.
We would like to use the Riesz representation theorem to define a measure on the real line \mathbb R that gives the usual integral expected from elementary calculus. To apply the Reisz representation theorem, we need a sensible definition of the integral of a continuous compactly supported function.
Let us consider a continuous function f\colon [a,b]\rightarrow {\mathbb R}. Let n be a positive integer. Then the interval [a,b] can be divided into S^n equal-sized pieces:
a < a+ 2^{-n}(b-a) < a+ 2(2^{-n})(b-a) < \cdots < a+ (2^n -1)(2^{-n})(b-a) < b
Let us define
\mu_{n,r} = \inf \{ f(x) |a+r2^{-n}(b-a) \leq f(x) < a+ (r+1)2^{-n}(b-a)
and
I_n (f) = \sum_{r=0}^{2^n -1} 2^{-n}(b-a) \mu_{n,r}
The following observations are clear.
\beginitemize
\item The sequence (I_n (f)) is monotonically increasing
\item Since the interval [a,b] is compact, and the function f is continuous, there is a constant C such that f(x)\leq C for all x\in [a,b]. Hence I_n (f) \leq C(b-a) for all n.
\enditemize
It follows that we have a well-defined limit
\Lambda (f) := \lim_{n\rightarrow \infty} I_n (f)
We would like to extend the definition of the function \Lambda. There are two stages to this extension.
\beginitemize
\item Let f\colon [a,b]\rightarrow {\mathbb C} be a continuous function. Write f(x) = u(x) + iv(x), where u,v\colon [a,b]\rightarrow {\mathbb R}, and define
\Lambda (f) = \Lambda (u) + i \Lambda (v)
\item Let f\in C_c ({\mathbb R}). Let [a,b]\supseteq \operatorname{supp} (f). Then we define
\Lambda (f) = \Lambda (f|_{[a,b]})
\enditemize
The following result is straightforward to check.
Proposition 3.8.1
The map \Lambda is a positive linear functional on the space C_c ({\mathbb R}).
proof to be filled in!
Definition 3.8.2
Let f\in C_c ({\mathbb R}). Then the number \Lambda (f) is called the \em Riemann integral of f.
Definition 3.9.1
Let \Lambda \colon C_c ({\mathbb R})\rightarrow {\mathbb C} be the Riemann integral. Then the \em Lebesgue measure on \mathbb R is the unique measure such that
\int_{\mathbb R} fd\mu = \Lambda (f)
whenever f\in C_c ({\mathbb R}).
By the Riesz representation, the Lebesgue measure exists and is unique on the collection of all Borel sets. The integral of a Borel measurable function with respect to the Lebesgue measure is termed the \em Lebesgue integral. We will normally write
\int_a^b fd\mu := \int_{\mathbb R} f\chi_{(a,b)}d\mu
Proposition 3.9.2
Let a < b be real numbers. Then \mu (a,b) = b-a.
Proof.
Let [c,d]\subseteq (a,b) be a compact interval. By Urysohn's lemma, there is a function f\in C_c ({\mathbb R}) such that \chi_{[c,d]} \leq f \leq \chi_{(a,b)}.
By definition of the Riemann integral:
d-c \leq \int_{\mathbb R} f \leq b-a
Let c\rightarrow a and d\rightarrow b. Then f\rightarrow \chi_{(a,b)} and by the dominated convergence theroem,
\int_{\mathbb R} f \rightarrow \mu (a,b)
It follows that \mu (a,b) = b-a, and we are done.
A similar computation tells us that
\mu [a,b] = \mu [a,b) = \mu (a,b] = b-a
whenever a < b.
The next fundamental property of the Lebesgue measure follows from a topological property of the real line, which we will state without proof.

Corollary 3.9.4
Let E\subseteq {\mathbb R} be a Borel set. Then \mu (X+E) = \mu (E) whenever x\in {\mathbb R}.
proof to be filled in!
We conclude with a general characterisation of sets of measure zero, or \em null sets.
Theorem 3.9.5
Let E\subseteq {\mathbb R} be a set such that every subset of A is measurable. Then \mu (A) =0.
Proof.
The set \mathbb R is an Abelian group under the operation of addition, and the set \mathbb Q is a subgroup. Let E be a set of real numbers containing precisely one element of each coset x+ {\mathbb Q} \in {\mathbb R}/{\mathbb Q}.
We claim:
\beginitemize
\item (r+E)\cap (s+E) = \emptyset whenever r,s\in {\mathbb Q}, r\neq s.
\item Let x\in {\mathbb R}. Then we can find an element r\in {\mathbb Q} such that x\in r+E.
\enditemize
To see the first claim, suppose that x\in (r+E)\cap (s+E), where r,s\in {\mathbb Q}. Then there are elements y,z\in E such that r+y = s+z, and so y-z\in {\mathbb Q}. But the definition of the set E means that r=s.
As for the second claim, let x\in {\mathbb R}. Construction of the set E means that we can find a point y\in E such that x-y \in {\mathbb Q}. But x = y + (x-y) so the claim is established.
We now use the above to claims to prove the theorem. Let t\in {\mathbb Q}, and define A_t := A\cap (t+E). The set A_t is measurable since it is a subset of the set A. Consider a compact subset K\subseteq A_t, and let
H = \bigcup_{r\in {\mathbb Q}\cap [0,1]} (r+K)
Then the set H is bounded and measurable, so \mu (H) < \infty. The first of the above claims tells us that the sets r+K are pair-wise disjoint, so
\mu (H) = \sum_{r\in {\mathbb Q}\cap [0,1]} \mu (r+K) = \sum_{r\in {\mathbb Q}\cap [0,1]} \mu (K)
by corollary 3.9.4. It follows that \mu (K)=0 whenever K\subseteq A_t is compact.
So \mu (A_t) =0. But
A = \bigcup_{t\in {\mathbb Q}} A_t
and it follows that \mu (A)=0.
Corollary 3.9.6
Any countable subset of the space {\mathbb R} has measure zero.
proof to be filled in!
Corollary 3.9.7
There are non-measurable subsets of the space \mathbb R.
proof to be filled in!
By convention, when a < b are real numbers, and \mu is the Lebesgue measure on the space \mathbb R, we simplfy our notation slightly and write just
\int_a^b f(x)dx := \int_a^b fd\mu
If b < a, we write
\int_a^b f(x)dx := - \int_b^a f(x)dx
Linearity of the integral gives us the equation
\int_a^b f(x)dx = \int_a^c f(x)dx + \int_c^b f(x)dx
whenever a,b,c\in {\mathbb R}.
This new notation is convenient when integating a concrete function given by some definite formula.
In this section we will focus on one major result, which is of absolutely vital importance when trying to calculate integrals. This result is termed the em fundamental theorem of calculus.
Theorem 3.10.1
Let f\colon [a,b] \rightarrow {\mathbb C} be a continuous function. Define a function F\colon [a,b]\rightarrow {\mathbb C} by the formula
F(x) = \int_a^x f(y)dy
Then the function F is differentiable on the open interval (a,b), and has a one-sided derivative at the end-points a and b. In all cases, the derivative is given by the formula
F'(x) = f(x)
Proof.
Let \varepsilon > 0, and let x\in [a,b]. Since the function f is continuous, we can choose \delta > 0 such that |f(x+h) -f(x)| < \varepsilon whenever |h| < \delta and x+h\in [a,b].
Let x\in [a,b], and x+h\in [a,b]. Observe:
F(x+h)- F(x)= \int_x^{h+h} f(y)dy
and
hf(x) = f(x)\mu (x,x+h) = \int_x^{x+h} f(x)dy
Suppose that |h| < \delta. Then |f(y)-f(x)| < \varepsilon whenever y\in [x,x+h], and so:
\left| \int_x^{x+h} f(y)-f(x)dy \right| \leq \int_x^{x+h} |f(y) -f(x)|dy \leq \varepsilon |h|
Thus:
|F(x+h) -F(x) - hf(x)| \leq \varepsilon |h|
whenver |h| < \delta. It follows that the function F is differentiable, and F'(x) =f(x) as claimed.
In actual fact, the more useful form of the fundmantal theorem of calculus is a variation of the above formula.
Corollary 3.10.2
Let F\colon [a,b]\rightarrow {\mathbb C} be a function with a continuous derivative f. Then
\int_a^b f(x)dx = F(b)-F(a)
Proof.
Define
F_0 (x) = \int_a^x f(y)dy
Then by the above version of the fundamental theorem of calculus, F_0'(x) = f(x) whenever x\in [a,b]. Hence F_0'(x) = F'(x) whenever x\in [a,b], so there is a constant C such that F_0(x) = F(x) +C for all x\in [a,b].
We know that F_0 (a)=0. Therefore C=-F(a). We see that
int_a^b f(x)dx = F_0 (b) = F(b) - F(a)
as claimed.
The various integration formulae, such as integration by parts and the change of variable formula, come from the fundamental theorem of calculus along with the corresponding formulae for differentives, such as the derivative of a product and the derivative of a composition.
Let \Omega_1 and \Omega_2 be measure spaces, with measures \mu_1 and \mu_2 on \sigma-algebras {\mathcal M}_1 and {\mathcal M}_2 respectively.
Definition 3.11.1
We call a subset of the form A\times B\subseteq X\times Y, where A\in {\mathcal M}_1 and B\in {\mathcal M}_2 a \em measurable rectangle. A finite union of measurable rectangles is called an \em elementary set.
We write {\mathcal M}_{12} to denote the smallest \sigma-algebra in the set \Omega_1\times \Omega_2 that contains every measurable rectangle.
We want to define a measure on the \sigma-algebra {\mathcal M}_{12}. Before we can do this, we need some technical constructions.
Definition 3.11.2
Let \mathcal C be a collection of subsets of some set. Suppose that the following two conditions hold:
\beginitemize
\item Let (A_n) be a sequence of sets in the collection \mathcal C such that A_n\subseteq A_{n+1} for all n. Then \bigcup_{n=1}^\infty A_n \in {\mathcal C}.
\item Let (B_n) be a sequence of sets in the collection \mathcal C such that B_n\supseteq B_{n+1} for all n. Then \bigcup_{n=1}^\infty B_n \in {\mathcal C}.
\enditemize
Then we call the collection \mathcal C a \em monotone class.
The proof of the following lemma is elementary, but rather abstract. We omit it.
Lemma 3.11.3
The \sigma-algebra {\mathcal M}_{12} is the smallest monotone class in the product \Omega_1 \times \Omega_2 which contains all elementary sets.
proof to be filled in!
Given a subset E\subseteq \Omega_1 \times \Omega_2, and points x\in Omega_1 and y\in \Omega_2, let us write
E_x = \{ y\in \Omega_2 |(x,y)\in E \} \qquad E^y = \{ x\in \Omega_1 |(x,y)\in E \}
Proposition 3.11.4
Let E\in {\mathcal M}_{12}. Then E_xin {\mathcal M}_1 and E^y\in {\mathcal M}_2 whenever x\in \Omega_1 and y\in \Omega_2.
Proof.
Let x\in \Omega_1. Let \mathcal M be the collection of all elements E\in \Omega_1 \times \Omega_2 such that E_x\in \Omega_2. It is straightforward to check that \mathcal M is a \sigma-algebra that contains every measurable rectangle. Therefore {\mathcal M}_{12}\subseteq {\mathcal M}, and we see that E_x\in {\mathcal M}_2 for every measurable set E\subseteq \Omega_1\times \Omega_2 and point x\in \Omega_2.
The corresponding statement concerning sets of the form E^y is proved in the same way.
Corollary 3.11.5
Let X be a topological space, and let f\colon \Omega_1 \times \Omega_2 \rightarrow X be a measurable function. Choose points x\in \Omega_1 and y\in \Omega_2.
Definition 3.11.6
A measure space \Omega is called \em \sigma-finite if it is a countable union of spaces of finite measure.
Example 3.11.7
The space \mathbb R, equipped with the standard Lebesgue measure, is \sigma-finite.
The following result lets us define measures on products of \sigma-finite measure spaces.
Theorem 3.11.8
Let \Omega_1 and \Omega_2 be \sigma-finite measure spaces. Let E\subseteq \Omega_1\times \Omega_2 be a measurable subset. Then we can define measurable functions f\colon \Omega_1\rightarrow [0,\infty] and g\colon \Omega_1\rightarrow [0,\infty] by the formulae
f_E(x) = \mu_2 (E_x) \qquad g_E(y) = \mu_1 (E^y)
respectively. Further,
\int_{\Omega_1} f_E = \int_{\Omega_2} g_E
Proof.
Measurability of the functions f_E and g_E associated as above to a measurable set E\subseteq X\times Y follows from the above proposition and corollary; all that remains it to prove the main equation.
Let {\mathcal M} be the set of all measurable subsets E\subseteq \Omega_1 \times \Omega_2 such that the equation
\int_{\Omega_1} f_E = \int_{\Omega_2} g_E
holds.
Let E=A\times B be a measurable rectangle. Then f_E = \mu_2 (B)\chi_A and g_E = \mu_1 (A)\chi_B. It follows that
\int_{\Omega_1} f_E = \int_A \mu_2 (B) = \mu_1 (A) \mu_2 (B) \qquad \int_{\Omega_1} g_E = \int_B \mu_1 (A) = \mu_1 (A) \mu_2 (B)
so E\in {\mathcal M}.
Let (E_n) be a sequence of sets in the collection \mathcal M such that E_n\subseteq E_{n+1} for all n. Write
E= \bigcup_{n=1}^\infty E_n
Then the sequences of functions (f_{E_n}) and (g_{E_n}) are monotonic increasing, with limits f_E and g_E respectively. We know that E_n \in {\mathcal M} for all n, so that the equation
\int_{\Omega_1} f_{E_n} = \int_{\Omega_2} g_{E_n}
holds for all n. The monotone convergence theorem gives us the equation
\int_{\Omega_1} f_E = \int_{\Omega_2} g_E
and so tells us that E\in {\mathcal M}.
As a consequence of the above calculation, we can easily show that the union of a discrete sequence of measurable sets in the set \mathcal M also belongs to the set \mathcal M. Let (E_n) be a sequence of sets in the collection \mathcal M such that E_1\subseteq A\times B, where \mu_1 (A) < \infty, \mu_2 (B) < \infty, and E_n\supseteq E_{n+1} for all n. Write
E= \bigcup_{n=1}^\infty E_n
Then an argument similar to the above one, only using the dominated convergence theorem rather than the monotone convergence theorem, tells us that the set E belongs to the collection \mathcal M.
Now, let \Omega_1 = \cup_{n=1}^\infty \Omega_1^{(n)} and \Omega_2 = \cup_{n=1}^\infty \Omega_2^{(n)}, where \mu_1 (\Omega_1^{(m)}) < \infty and \mu_2 (\Omega_2^{(n)}) < \infty for all m,n\in {\mathbb N}. Given a set E\subseteq \Omega_1\times \Omega_2, let us write
E_{mn} = E \cap (\Omega_1^{(m)} \times \Omega_2^{(n)}
Let {\mathcal C} be the collection of all measurable sets E\subseteq \Omega_1 \times \Omega_2 such that E_{mn}\in {\mathcal M} for all natural numbers m and n. Then the above calculations tell us that the collection \mathcal C is a monotone class that contains every elementary rectangle. It follows from lemma 3.11.3 {\mathcal M}_{12}\subseteq {\mathcal C}, and we are done.
To paraphrase the above theorem, the equation
\int_{\Omega_1} \left( \int_{\Omega_2} \chi_E (x,y) d\mu_2 (y) \right) d\mu_1 (x)= \int_{\Omega_2} \left( \int_{\Omega_1} \chi_E (x,y) d\mu_1 (x) \right) d\mu_2 (y)
holds for every measurable set E\subseteq \Omega_1 \times \Omega_2.
Definition 3.11.9
Let \Omega_1 and \Omega_2 be \sigma-finite measure sets. Then we define a measure \mu on the product \Omega_1 \times \Omega_2 by writing
\mu (E) := \int_{\Omega_1} \left( \int_{\Omega_2} \chi_E (x,y) d\mu_2 (y) \right) d\mu_1 (x)= \int_{\Omega_2} \left( \int_{\Omega_1} \chi_E (x,y) d\mu_1 (x) \right) d\mu_2 (y)
whenever the set E\subseteq \Omega_1 \times \Omega_2 is measurable.
It is easy to check that the above definition satisfies the axioms required of a measure. As a special case of the above definition, we can now define a Lebesgue measure on the space {\mathbb R}^n by viewing it as a product of copies of the space \mathbb R. This measure is defined on every Borel set, and the measure of the n-dimensional cuboid
[a_1 , b_1] \times \cdots \times [a_n , b_n]
is the product
(b_1 - a_1)(b_2-a_2) \cdots (b_n-a_n)
In the previous section, we saw how to define measures on products of \sigma-finite measure spaces. We can therefore integrate on such spaces. The purpose of this section is two state two results on the integrability of such functions, and how they are integrated. These results are usually put together, and referred to in one piece as \em Fubini's theorem.
Theorem 3.12.1
Let \Omega_1 and \Omega_2 be \sigma-finite measure spaces, and let f\colon \Omega_1 \times \Omega_2 \rightarrow {\mathbb C} be an integrable function. Then the functions f(x,-) and f(-,y) are integrable almost everywhere, and the functions
x\mapsto \int_{\Omega_2} f(x,y) d\mu_2 (y) \qquad y\mapsto \int_{\Omega_2} f(x,y) d\mu_2 (y)
are integrable.
Let s\colon \Omega_1 \times \Omega_2 \rightarrow {\mathbb C} be a simple function. Then the functions s(x,-) and s(-,y) are integrable almost everywhere, the functions
x\mapsto \int_{\Omega_2} s(x,y) d\mu_2 (y) \qquad y\mapsto \int_{\Omega_2} s(x,y) d\mu_2 (y)
are integrable, and the equation
\int_{\Omega_1 \times \Omega_2} s(x,y) d\mu (x,y)
=\int_{\Omega_1} \left( \int_{\Omega_2} s (x,y) d\mu_2 (y) \right) d\mu_1 (x)= \int_{\Omega_2} \left( \int_{\Omega_1} s (x,y) d\mu_1 (x) \right) d\mu_2 (y)
holds by theorem 3.11.8 and the definition of the product measure.
Now, suppose that f(x,y)\geq 0 for all points (x,y)\in \Omega_1 \times \Omega_2. Since the function f is measurable, by proposition 3.3.2 there is a monotonically increasing sequence, (s_n), of simple functions, with point-wise limit f. The result therefore follows in this case by the monotone convergence theorem.
By splitting a real-valued function into positive and negative parts, we see that the result holds for all real-valued functions. We can deduce the result for complex-valued functions by splitting such a function into real and imaginary parts.
For the above theorem to be useful, we would like a criterion for a function f\colon \Omega_1 \times \Omega_2 \rightarrow {\mathbb C} to be integrable. Fortunately, such a condition forms the second half of Fubini's theorem, which is also sometimes referred to as Tonelli's theorem.
Theorem 3.12.2
Let \Omega_1 and \Omega_2 be \sigma-finite measure spaces, and let f\colon \Omega_1 \times \Omega_2 \rightarrow {\mathbb C} be an integrable function. Suppose that
\int_{\Omega_1} \left( \int_{\Omega_2} |f (x,y)| d\mu_2 (y) \right) d\mu_1 (x) < \infty
or
\int_{\Omega_2} \left( \int_{\Omega_1} |f (x,y)| d\mu_1 (x) \right) d\mu_2 (y) < \infty
Then the function f\colon \Omega_1\times \Omega_2 \rightarrow {\mathbb C} is integrable.
Proof.
The result is obvious if the function f is simple. A similar argument to the proof of Fubini's theorem gives us the result in general.
Combining the two theorems in this section (ie: the two halves of Fubini's theorem), we have the following handy result on swapping the order of integration.
Corollary 3.12.3
Let \Omega_1 and \Omega_2 be \sigma-finite measure spaces, and let f\colon \Omega_1 \times \Omega_2 \rightarrow {\mathbb C} be an integrable function.  This endows the ground field with a topology which will be needed in the
definition of a topological vector space. We therefore give here a brief introduction to topological division
rings and fields first.
 By an absolute value on R one understands a map |\cdot | : R \to \mathbb{R}_{\geq 0}
such that the following axioms hold true.
A division ring or field endowed with an absolute value is called a valued division ring respectively a valued field. An absolute value |\cdot| on a division ring R and the corresponding valued division ring (R,|\cdot|) are called non-archimedean if the strong triangle inequality is satisfied that is if
Otherwise |\cdot| and (R,|\cdot|) are called archimedean.
Lemma 4.1.3
Let (R,|\cdot|) be a valued division ring. Then
Proof.
(I) holds true since |1| = |1^2| = |1|^2 and |1| \neq 0 by 1 \neq 0.
To verify (II) it suffices to show that |-1| =1. But that holds true
since |-1|^2 = | (-1)^2| = 1 and |-1| \geq 0.
The last claim follows by
- |x-y| = |x| - (|y-x| +|x|) \leq |x| - |y| \leq |x-y| + |y | - |y| = |x-y|
and
|x - y| = | x + (-y)| \leq |x| + |-y| = |x| + |y| .
Examples 4.1.4
Proposition 4.1.5
A valued division ring (R,|\cdot|) is non-archimedean if and only if the image of \mathbb{Z} under
the canonical map \mathbb{Z}\to R is bounded.
Proof.
Assume that (R,|\cdot|) is a non-archimedean valued division ring.
Then, |0 \cdot 1| = |0|= 0 and, under the assumption that |(n-1)\cdot 1|\leq 1 for some n\in \mathbb{N}_{ > 0},
|n\cdot 1| = | (n-1)\cdot 1 + 1 | = \max \{ |(n-1)\cdot 1| , 1 \} = 1.
Hence by induction and since |-1| =1 one obtains that |n\cdot 1|\leq 1 for all n\in \mathbb{Z},
and the image of \mathbb{Z} in R is bounded.
To show the converse assume that the image of \mathbb{Z} in R is bounded by some constant C > 0. Then, for all x,y\in R
and n\in \mathbb{N}_{ > 0} by the binomial formula and the triangle inequality
|x+y|^n =\left|\sum_{k=0}^n {n \choose k} x^k y^{n-k} \right|
\leq (n+1) \,C \max \{|x|,|y|\}^n .
Taking the n-th root gives |x+y|\leq \big( (n+1)C\big)^{1/n} \max \{|x|,|y|\} which after passing to the limit n\to\infty
entails |x+y| \leq \max \{|x|,|y|\} since \lim\limits_{n\to\infty} \big( (n+1)C\big)^{1/n} = 1. Hence
(R,|\cdot|) is non-archimedean.
 Then for every \tau > 0 with \tau\leq 1 the
map |\cdot|^\tau: R\to\mathbb{R}_{\geq 0} is an absolute value on R as well.
It is archimedean if and only if |\cdot| is archimedean.
Proof.
To prove that |\cdot|^\tau is an absolute value it suffices to show that (a+b)^\tau \leq a^\tau + b^\tau
for all a,b\geq 0. Without loss of generality we may assume a \geq b > 0. By dividing through b^\tau
one sees that the claim is equivalent to (t +1 )^\tau \leq t^\tau + 1 for all t\geq 1.
For t=1 this is certainly true. The derivative of the function
h : {[ 1,\infty \rtsbrak} \to\mathbb{R}, t\mapsto (t +1 )^\tau - t^\tau now is
given by h'(t) = \tau \big( (t +1 )^{\tau-1} - t^{\tau-1} \big) which is negative
since \tau-1\leq 0 and 1+ t > t\geq 1. Hence h is monotone decreasing
and (t +1 )^\tau - t^\tau \leq 1 for all t\geq 1.
Since {\ltsbrak 0,\infty \rtsbrak} \to \mathbb{R}, t\mapsto t^\tau is strictly increasing and unbounded,
the image of \mathbb{Z} in R is unbounded with respect to |\cdot| if and only if
it is with respect to |\cdot|^\tau.
4.1.7 An absolute value |\cdot | : R \to \mathbb{R}_{\geq 0} on a division ring R induces the metric
d : R \times R \to \mathbb{R}_{\geq 0}, (x,y) \mapsto |x-y| which then gives
rise to a topology on R. This topology has the following properties:
Proof.
Addition is continuous since for all a,b,x,y \in R by the triangle inequality
d ( x + y , a+ b ) = | x + y -( a+ b)| \leq | x-a| + | y-b| = d(x,a) + d(y,b).
Actually, this even shows that addition is Lipschitz continuous.
Now fix a,b \in R and let C = \max \{ |a|,|b| \} + 1. Then for all x,y \in R
with d(y,b) < 1
d ( x \cdot y , a \cdot b ) = | (x \cdot y - a \cdot y) + (a \cdot y - a \cdot b)|
\leq | x-a| \, |y| + |a| \, | y-b| \leq C \big( d(x,a) + d(y,b) \big) .
Hence multiplication is continuous.
Finally, fix a \in R^\times and let x \in R^\times with d(x,a) < \frac{|a|}{2}. Then
|x| \geq |a| - d(x,a) > \frac{|a|}{2} > 0 and
d \left( x^{-1} , a^{-1} \right) = \left| x^{-1} - a^{-1} \right| = \left| x^{-1} \cdot a^{-1} \right|
\, | x-a | = \frac{1}{|x| \, |a|} d( x,a ) < \frac{2}{|a|^2} d( x,a ) .
So inversion is also continuous.
Definition 4.1.8
A division ring or field R which is equipped with a topology so that
(TDR1),
(TDR2) and
(TDR3) are satisfied is called a
topological division ring or a topological field, respectively.
Lemma 4.1.9
If |\cdot| is a non-trivial absolute value on the division ring R, then there exists an element
t\in R^\times such that the sequence (t^n)_{n\in\mathbb{N}} converges to 0. Furthermore in this case
every 0-neighborhood in R contains infinitely many elements.
Proof.
By non-triviality of |\cdot| there exists t\in R^\times such that |t|\neq 1. By possibly
passing to t^{-1} we can assume |t| < 1. Since then \lim\limits_{n\to\infty} |t|^n =0,
the sequence (t^n)_{n\in\mathbb{N}} converges to 0. This implies in particular that
for every \varepsilon > 0 the open ball \mathbb{B} (0,\varepsilon) =\{t \in R \mid |t| < \varepsilon \}
contains infinitely many elements. So the lemma is proved.

Then they are equivalent if and only if there exists e > 0 such that |\cdot|^\prime =|\cdot|^\tau.
In particular the trivial absolute value is the only one inducing the discrete topology on R.
Proof.
Let us first show the following proposition.
Since \left| x^{-1}\right| =\frac{1}{|x|} and \left| x^{-1}\right|^\prime =\frac{1}{|x|^\prime}
for all x \in R^\times, ( A)
implies that |x| > 1 if and only if |x|^\prime > 1 and that |x|= 1 if and only if |x|^\prime = 1.
To verify claim ( A) assume now that 0 < |x| < 1.
Then \lim\limits_{n\to\infty}|x^n|=0, hence (x^n)_{n\in\mathbb{N}} converges to 0. By assumption,
\lim\limits_{n\to\infty}|x^n|^\prime=0 then holds as well which implies that |x|^\prime < 1.
By switching |\cdot| and |\cdot|^\prime the converse holds true, so
( A) is proved.
Next we show that |\cdot| is trivial if and only if the induced topology on R is discrete.
Namely, if |\cdot| is non-trivial, then there exists x\in R^\times such that |x| \neq 1. After
possibly passing to \frac 1x we can achieve that |x| < 1. So \lim\limits_{n\to\infty}|x^n|=0, which
means that (x^n)_{n\in\mathbb{N}} is a sequence of non-zero elements of R converging to 0. But this implies that
the singleton \{ 0\} is not open in the topology induced by |\cdot|, hence this topology is non-discrete.
Since obviously the trivial absolute value induces the discrete topology on R the second claim of the theorem is
proved.
Now assume that |\cdot|^\prime =|\cdot|^\tau for some \tau > 0. Then a subset B \subset R is a metric
open ball with respect to |\cdot| if and only if it is one with respect to |\cdot|^\prime
since for x \in R and \varepsilon > 0
\big\{ y \in R \bigm\vert |y -x| < \varepsilon \big\} = \big\{ y \in R \bigm\vert |y-x|^\prime < \varepsilon^\tau \big\}
\text{ and }
\big\{ y \in R \bigm\vert |y- x|^\prime < \varepsilon \big\} = \big\{ y \in R \bigm\vert |y-x| < \varepsilon^{1/\tau} \big\} .
Hence the open sets with respect to the metric defined by |\cdot| coincide with those defined by |\cdot|^\prime and
the two absolute values are equivalent.
Let us finally show the other direction and assume that |\cdot| and |\cdot|^\prime are equivalent.
By the already proven second claim of the theorem we can restrict to the case where the induced topology is non-discrete
which means to the case where both |\cdot| and |\cdot|^\prime are non-trivial.
We show that there exists \tau > 0 such that |x|^\prime = |x|^\tau for all x\in R^\times with |x| > 1.
This is sufficient, since if |x|= 1, then |x|^\prime = 1 = |x|^\sigma for any \sigma > 0
by ( A), and since if x\in R^\times with |x| < 1 then
|x^{-1}| > 1 and
|x|^\prime = \frac{1}{\left|x^{-1}\right|^\prime} = \frac{1}{\left|x^{-1}\right|^\tau} = |x|^\tau .
The existence of a \tau > 0 with the claimed property is equivalent to the function
R^\times \to \mathbb{R},\: x \mapsto \frac{\ln |x|^\prime}{\ln |x|}
being constant. Assume that that is not the case. Then there exist x,y \in R^\times
with |x|,|y| > 1 such that
\frac{\ln |x|^\prime}{\ln |x|} \neq \frac{\ln |y|^\prime}{\ln |y|}. By possibly switching x and y we
can assume \frac{\ln |x|^\prime}{\ln |x|} < \frac{\ln |y|^\prime}{\ln |y|}.
But that implies \frac{\ln |x|^\prime}{\ln |y|^\prime} < \frac{\ln |x|}{\ln |y|} since the logarithms are positive
by assumptions on x and y and ( A). Hence there exists a
rational number \frac pq with p,q \in \mathbb{N}_{ > 0} such that
\frac{\ln |x|^\prime}{\ln |y|^\prime} < \frac pq < \frac{\ln |x|}{\ln |y|} .
Then |x^q|^\prime < |y^p|^\prime and |y^p| < |x^q| which entails
\left| \frac{x^q}{y^p} \right|^\prime < 1 \text{ and }
\left| \frac{x^q}{y^p} \right| > 1 .
This contradicts ( A) and the theorem is proved.

A topology \mathscr{T} on a vector space \mathrm{E} over R
is called a vector space topology if the following axioms hold true:
A vector space \mathrm{E} together with a vector space topology on it
is called a topological vector space (over R), for short a \textt.v.s.
Remark 4.2.2
Let us recall at this point some notation from linear algebra. Assume that \mathrm{V} is
a left vector space over the divison ring R.
If A,B\subset \mathrm{V} are two non-empty subsets, then A+B is
the set of all v\in \mathrm{V} for which there exist x\in A and y\in B such that
v = x+y. If A or B is empty, then A+B is defined as the empty set.
In case A is a singleton that is if A =\{x\}, then we often write x + B instead of
\{x\} +B. If \mathscr{B} \subset \mathscr{P}(\mathrm{V}) is a non-empty set of subsets of \mathrm{V},
then we denote by A + \mathscr{B} and x + \mathscr{B} the sets
\{ A + B \in \mathscr{P}(\mathrm{V}) \mid B \in \mathscr{B}\} and
\{ x + B \in \mathscr{P}(\mathrm{V}) \mid B \in \mathscr{B}\} , respectively.
If \mathscr{A} \subset \mathscr{P}(\mathrm{V}) is a second non-empty set of subsets of \mathrm{V},
then \mathscr{A} + \mathscr{B} stands for the set of all sets of the form A+B, where
A\in \mathscr{A} and B\in \mathscr{B}.
In case C is a subset of the ground ring R, then C \cdot A is defined as the set of all
v\in \mathrm{V} for which there exist
c\in C and x\in A such that v =c \cdot x. If C =\{c\} we write c\cdot A
for \{c\} \cdot A. Likewise, if A =\{x\}, C\cdot x stands for C\cdot\{x\}.
Analogously as for addition the sets \mathscr{C}\cdot A, C\cdot \mathscr{A}
and \mathscr{C}\cdot \mathscr{A} are defined when
\mathscr{C} \subset \mathscr{P}(R) and \mathscr{A} \subset \mathscr{P}(\mathrm{V})
are non-empty.
Proposition 4.2.3
Let \mathrm{E} be a \textt.v.s.~over a topological division ring R.
Then the following holds true:
Proof.
Definition 4.2.4
A subset C of a vector space \mathrm{E} over a valued division ring (R,|\cdot|) is called
Remark 4.2.5
Symmetry of a subset of a vector space of a division ring is even defined when the underlying division
ring does not carry an absolute value.
Lemma 4.2.6
Let C be a subset of a topological vector space \mathrm{E} over a
valued division ring (R,|\cdot|) and r\in R.
Proof.
Without loss of generality we can assume C \neq \emptyset.
Claim (I) then follows immediately since multiplication
by -1 is a homeomorphism.
To prove claim (II) assume that C is circled.
Let t\in R with |t|\leq 1. Assume v \in \overline{C} and consider tv. We have to show
that tv \in \overline{C}. If t=0 then tv =0\in C \subset \overline{C} since C is circled.
So we can assume t\neq 0 and need to show that for every neighborhood V of tv the intersection
C\cap V is non-empty. Since |t| > 0, the homothety \ell_t:\mathrm{E}\to \mathrm{E}, w\mapsto tw is a
homeomorphism with inverse
\ell_{t^{-1}}. Hence t^{-1}V is a neighborhood of v. Since v lies in the closure of C
there exists an element w\in C\cap t^{-1}V. Hence tw \in C \cap V by assumption on C
and \overline{C} is circled.
If v \in \mathring{C}\cup\{0\} then 0 = 0 \cdot v \in \mathring{C}\cup\{0\}. It remains to show that
tv \in \mathring{C}\cup\{0\} for t\in R with 0 < |t|\leq 1 and v \in \mathring{C} \setminus \{ 0\}.
Under this assumption the homothety \ell_t is a homeomorphism, so t \mathring{C} is an open subset
of C since C is circled. Hence t v \in t \mathring{C} \subset \mathring{C}, and \mathring{C}\cup\{0\}
is circled as well.
Claim (III) follows immediately from the observation that
for v\in C and t\in R the relation trv \in rC holds true if tv \in C.
Proposition and Definition 4.2.7
The intersection of a non-empty family (C_i)_{i\in I} of symmetric (respectively circled) subsets C_i \subset \mathrm{E}, i\in I
of a topological vector space \mathrm{E} valued division ring (R,|\cdot|)
is symmetric (respectively circled).
In particular, if A \subset \mathrm{E} is a subset, then the sets
\operatorname{Sym} A = \bigcap\limits_{A \subset B \subset \mathrm{E} \atop {B \text{ is symmetric}} } B \quad \text{and} \quad
\operatorname{Circ} A = \bigcap\limits_{A \subset B \subset \mathrm{E} \atop {B \text{ is circled}} } B
are symmetric and circled, respectively. They have the property that
\operatorname{Sym} A is the smallest symmetric and \operatorname{Circ} A the smallest circled subsets of \mathrm{E} containing A.
They are called the symmetric and the circled hull of A, respectively.
Analogously,
\overline{\operatorname{Sym}}\, A = \bigcap\limits_{A \subset B =\overline{B} \subset \mathrm{E} \atop {B \text{ is symmetric}} } B \quad \text{and} \quad
\overline{\operatorname{Circ}}\, A = \bigcap\limits_{A \subset B =\overline{B} \subset \mathrm{E} \atop {B \text{ is circled}} } B
are called the closed symmetric and the closed circled hull of A, respectively.
They have the property that \overline{\operatorname{Sym}}\, A is the smallest closed symmetric and \overline{\operatorname{Circ}}\, A the smallest closed
circled subset of \mathrm{E} containing A.
Proof.
Note first that all the hulls in the proposition are well-defined since \mathrm{E} is
closed and circled.
Let C denote the intersection of the family (C_i)_{i\in I}. Assume that for some t\in R with |t|\leq 1
the inclusion tC_i \subset C holds true for all i\in I. Then tC \subset C, hence if all
C_i are symmetric (respectively circled), so is C.
This observation now entails that \operatorname{Sym} A is symmetric, \operatorname{Circ} is circled, \overline{\operatorname{Sym}}\, A is closed
and symmetric, and finally that \overline{\operatorname{Circ}}\, A is closed and circled. Moreover, all those sets contain A.
The minimality properties of these sets are clear by construction.
Remark 4.2.8
Observe that by the proposition A is symmetric if and only if \operatorname{Sym} A = A and
circled if and only if \operatorname{Circ} A = A. Analogously,
\overline{\operatorname{Sym}}\, A = A if and only if A is closed symmetric and
\overline{\operatorname{Circ}}\, A = A if and only if A is closed and circled.
Lemma 4.2.9
Let \mathrm{E} be a topological vector space over the valued division ring (R,|\cdot|)
and A \subset \mathrm{E} non-empty. Then
\operatorname{Sym} A = A \cup -A \quad \text{and} \quad \operatorname{Circ} A = \bigcup_{t\in R, \: |t|\leq 1} t A .
For the closed hulls one has
\overline{\operatorname{Sym}}\, A = \overline{\operatorname{Sym} A} \quad \text{and} \quad \overline{\operatorname{Circ}}\, A = \overline{\operatorname{Circ} A} .
Proof.
Since A \cup -A is symmetric by definition, contains A, and is contained in
\operatorname{Sym} A, the equality \operatorname{Sym} A = A \cup -A holds true.
Similarly, \bigcup_{t\in R, \: |t|\leq 1} t A is circled by definition, contains A, and is contained
in \operatorname{Circ} A by definition of the circled hull. Hence \operatorname{Circ} A = \bigcup_{t\in R, \: |t|\leq 1} t A .
The remainder of the claim follows from \Crefthm:closure-interior-circled-set-circled.
Definition 4.2.10
Assume that B,C are subsets of a vector space \mathrm{E} over the valued division ring
(R,|\cdot|). Then one says that
If the vector space \mathrm{E} carries in addition a vector space topology, then one says that
Proposition 4.2.11
The filter of zero neighborhoods of a topological vector space \mathrm{E} over
(R,|\cdot|) has a filter base \mathscr{B} with the following properties:
Conversely, if \mathscr{B} is a filter base on an R-vector space \mathrm{E} such that
(I) to (III)
hold true, then there exists a unique vector space topology on \mathrm{E} such that \mathscr{B} is a neighborhood base
at the origin. In case the ground ring R is archimedean, a filter base on \mathrm{E} which satisfies
(I) and (II)
already induces a unique vector space topology having \mathscr{B} as a neighborhood base
at 0.
Proof.
Assume that \mathrm{E} is a \textt.v.s. Let \mathscr{B} be the set of circled neighborhoods of 0. We show first that
\mathscr{B} is a base of the filter \mathscr{U}_0 of zero neighborhoods. Let W \in \mathscr{U}_0.
By Axiom (TVS2) there exists an \varepsilon > 0 and an open
zero neighborhood U such that tU \subset W for all t\in R with |t| < \varepsilon.
Then V = \bigcup\limits_{t\in R^\times \:\&\: |t| < \varepsilon} tU is a zero neighborhood since
by \Crefthm:zero-neighborhood-topological-division-ring-infinite the set of t\in R^\times
with |t| < \varepsilon is non-empty. By construction V is contained in W and circled, so V \in \mathscr{B}.
Hence \mathscr{B} is a filter base of \mathscr{U}_0.
Next recall that there exists t\in R^\times with 0 < |t| < 1 since the absolute value |\cdot| is non-trivial.
Let V\in \mathscr{B}. Then sV\subset V for all s\in R with |s|\leq 1 which entails stV \subset tV for all
such s. Hence tV is circled and an element of \mathscr{B} as well. This proves
(III).
Since addition on \mathrm{E} is continuous, there exist for given V\in \mathscr{B} open neighborhoods
U_1,U_2 of the origin such that U_1+U_2\subset V. Choose U\in \mathscr{B} such that U \subset U_1 \cap U_2.

To show that any V\in\mathscr{B} is absorbing let v\in \mathrm{E}. By continuity of scalar multiplication
there exists \varepsilon > 0 such that tv \in V for all t\in R with |t| < \varepsilon.
By \Crefthm:basic-properties-sets-maps-topological-vector-spaces (I)
this entails v \in tV for all t\in R with |t| > \varepsilon and V is absorbing.
Now assume that \mathrm{E} is an R-vector space and that \mathscr{B} is a filter base that satisfies
(I), (II) and,
if |\cdot| is non-archimedean, (III).
Since \mathscr{B} consists of non-empty circled sets, 0\in V for all V\in\mathscr{B}. Let
\mathscr{T} \subset \mathscr{P}(\mathrm{E}) be the set of all U\subset \mathrm{E} such that
for each v\in U there exists V\in \mathscr{B} with v+V \subset U. By definition and since \mathscr{B} is a
filter base, \mathscr{T} is a topology on \mathrm{E}. We show that \mathscr{B} is a base of the filter
\mathscr{U}_0 of zero neighborhoods. By definition of \mathscr{T} there exists for each U \in \mathscr{U}_0
a V\in \mathscr{B} such that V\subset U. So it remains to show that each V\in \mathscr{B} is a zero neighborhood.
To this end let U be the set of all v\in V for which there exists
a W\in \mathscr{B} with v + W \subset V. Since 0 +V \subset V one has 0\in U. The relation
U \subset V holds because 0\in W for all W\in \mathscr{B}. Now let v \in U.
By (I) there exists W^\prime such that
v + W^\prime + W^\prime \subset V which entails v + W^\prime \subset U. Hence U\in\mathscr{T}
and V is a zero neighborhood.
Next we verify that \mathscr{T} is a vector space topology. We start with continuity of addition.
Let W be an open neighborhood of v+w, where v,w\in \mathrm{E}. Then there exists V\in \mathscr{B} such that
v+w + V \subset W. Choose U\in \mathscr{B} such that U+U \subset V. Then
v+U and w +U are neighborhoods of v and w, respectively, and (v+U) + (w +U) \subset v+w +V\subset W.
So addition is continuous. We continue with scalar multiplication. Let W be an open neighborhood of rv,
where r\in R and v\in \mathrm{E}. Then there exists V \in \mathscr{B} such that rv + V + V \subset W.
Since V is absorbing by (II) there exists \varepsilon > 0 such that
(s-r)v \in V for all s\in R with |s-r| < \varepsilon. Now if |\cdot| is non-archimedean
choose t\in R^\times according to (III),
and put V_n = t^n V for all n\in\mathbb{N}.
In the archimedean case let t = \frac 12 and use (I) to construct
recursively a sequence (V_n)_{n\in \mathbb{N} } of elements of \mathscr{B} such that
2^nV_n = V_n + \ldots + V_n \subset V, where the sum has 2^n summands. In either of these cases, choose
N\in \mathbb{N} large enough so that |t|^N < \frac{1}{|r| + \varepsilon}. Then V_N \in \mathscr{B}
and v + V_N is a neighborhood of v. Moreover, for w \in v + V_N there exists an element
x\in V such that w-v = t^N x. Then the relation
s (w-v) = s t^N x \in V holds whenever |s-r| < \varepsilon since V_N is circled. Hence for such w and s
sw = rv + s (w-v) + (s-r)v \in rv + V + V \subset W .
This means that scalar multiplication is continuous, and the proof is finished.
\addcontentslinetocsubsectionMorphisms of topological vector spaces
Definition 4.2.12
By a morphism of topological vector spaces over the topological division
ring R one understands a continuous R-linear map f:\mathrm{E} \to\mathrm{F}
between two topological vector spaces \mathrm{E} and \mathrm{F}
over R. The space of morphisms between \mathrm{E} and \mathrm{F} will be denoted
\operatorname{Hom}\nolimits_{R\text{-}\mathsf{TVS}} (\mathrm{E},\mathrm{F}) or just
\operatorname{Hom}\nolimits_R (\mathrm{E},\mathrm{F}) or \operatorname{Hom}\nolimits (\mathrm{E},\mathrm{F})
if now confusion can arrise.
Theorem 4.2.13
The topological vector spaces over a topological division ring R as objects together
with their morphisms form an additive category which we denote by R\text{-}\mathsf{TVS}.

Proof.
Observe first that the identity map \mathrm{id}_\mathrm{E} on a topological vector space \mathrm{E}
is linear and continuous and so is the composition g\circ f of two
morphisms of topological vector spaces f:\mathrm{E} \to\mathrm{F} and
g:\mathrm{F} \to\mathrm{G}. Hence topological vector spaces over R together with linear
and continuous maps between them form a category.
Next check that the hom-space \operatorname{Hom}\nolimits (\mathrm{E},\mathrm{F}) is an abelian group.
Associativity and commutativity of addition
follow from the respective properties on \mathrm{F}. The zero element is the constant map
\mathrm{E} \to\mathrm{F}, v \mapsto 0 and the inverse of a morphism f:\mathrm{E} \to\mathrm{F}
is given by -f:\mathrm{E} \to\mathrm{F}, v \mapsto -f(v).
Similarly one checks that multiplication by scalars on \operatorname{Hom}\nolimits (\mathrm{E},\mathrm{F}) is associative
and distributes from the left and from the right over addition since scalar multiplication on
\mathrm{F} has these properties. Finally, the unit of R acts as identity on
\operatorname{Hom}\nolimits (\mathrm{E},\mathrm{F}) since it does so on \mathrm{F}. Hence \operatorname{Hom}\nolimits (\mathrm{E},\mathrm{F})
carries the structure of an R left vector space.
Composition of morphisms
\operatorname{Hom}\nolimits (\mathrm{E},\mathrm{F}) \times \operatorname{Hom}\nolimits (\mathrm{F},\mathrm{G}) \to \operatorname{Hom}\nolimits (\mathrm{E},\mathrm{G}),
(f,g)\to g \circ f is an R-bilinear map as the following equalities for
f,f_1,f_2\in \operatorname{Hom}\nolimits (\mathrm{E},\mathrm{F}), g,g_1,g_2 \in \operatorname{Hom}\nolimits (\mathrm{F},\mathrm{G}),
r\in R, and v\in \mathrm{E} show:
\tag{4.2.3}
(f \circ (g_1+g_2)) (v) = f( (g_1+g_2) (v) ) = f( g_1(v) +g_2(v) ) =
\quad = f\circ g_1(v) + f\circ g_2(v) = (f\circ g_1+ f\circ g_2) (v),
(f \circ (r g)) (v) = f( (rg) (v) ) = f(r g(v)) = r f( g(v)) = (r(f\circ g)) (v),
((f_1+f_2) \circ g) (v) = (f_1+f_2) ( g (v)) = f_1(g(v)) + f_2 (g(v)) =
\quad = f_1\circ g(v) + f_2\circ g(v) = (f_1\circ g+ f_2\circ g) (v),
((rf) \circ g) (v) = (rf)( g (v)) = r (f(g(v))) = r (f \circ g (v)) = (r(f\circ g)) (v) .
Hence R\text{-}\mathsf{TVS} is a category enriched over the category of R-vector spaces.
In particular, R\text{-}\mathsf{TVS} then is an additive category.
Example 4.2.14
For every \textt.v.s.~\mathrm{E} and non-zero element t of the ground ring R the
map \ell_t :\mathrm{E} \to \mathrm{E}, v \mapsto tv is an isomorphism of topological vector spaces
by \Crefthm:basic-properties-sets-maps-topological-vector-spaces (I).
Proposition and Definition 4.2.15
A linear map f: \mathrm{E} \to \mathrm{F} between topological
vector spaces over a valued division ring (R,|\cdot|)
maps symmetric sets to symmetric sets and circled sets to circled sets.
If in addition f is continuous, then f is bounded that means
it maps bounded subsets of \mathrm{E} to bounded subsets of
\mathrm{F}.
Proof.
Since by linearity f(tv) = t f(v) for all v\in \mathrm{E} and t\in R,
f (C) is symmetric (respectively circled) if the subset C\subset \mathrm{E} is.
To verify the second claim let B\subset \mathrm{E} be bounded and
V\subset \mathrm{F} a zero neighborhood. Then f^{-1} (V) is a zero neighborhood
in \mathrm{E} by continuity of f. Hence there exists an r\in \mathbb{R}_{\geq 0}
such that B \subset tf^{-1} (V) for all t\in R with |t|\geq r. By linearity
of f one obtains f(B) \subset tV for all such t, so f is bounded.
Remark 4.2.16
By the proposition continuity of a linear map between topological vector spaces
implies the map to be bounded. As we will see later in this monograph,
the converse does in general not hold true unless the underlying
topological vector spaces are for example normable.
\addcontentslinetocsubsectionNormed real division algebras and local convexity
4.2.17
We give valued division rings (R,|\cdot|) which carry the structure of an \mathbb{R}-algebra such that
for all r\in \mathbb{R} and x\in R
| rx | = |r|_\infty\cdot |x|
a particular name and call them normed real division algebras. Since \mathbb{R} with its standard absolute value
is archimedean, so is every normed real division algebra.  A subset C of an
R-vector space \mathrm{E} then is absolutely convex if and only
if it is circled and convex.
Proof.
The claim is trivial when C=\emptyset, so we assume that C is nonempty.
Let C be absolutely convex. Since C contains at least one element v
one has 0 = 0 \cdot v + 0 \cdot v \in C. Hence
tv = (1-|t|) \cdot 0 + t v \in C for all v\in C and t\in R with |t|\leq 1.
So C is circled. By definition of absolute convexity C is convex.
If C is circled and convex, then it contains with elements v,w also t v + s w
if |t| + |s|\leq 1. To see this observe first that \tau v \in C and \sigma w \in C
where the elements \tau,\sigma \in R have been chosen so that
|\tau|=|\sigma|=1, t = |t| \cdot\tau and s = |s| \cdot\sigma. Now if |t| + |s| =0, then
t v + s w = 0 \in C since C is circled. If |t| + |s| > 0, then
t v + s w =
(|t| + |s|) \left( \frac{|t|}{|t|+|s|} \tau v + \frac{|s|}{|t|+|s|} \sigma w\right) \in C
since C is convex and circled. Hence C is absolutely convex.
Lemma 4.2.20
A linear map f:\mathrm{E} \to \mathrm{F} between vector spaces over a normed
real divison algebra R maps convex sets to convex sets,
absolutely convex sets to absolutely convex sets, and cones to cones.
Proof.
This an immediate consequence of the linearity of f.
 Then the following holds true.
Proof.
We consider only the cases C,D\neq \emptyset because otherwise the claim is trivial.
Proposition and Definition 4.2.22
The intersection of a non-empty family (C_i)_{i\in I} of convex (respectively absolutely convex) subsets
C_i \subset \mathrm{E}, i\in I of a topological vector space \mathrm{E} over a normed real division algebra R
is convex (respectively absolutely convex).
In particular, if A \subset \mathrm{E} is a subset, then the sets
\operatorname{Conv} A = \bigcap\limits_{A \subset B \subset \mathrm{E} \atop {B \text{ is convex}} } B \quad \text{and} \quad
\operatorname{AConv} A = \bigcap\limits_{A \subset B \subset \mathrm{E} \atop {B \text{ is absolutely convex}} } B
are convex and absolutely convex, respectively. The set \operatorname{Conv} A is called the convex hull of A
and is the smallest convex set containing A. Similarly, \operatorname{AConv} A is the smallest absolutely convex set
containing A. It is called the absolutely convex hull of A.
The closed convex hull \overline{\operatorname{Conv}}\, A and the closed absolutely convex hull \overline{\operatorname{AConv}}\, A
of A are defined by
\overline{\operatorname{Conv}}\, A = \bigcap\limits_{A \subset B = \overline{B} \subset \mathrm{E} \atop {B \text{ is convex}} } B
\quad \text{and} \quad
\overline{\operatorname{AConv}}\, A = \bigcap\limits_{A \subset B = \overline{B} \subset \mathrm{E} \atop {B \text{ is absolutely convex}} } B .

Let C be the intersection \bigcap\limits_{i\in I} C_i and assume that each C_i is absolutely convex.
Let v,w\in C and s,t \in R. Then v,w\in C_i, hence tv + sw \in C_i for all i\in I. Therefore
tv +sw \in C and C is absolutely convex. This argument also shows that C is convex if all C_i are convex.
The rest of the claim follows as in the proof of \Crefthm:symmetric-circled-hulls.
Remark 4.2.23
The proposition in particular entails that A is convex if and only if \operatorname{Conv} A = A and
absolutely convex if and only if \operatorname{AConv} A = A. Analogously,
\overline{\operatorname{Conv}}\, A = A if and only if A is closed and convex,
and \overline{\operatorname{AConv}}\, A = A if and only if A is closed and absolutely convex.
Lemma 4.2.24
Let A \subset \mathrm{E} be a non-empty subset of a \textt.v.s.~\mathrm{E}
over a normed real division algebra R. Then
\tag{4.2.4}
\operatorname{Conv} A = \left\{ \sum_{i=1}^k t_i v_i\in
\mathrm{E} \bigm\vert k \in \mathbb{N}_{ > 0},\: v_1,\ldots v_k\in A,\: t_1\ldots ,t_k \in \mathbb{R}_{\geq 0}, \: \sum_{i=1}^k t_i = 1 \right\}
,
\operatorname{AConv} A = \left\{ \sum_{i=1}^k t_i v_i\in
\mathrm{E} \bigm\vert k \in \mathbb{N}_{ > 0},\: v_1,\ldots v_k\in A,\: t_1\ldots ,t_k \in R, \: \sum_{i=1}^k |t_i| \leq 1 \right\}
.
For the closed hulls one has
\overline{\operatorname{Conv}}\, A = \overline{\operatorname{Conv} A} \quad \text{and} \quad \overline{\operatorname{AConv}}\, A = \overline{\operatorname{AConv} A} .
Finally, if A is circled, then
\operatorname{AConv} A = \operatorname{Conv} A .
Proof.
By definition, the right hand side of Eq.~\eqrefeq:convex-hull-expression is convex and contains A,
hence it contains \operatorname{Conv} A. Conversely, one shows by induction on k\in \mathbb{N}_{ > 0} and convexity of \operatorname{Conv} A
that each element of the form \sum_{i=1}^k t_i v_i with v_1,\ldots , v_k\in A and
t_1,\ldots ,t_k \in \mathbb{R}_{\geq 0} such that \sum_{i=1}^k t_i = 1 is in \operatorname{Conv} A. This proves
Eq.~\eqrefeq:convex-hull-expression.
The proof of Eq.~\eqrefeq:absolutely-convex-hull-expression is similar. Observe that the right hand side of
Eq.~\eqrefeq:absolutely-convex-hull-expression is absolutely convex and contains A.
Hence it contains \operatorname{AConv} A. An argument using induction on k\in \mathbb{N}_{ > 0} and absolute convexity of
\operatorname{AConv} A shows that each element of the form \sum_{i=1}^k t_i v_i with v_1,\ldots v_k\in A and
t_1\ldots ,t_k \in R such that \sum_{i=1}^k |t_i| \leq 1 is in \operatorname{Conv} A.
So Eq.~\eqrefeq:absolutely-convex-hull-expression holds true as well.
The claim about the closed hulls is a consequence of \Crefthm:closure-interior-convex-set-convex.
For the proof of the last claim it suffices to show that \operatorname{Conv} A is circled if A is.
To this end let v \in \operatorname{Conv} A and t\in R with |t|\leq 1. Then
one can write v in the form v = \sum_{i=1}^k t_i v_i with
v_1,\ldots, v_k\in A and t_1, \ldots ,t_k \in \mathbb{R}_{\geq 0}, where \sum_{i=1}^k t_i = 1.
Hence rv = \sum_{i=1}^k t_i (rv_i), which is in \operatorname{Conv} A, since rv_i\in A for all i because
A is circled.
Lemma 4.2.25
Let A \subset \mathrm{E} be a non-empty subset of a \textt.v.s.~\mathrm{E}
over a normed real division algebra R.
Proof.
 Then a convex subset C\subset \mathrm{E} is
absorbent in \mathrm{E} if and only if it is absorbent in the realification \mathrm{E}^\mathbb{R}.
Proof.
It suffices to show the non-trivial direction. So assume that C is convex and absorbent in the
realification \mathrm{E}^\mathbb{R}. Denote by u_1,\ldots , u_n the standard basis of \mathbb{K}
over \mathbb{R} with n=2 or n=4 depending on \mathbb{K}. In particular this means u_1 =1.
For given v\in\mathrm{E} there now exists r\in\mathbb{R}_{\geq 0} such that
\pm \frac{1}{u_1} v , \ldots , \pm \frac{1}{u_n} v \in tC \quad \text{for all } t\geq r .
Without loss of generality we can assume r\geq 1. Let z\in \mathbb{K} with |z| \geq n r.
Then the vectors
c_1 = \operatorname{sgn} z_1\frac{n}{|z| \, u_1} v, \ldots , c_n = \operatorname{sgn} z_n\frac{n}{|z| \, u_n} v
are elements of C. By convexity of C and since 0\in C one has
\frac{|z_1|}{|z|} c_1 , \ldots , \frac{|z_n|}{|z|} c_n \in C. Again by convexity one concludes
\frac{1}{z} v = \sum_{i=1}^n \frac{z_i}{|z|^2 \, u_i} v =
\sum_{i=1}^n \frac{|z_i|}{n |z|} c_i \in C .
Hence C is absorbing and the claim is proved.
Definition 4.2.27
A topological vector space \mathrm{E} over a normed real division algebra R for which
Axiom \hyperref[axiom:tvs-local-convexity]LCVS below holds true is called a
locally convex topological vector space, a locally convex vector space or shortly a
locally convex \textt.v.s.
Remark 4.2.28
For better readability, we often say locally convex topology instead of locally convex vector space topology.
Proposition 4.2.29
The locally convex topological vector spaces over a normed real division algebra R together with the continuous
linear maps between them form a full subcategory of the category R\text{-}\mathsf{TVS} of topological
R-vector spaces. It is denoted R\text{-}\mathsf{LCVS}.
Proof.
This is clear by definition.
Proposition and Definition 4.2.30
The filter of zero neighborhoods of a locally convex topological vector space \mathrm{E} over a
normed real divison algebra R has a filter base \mathscr{B} with the following properties:
Conversely, if \mathscr{B} is a filter base on an R-vector space \mathrm{E} such that
(I) holds true and such that each
element of \mathscr{B} is absolutely convex and absorbing, then there exists a unique locally
convex vector space topology on \mathrm{E} such that \mathscr{B} is a neighborhood base
at the origin. One calls this topology the locally convex topology generated by \mathscr{B}.
Proof.
Let \mathrm{E} be a locally convex \textt.v.s.Let \mathscr{B} be the collection of all barrels
which are at the same time zero neighborhoods. Let V be an element of \mathscr{U}_0,
the filter of zero neighborhoods. Since \mathrm{E} is \textsf(T3) by
\Crefthm:basic-properties-sets-maps-topological-vector-spaces, there exists a closed
zero neighborhood V_a such that V_a \subset V . By local convexity of \mathrm{E} there exists
a convex zero neighborhood V_b with V_b\subset V_a. By
\Crefthm:topological-vector-space-zero-neighborhood-filter-base-circled-absorbing there exists a circled
zero neighborhood V_c with V_c\subset V_b. The closed convex hull U = \overline{\operatorname{Conv}}\, V_c then is
a barrel contained in V. Since it is a zero neighborhood it is an element of \mathscr{B}, and \mathscr{B}
is a filter base of \mathscr{U}_0.
To verify (I), let V \in \mathscr{B} and observe that
by continuity of addition there exist zero neighborhoods U_1 and U_2 such that U_1 + U_2 \subset V.
Choose U\in \mathscr{B} such that U \subset U_1\cap U_2. Then U + U \subset V.
Claim (III) holds true since multiplication
by an element t\in R^\times is a homeomorphism which preserves circled and convex sets.
The remaining claim follows immediately from
\Crefthm:topological-vector-space-zero-neighborhood-filter-base-circled-absorbing
and the observation that a real division algebra is archimedean.
Corollary 4.2.31
Let \mathscr{S} be a non-empty set of absolutely convex and absorbent subsets of an vector space \mathrm{E}
over a normed real divison algebra R.
Then the set
\mathscr{B} = \Big\{ t\, \bigcap\limits_{B\in\mathscr{F}} B \in\mathscr{P}(\mathrm{E}) \bigm\vert
\mathscr{F}\subset \mathscr{S}\text{ finite, } \mathscr{F} \neq \emptyset \:\: \ \:\: t > 0
\Big\}
is a base of the filter of zero neighborhoods of a locally convex topology
\mathscr{T} on \mathrm{E}. This topology is uniquely determined by that property and the coarsets among all
vector space topologies for which \mathscr{S} is a set of zero neighborhoods. The topology \mathscr{T} is called the
locally convex topology generated by \mathscr{S}.
4.3.1 Throughout the rest of this chapter the symbol \mathbb{K} will always
stand for the field of real numbers \mathbb{R}, the field of complex numbers \mathbb{C} or the
division algebra of quaternions \H. We assume these division algebras to be equipped
with their standard absolute values |\cdot |.
Moreover, vector spaces are assumed to be defined over the ground field \mathbb{K}
unless mentioned differently and are always assumed to be left vector spaces.
\addcontentslinetocsubsectionSeminorms and induced vector space topologies
Definition 4.3.2
By a seminorm on a vector space \mathrm{E} one
understands a map p:\mathrm{E} \to \mathbb{R} with the following properties:
A seminorm is called a norm if in addition the following axiom is satisfied:
A vector space \mathrm{E} equipped with a norm
\| \cdot \| : \mathrm{E} \to \mathbb{R}_{\geq 0} is called a normed vector space.
4.3.3
Let us introduce some useful further properties a map p:\mathrm{E} \to \mathbb{R}
can have. One calls such a map p
Lemma 4.3.4
For a real-valued map p:\mathrm{E} \to \mathbb{R} on a vectorspace \mathrm{E} the following are equivalent:
Proof.
Let p be sublinear. Then p is subadditive by definition. Subadditivity implies
p(0) \leq p(0) + p(0), hence p(0)\geq 0.
By sublinearity
p(0) = p(0\cdot 0 + 0\cdot 0) \leq 0 \cdot p(0) + 0\cdot p(0) = 0 ,
so p(0)=0. We show that p is positively homogeneous.
Applying sublinearity again one checks for v\in \mathrm{E} and t\geq 0 that
p(tv)=p(tv+ 0\cdot 0) \leq t p(v) + 0 \cdot p(0)= tp(v) ,
so p is positively homogeneous and the implication
(I) \implies (III) follows.
If p is positively homogeneous and subadditive, then for v,w \in \mathrm{E} and
t,s > 0 with t +s =1
p(tv +sw) \leq p(tv) + p(sw) \leq t p(v) + s p(w),
so p is convex. This gives the implication
(III) \implies (II).
If p is positively homogeneous and convex, then one computes for v,w \in \mathrm{E} and t,s\geq 0
with t+s > 0
p( tv +sw ) = (t+s)\, p \left( \frac{t}{t+s} v+ \frac{s}{t+s} w \right) \leq
(t+s) \left( \frac{t}{t+s} p(v) + \frac{s}{t+s} p(w) \right) = t p(v)+ sp(w) .
Since p(0)=\lim\limits_{t\searrow 0} p(t0) = \lim\limits_{t\searrow 0} t\, p(0) = 0 by positive homogeneity,
p then has to be sublinear and one obtains the implication
(II) \implies (I).
Lemma 4.3.5
Let p:\mathrm{E} \to \mathbb{R} be a real-valued map defined on a vectorspace \mathrm{E} over \mathbb{K}.
Proof.
Definition 4.3.6
If p :\mathrm{E} \to \mathbb{R} is a seminorm on a vector space \mathrm{E}, we denote for every v \in \mathrm{E}
and \varepsilon > 0 by \mathbb{B}_p (v,\varepsilon) the (open) \varepsilon-ball associated with p and
with center v that is the set
\mathbb{B}_p (v,\varepsilon) = \big\{ w \in \mathrm{E} \bigm\vert p(w-v) < \varepsilon \big\} .
The closed \varepsilon-ball associated with p and with center v
is defined as
\overline{\mathbb{B}}_p (v,\varepsilon) = \big\{ w \in \mathrm{E} \bigm\vert p(w-v) \leq \varepsilon \big\} .
The positive number \varepsilon is sometimes called the radius of the ball.
When by the context it is clear which seminorm p a ball is associated with
we often do not mention p explicitely. This is in particular the case
when the underlying vector space is a normed vector space.

If P is a finite set or a finite family of seminorms on \mathrm{E} we define the
open and closed \varepsilon-multiballs with center v by
\mathbb{B}_P (v,\varepsilon) =
\big\{ w \in \mathrm{E}\bigm\vert p(w-v) < \varepsilon \text{ for all } p\in P\big\}
and
\overline{\mathbb{B}}_P (v,\varepsilon) = \big\{ w \in \mathrm{E} \bigm\vert p(w-v) \leq \varepsilon
\text{ for all } p\in P \big\} ,
respectively.
Remark 4.3.7
For convenience, we will also use the symbols \mathbb{B}_p (0,\varepsilon) and \overline{\mathbb{B}}_p (0,\varepsilon)
to denote the sets \big\{ v \in \mathrm{E} \bigm\vert p(v) < \varepsilon \big\}
and \big\{ v \in \mathrm{E} \bigm\vert p(v) \leq \varepsilon \big\}, respectively, when p:\mathrm{E} \to \mathbb{R}
is just a real-valued convex map on the vector space \mathrm{E}. Note that for such a p the
set \big\{ v \in \mathrm{E} \bigm\vert p(v) < 0 \big\} might be non-empty. But as we have shown in
\Crefthm:properties-subadditive-convex-real-valued-maps-vector-space
the sets \mathbb{B}_p (0,\varepsilon) and \overline{\mathbb{B}}_p (0,\varepsilon) associated to a convex p share with the
the balls associated to a seminorm several nice properties like convexity.
Proposition 4.3.8
Let \mathrm{E} be a \mathbb{K}-vector space, and P a finite set of seminorms on \mathrm{E}.
Then, for every \varepsilon > 0 and v\in \mathrm{E}, the \varepsilon-multiballs
\mathbb{B}_P (v,\varepsilon) and \overline{\mathbb{B}}_P (v,\varepsilon) are convex.
The \varepsilon-multiballs
\mathbb{B}_P (0,\varepsilon) and \overline{\mathbb{B}}_P (0,\varepsilon) are absolutely convex and absorbent.
Proof.
Axiom (N2) immediately entails that \mathbb{B}_p (0,\varepsilon) and
\overline{\mathbb{B}}_P (0,\varepsilon) are circled. Axiom (N3)
together with (N2) entails that the sets
\mathbb{B}_p (v,\varepsilon) and \overline{\mathbb{B}}_P (v,\varepsilon)
are convex. Namely, if w_1, w_2 \in \mathbb{B}_p (v,\varepsilon)
and t \in [0,1], then one has for all seminorms p\in P
p\left( tw_1 + (1-t) w_2 - v \right) \leq
t \, p \left( w_1 - v \right) + (1-t) \, p \left( w_2 - v \right) <
t \, \varepsilon + (1-t) \, \varepsilon = \varepsilon
and likewise p\left( tw_1 + (1-t) w_2 - v \right) \leq \varepsilon for all
w_1, w_2 \in \overline{\mathbb{B}}_P(v,\varepsilon) and p\in P.
Now let v\in \mathrm{E} and \varepsilon > 0 be given.
Put t_p = \frac{p(v)+1}{\varepsilon} for every p\in P and t_0 = \max \{t_p\mid p\in P\}.
Then one has for all t\in \mathbb{K} with |t|\geq t_0 and for all p\in P
p\left( \frac{1}{t} v \right) \leq \frac{\varepsilon}{p(v)+1} \, p(v) < \varepsilon ,
hence v \in t \, \mathbb{B}_p (0,\varepsilon). So \mathbb{B}_p (0,\varepsilon) is absorbing.
Since \overline{\mathbb{B}}_P (0,\varepsilon) contains the absorbing set \mathbb{B}_p (0,\varepsilon)
it is absorbing as well.
Proposition and Definition 4.3.9
Assume to be given a set P of seminorms on a vector space \mathrm{E}.
Let \mathscr{F} (P) be the collection of all finite subsets of P.
A topological base on \mathrm{E} then is given by
\mathscr{B}_P = \big\{ \mathbb{B}_Q (v,\varepsilon) \bigm\vert
Q \in \mathscr{F} (P) , \: v \in \mathrm{E} , \: \varepsilon > 0 \big\} .
The topology generated by this base is denoted \mathscr{T}_P and called the topology
generated or defined by P. Moreover, \mathscr{T}_P is a locally convex
vector space topology on \mathrm{E}.
Proof.
We prove continuity of addition first. Let v_1,v_2\in \mathrm{E}, Q \in \mathscr{F} (P), and \varepsilon > 0.
Since the triangle inequality holds for every seminorm in F, one has
\mathbb{B}_Q \left(v_1,\frac\varepsilon 2\right) + \mathbb{B}_Q \left(v_2,\frac\varepsilon 2\right)
\subset \mathbb{B}_Q (v_1+v_2, \varepsilon) ,
which entails continuity of addition at each (v_1v_2) \in \mathrm{E} \times \mathrm{E}.
Next consider multiplication by scalars and let \lambda \in \mathbb{K} and v\in \mathrm{E}.
Again let Q= (p_j)_{j\in J} \in \mathscr{F} (P) and \varepsilon > 0.
Let C_1 = \sup \{ p_j (v) \mid j \in J \} + 1 and C_2 = |\lambda| + 1
and put \delta_1 = \min \{1,\frac{\varepsilon}{2 \, C_1} \}
and \delta_2 = \frac{\varepsilon}{2 \, C_2}.
Then one obtains by absolute homogeneity and subadditivity of each seminorm
p_j ( \mu w -\lambda v) \leq | \mu | \, p_j( w-v) + | \mu -\lambda| \, p_j(v) \quad \text{for all } \lambda \in \mathbb{K}
\text{ and } w \in \mathrm{E},
hence
\mathbb{B} (\lambda,\delta_1 ) \cdot \mathbb{B}_Q (v,\delta_2) \subset
\mathbb{B}_Q (\lambda \cdot v ,\varepsilon) ,
where \mathbb{B} (\lambda, \delta_1) = \{ \mu \in \mathbb{K} \mid |\mu -\lambda| < \delta_1\} .
This shows continuity of scalar multiplication at each (\lambda,v) \in \mathbb{K} \times \mathrm{E}.
Since each of the base elements \mathbb{B}_Q(v,\varepsilon) \in \mathscr{B}_P is convex by
the preceding proposition, Axiom \hyperref[axiom:tvs-local-convexity]LCVS holds true as well
and the topology \mathscr{T}_P is locally convex.
\addcontentslinetocsubsectionGauge functionals and induced seminorms
4.3.10
As we have seen, any vector space with a topology defined by a family of seminorms
on it is a locally convex topological vector space. The converse also holds true. The fundamental notion
needed for the proof of this is the following.
Definition 4.3.11
Let \mathrm{E} be a vector space and A \subset \mathrm{E} absorbent. Then the map
p_A : \mathrm{E} \to \mathbb{R}_{\geq 0} , \: v \mapsto p_A (v)= \inf \big\{ t \in \mathbb{R}_{ > 0} \bigm\vert v \in t A \big\}
is called the gauge functional, the Minkowski functional or the Minkowski gauge
of A.
Remark 4.3.12
By definition of an absorbent set, \big\{ t \in \mathbb{R}_{ > 0} \bigm\vert v \in t A \big\} is non-empty
whenever A\subset \mathrm{E} is absorbent. Hence p_A is well-defined for such A.
Proposition 4.3.13
The Minkowski gauge p_A :\mathrm{E} \to \mathbb{R}_{\geq 0} of an absorbent subset A
of a vector space \mathrm{E} has the following properties.
Proof.
If t > 0, then tv \in sA for some s > 0 if and only if v \in \frac st A.
Hence \big\{ s \in \mathbb{R}_{ > 0} \bigm\vert t v \in s A \big\} and t \big\{ s \in \mathbb{R}_{ > 0} \bigm\vert v \in s A \big\}
coincide for all t > 0, so (I) follows.
Assume that A is convex. Let v,w\in \mathrm{E} and \varepsilon > 0. Then there exist
t > p_A(v) and s > p_A(w) such that
v \in tA, w\in sA, t < p_A(v) +\frac \varepsilon 2 and
s < p_A(w) +\frac \varepsilon 2. By convexity of A and
\Crefthm:weighted-sum-convex-absolutely-convex-sets,
v+w \in tA+ sA = (t+s)A.
Hence p_A(v+w) \leq (t+s) < p_A(v) + p_A (w) + \varepsilon.
Since \varepsilon > 0 was arbitrary, p_A(v+w) \leq p_A(v) + p_A (w) and p_A is subadditive.
If v \in tA for some t with 0 < t < 1, then p_A(v) \leq t < 1 by definition.
Conversely, if p_A(v) < 1, then there exists a t > 0 such that t < 1 and v \in tA.
Hence the equality \mathbb{B}_p (v,1) = \bigcup_{0 < t < 1} tA follows.
Since A is absorbing, 0 is an element of A. By convexity of A
one therefore concludes tA = (1-t) \{ 0 \} + tA \subset A whenever 0 < t < 1.
For t > 1 this shows \frac 1t A \subset A, hence A \subset tA.
So the relation \bigcup_{0 < t < 1} tA \subset A \subset \bigcap_{1 < t} tA is proved.
Now assume that v \in tA for all t > 1. Then p_A(v)\leq 1 by definition.
If conversely p_A(v)\leq 1, then there exists for each \varepsilon > 0 an s\geq 0
such that p_A(v) \leq s, v \in sA and s < 1 +\varepsilon.
Hence, for t\geq 1 +\varepsilon by \Crefthm:weighted-sum-convex-absolutely-convex-sets
and 0\in A,
v \in sA = s A + (t-s) \{0\} \subset s A + (t-s) A = tA .
Since \varepsilon > 0 was arbitrary, v \in tA for all t > 1 follows. So one obtains
the equality \bigcap_{1 < t} tA = \overline{\mathbb{B}}_p (v,1),
and (II) is proved.
To verify (III) recall that A is circled whenever
A is absolutely convex. This entails for r\in \mathbb{K}, v\in \mathrm{E}
and absolutely convex A
p_A (rv)= \inf \big\{ t \in \mathbb{R}_{ > 0} \bigm\vert r v \in tA \big\} =
\inf \big\{ t \in \mathbb{R}_{ > 0} \bigm\vert |r|v \in t A \big\} = p_A (|r|v) = |r| p_A(v) ,
where for the last equality we have used (I).
Lemma 4.3.14
Let A and B be absorbent subsets of a vector space \mathrm{E}. Then
the following holds true.
Proof.
Lemma 4.3.15
Let p:\mathrm{E} \to\mathbb{R} be a sublinear map on a vector space \mathrm{E} and
A\subset \mathrm{E} convex. If
\mathbb{B}_p (0,1) \subset A \subset \overline{\mathbb{B}}_p (0,1) ,
then the gauge functional p_A coincides with \sup \{p,0\} . If p is even
a seminorm, then p=p_A.
Proof.
Let p:\mathrm{E}\to\mathbb{R} be sublinear. Observe that then \mathbb{B}_p (0,1) is absorbent by
\Crefthm:properties-subadditive-convex-real-valued-maps-vector-space
(IV).
Hence A must also be absorbent by assumption, so the associated Minkowski gauge p_A is
positively homogeneous by
\Crefthm:properties-gauge-functional-absorbent-set
(I).
Assume now that there exists v\in\mathrm{E} such that \max\{p(v),0 \} < p_A(v). By positive
homogeneity of p and p_A
one can achive by possibly multiplying v by a positive real number that \max\{p(v),0 \} < 1 < p_A(x). The first
inequality entails v \in \mathbb{B}_p (0,1), the second v\notin \overline{\mathbb{B}}_p (0,1) which is a contradiction.
Next assume that there exists v\in\mathrm{E} with p_A(v) < \max\{p(v),0 \}. As before one can then achieve that
p_A(v) < 1 < \max\{p(v),0 \} for some v\in\mathrm{E}. By the first inequality one concludes v \in A,
by the second v \notin A. This is a contradiction. So the equality \max\{p(v),0 \} = p_A(v) holds for all v\in \mathrm{E}.
In case p is a seminorm, then p(v) \geq 0 for all v\in \mathrm{E} and the second claim follows by the first.
Proposition 4.3.16
Let \mathrm{E} be a topological vector space, and p:\mathrm{E} \to \mathbb{R} be sublinear. Then the following are equivalent.
Proof.
Let us first show (I) \implies (II).
To this end fix \varepsilon > 0. By assumption there exists a zero neighborhood V\subset \mathrm{E} such that
|p(v)| < \varepsilon for all v\in V. By possibly passing to V \cap (-V) one can assume that
V is symmetric.
\Crefthm:properties-subadditive-convex-real-valued-maps-vector-space (II)
now implies
| p(v) - p(w)| < \varepsilon \quad \text{for all } v,w \in V .
Hence p is uniformly continuous.
The implications (II) \implies (III)
and (III) \implies (IV) are trivial.
It remains to prove
(IV) \implies (I).
Assume that \mathbb{B}_p (0,1) is a zero neighborhood.
Then there exists a symmetric zero neighborhood V contained in \mathbb{B}_p (0,1). Since p(0)=0 one concludes by
\Crefthm:properties-subadditive-convex-real-valued-maps-vector-space (II)
| p(v) | < \max \{ p(v), p(-v )\} < 1 \quad \text{for all } v \in V .
But this implies |p(v)| < \varepsilon for all v \in \varepsilon V and \varepsilon > 0,
so p is continuous at the origin.
\addcontentslinetocsubsectionNormability
Definition 4.3.17
A topological vector space \mathrm{E} is called seminormable if its topology is generated by a single
seminorm p:\mathrm{E}\to \mathbb{R}_{\geq 0}.
If the topology on \mathrm{E} coincides with the vector space topology generated by a norm \|\cdot\|,
then one calls \mathrm{E} normable.
Theorem 4.3.18[Kolmogorov's normability criterion]
A topological vector space \mathrm{E} is normable if and only if it is a ref_error space and
possesses a bounded convex neighborhood of the origin.
Proposition 4.5.1
Let X be a topological space and (Y,d) a metric space.
Then the following holds true.
Proof.
Note first that by the triangle inequality there exists for every f\in \mathscr{B} (X,Y) and y\in Y
a real number C_{f,y} > 0 such that
d\big(f(x),y\big) \leq C_{f,x} \quad \text{for all } x\in X .
\mbox
Proposition 4.5.2
Let X be a topological space and \mathbb{K} the division algebra of real or complex numbers or of quaternions.
Then the following holds true.
Proof.
Eq.~\eqrefeq:algebra-bounded-functions-values-real-complex-numbers-quaternions is obvious since
the distance of two elements a,b\in \mathbb{K} is given by d(a,b) = |a-b|, so in particular
d(a,0) = |a|. Let f,g \in \mathscr{B} (X,\mathbb{K}) and choose C_f,C_g \geq 0 so that
|f(x)| \leq C_f and |g(x)| \leq C_g for all x \in X. Then, by the triangle inequality
and absolute homogeneity of the absolute value,
| f(x) + g (x)| \leq C_f + C_g , \quad | a\, f(x) | \leq |a| \, C_f , \quad \text{and} \quad
| f(x) \cdot g (x)| \leq C_f \cdot C_g .
Hence the sum and the product of two bounded functions are bounded and so is any scalar multiple of a bounded function.
Therefore, \mathscr{B} (X,\mathbb{K}) is an algebra over \mathbb{K}. Using the triangle inequality and absolute homogeneity of the
absolute value again one verifies that \|f \|_\infty is a norm on \mathscr{B} (X,\mathbb{K}) indeed and that
it fulfills \|fg \|_\infty\leq \|f \|_\infty \cdot \|g \|_\infty for all f,g \in \mathscr{B} (X,\mathbb{K}).
Furthermore, by definition, \|f \|_\infty = \varrho (f,0) for all f \in \mathscr{B} (X,\mathbb{K}), where \varrho is
defined as in \Crefthm:metric-structure-space-bounded-maps.
Since (\mathscr{B} (X,\mathbb{K}),\varrho) is a complete metric space,
(\mathscr{B} (X,\mathbb{K}),\|\cdot \|_\infty) therefore is a Banach algebra. This proves the first claim.
For the second observe that for f, g \in \mathscr{C}_{b} (X,\mathbb{K}) and a\in \mathbb{K} the
sum f+g, the scalar multiple af, and the product f\cdot g are elements of \mathscr{C}_{b} (X,\mathbb{K}) again.
To verify this let x\in X and \varepsilon > 0. Choose neighborhoods
U_1 and U_2 of x so that
| f (y) - f(x) | < \min \left\{\frac{\varepsilon}{2}, \frac{\varepsilon}{|a|+1}, \frac{\varepsilon}{2(|g(x)|+1)} \right\}
\quad \text{for } y\in U_1
and
| g(y) - g(x) | < \left\{1,\frac{\varepsilon}{2},\frac{\varepsilon}{2(|f(x)|+1)} \right\} \quad\text{for } y\in U_2 .
Then for all y\in U_1\cap U_2
| (f+g) (y) - (f+g) (x) | \leq | f (y) - f(x) | + | g (y) - g (x) | < \varepsilon ,
| (af) (y) - (af) (x) | \leq |a| \cdot | f (y) - f(x) | < \varepsilon ,
| (f\cdot g) (y) - (f\cdot g) (x) | \leq |g(y)| \cdot | f (y) - f(x) | + |f(x) | \cdot | (g (y) - g (x) |
< \varepsilon .
This means that f+g, af and fg are continuous in x, hence elements of \mathscr{C}_{b} (X,\mathbb{K})
since x \in X was arbitrary. So \mathscr{C}_{b} (X,\mathbb{K}) is a subalgebra of \mathscr{B} (X,\mathbb{K}).
By \Crefthm:metric-structure-space-bounded-maps one knows that \mathscr{C}_{b} (X,\mathbb{K})
is a closed subspace of \mathscr{B} (X,\mathbb{K}). The rest of the claim is obvious.
4.5.3
As the next step, we introduce seminorms and their topologies on spaces of differentiable functions defined over an
open set \Omega\subset\mathbb{R}^n. We agree that from now on \Omega will always denote in this section an open subset
of \mathbb{R}^n. For any differentiability order m\in \mathbb{N} \cup \{ \infty\}
the symbol \mathscr{C}^m (\Omega) stands for the space of m-times continuously differentiable complex valued
functions on \Omega. For i=1,\ldots,n we denote by x^i : \mathbb{R}^n \to \mathbb{R} the i-th coordinate function
and, if m\geq 1, by \partial_i : \mathscr{C}^{m} (\Omega) \to \mathscr{C}^{m-1} (\Omega)
the operator which maps f \in \mathscr{C}^m (\Omega) to the partial derivative
\frac{\partial f}{\partial x^i}. More generally, if \alpha \in \mathbb{N}^n is a multiindex satisfying
|\alpha| = \alpha_1+\ldots \alpha_n \leq m, then we write
\partial^\alpha : \mathscr{C}^{m} (\Omega) \to \mathscr{C}^{m-|\alpha|} (\Omega) for the higher order partial derivative
which maps f \in \mathscr{C}^{m} (\Omega) to
\frac{\partial^{|\alpha|} f}{\partial x_1^{\alpha_1} \cdot \ldots \cdot \partial x_n^{\alpha_n}}. Recall that the
sum and the product of two m-times differentiable functions and scalar multiples of m-times differentiable functions
are again m-times differentiable, hence \mathscr{C}^{m} (\Omega) forms a \mathbb{C}-algebra.
Now we define \overline{\mathscr{C}}^{m} (\Omega) to be the space of continuous functions on the closure \overline{\Omega}
which are m-times continuosly differentiable on \Omega so that each of its partial derivatives of order \leq m
has a continuos extension to \overline{\Omega}. Since the operators \partial_i are linear and also derivations
by the Leibniz rule, \overline{\mathscr{C}}^{m} (\Omega) is a subalgebra of \mathscr{C}^{m} (\Omega). In general, these
algebras do not coincide as for example the function \frac 1x on \mathbb{R}_{ > 0} shows. It is an element
of \mathscr{C}^{\infty} (\mathbb{R}_{ < 0}) but can not be extended to a continuous function on \mathbb{R}_{\geq 0},
so is not an element of \overline{\mathscr{C}}^{\infty} (\mathbb{R}_{ < 0}).
If X \subset \mathbb{R}^n is locally closed which means that X is the intersection of an open and a closed susbet of \mathbb{R}^n,
then define \mathscr{C}^{m} (X) as the quotient space \mathscr{C}^{m} (\Omega) / \mathscr{J}_{X} (\Omega), where
\Omega \subset \mathbb{R}^n open is chosen so that X = \overline{X} \cap \Omega and where
\mathscr{J}_{X} denotes the ideal sheaf of all m-times continuously differentiable functions
vanishing on X that is
\mathscr{J}_{X} (\Omega) = \big\{ f \in \mathscr{C}^{m} (\Omega) \bigm\vert f|_X = 0 \big\} .
Using a smooth partition of unity type of argument one shows that \mathscr{C}^{m} (X) does not depend on the particular
choice of the neighborhood \Omega in which X is relatively closed and that \mathscr{C}^{m} (X) can be naturally
identified with the space of continuous functions on X which have an extension to an element of
\mathscr{C}^{m} (\Omega).
Proposition 4.5.4
Let \Omega \subset \mathbb{R}^n be open and bounded and m\in \mathbb{N}_{ > 0}. Then \overline{\mathscr{C}}^{m} (\Omega)
equipped with the norm