# Measure Theory/Riesz' representation theorem

## Theorem (Riesz' representation theorem)

Let $X$  be a locally compact Hausdorff space and let $\Lambda$  be a positive linear functional on $C_{c}(X)$ . Then, there exists a $\sigma$ -field $\Sigma$  containing all Borel sets of $X$  and a unique measure $\mu$  such that

1. $\Lambda f=\displaystyle \int _{X}fd\mu$  for all $f\in C_{c}(X)$
2. $\mu (K)<\infty$  for all compact $K\in \Sigma$
3. If $E\in \Sigma$ , and $\mu (E)<\infty$  then $\mu (E)=\inf\{\mu (V)|E\subset V,V{\text{ open}}\}$
4. If $E\in \Sigma$ , and $\mu (E)<\infty$  then $\mu (E)=\sup\{\mu (K)|E\supset K,K{\text{ compact}}\}$
5. The measure space $(X,\Sigma ,\mu )$  is complete

Proof

Recall the Urysohn's lemma:

If $X$  is a locally compact Hausdorff space and if $V$  is open and $K$  is compact with $K\subset V$  then

there exists $f:X\to [0,1]$  with $f\in C_{c}(X)$  satisfying $f(K)=\{1\}$  and ${\text{supp}}f\subset V$ . This is written in

short as $K\prec f\prec V$

We shall first prove that if such a measure exists, then it is unique. Suppose $\mu _{1},\mu _{2}$  are measures that satisfy (1) through (5)

It suffices to show that $\mu _{1}(K)=\mu _{2}(K)$  for every compact $K$

Let $K$  be compact and let $\epsilon >0$  be given.

By (3), there exists open $V\subset X$  with $V\supset K$  such that $\mu _{1}(V)<\mu _{1}(K)+\epsilon$

Urysohn's lemma implies that there exists $f\in C_{c}(X)$  such that $K\prec f\prec V$

(1) implies that $\mu _{2}(K)\leq \displaystyle \int _{X}fd\mu _{2}=\Lambda f=\int _{X}fd\mu _{1}$ . But $\mu _{1}(V)<\mu _{1}(K)+\epsilon$ , that is $\mu _{2}(K)<\mu _{1}(K)+\epsilon$ . We can similarly show that $\mu _{1}(K)<\mu _{2}(K)+\epsilon$ . Thus, $\mu _{2}(K)=\mu _{1}(K)$

Suppose $V$  is open in $X$ , define $\mu (V)=\sup\{\Lambda f|f\prec V\}$

If $V_{1}\subset V_{2}$  are open , then $\mu (V_{1})\leq \mu (V_{2})$

If $E$  is a subset of $X$  then define $\mu (E)=\inf\{\mu (V)|E\subset V,V{\text{ open}}\}$

Define $\Sigma _{F}=\{E\subset X|\mu (E)<\infty ,\mu (E)=\sup\{\mu (K):K\subset E,K{\text{ compact}}\}\}$

Let $\Sigma =\{E\subset X|E\cap K\in \Sigma _{F}{\text{ for all }}K{\text{ compact in }}X\}$

monotonicity of $\mu$  is obvious for all subsets of $X$

Let $E\subset X$  with $\mu (E)=0$

It is obvious that $E\in \Sigma _{F}$  which implies that $E\in \Sigma$ . Hence, we have that $\{X,\Sigma ,\mu \}$  is complete.

### Step 1

Suppose $\displaystyle \{E_{i}\}_{i=1}^{\infty }$  is a sequence of subsets of $X$  then, $\displaystyle \mu (\bigcup E_{i})\leq \sum _{i=1}^{\infty }\mu (E_{i})$

Proof

Let $V_{1},V_{2}$  be open subsets of $X$ . We wish to show that $\mu (V_{1}\cup V_{2})\leq \mu (V_{1})+\mu (V_{2})$

Given $\epsilon >0$ , let $g\in C_{c}(X)$  be such that $g\prec (V_{1}\cup V_{2})$  (so that $\displaystyle {\text{supp }}g=K\subset (V_{1}\cup V_{2})$ ) and $\mu (V_{1}\cup V_{2})-\epsilon \leq \Lambda g$ . This is possible because $\mu (V_{1}\cup V_{2})=\sup\{\Lambda g|g\prec V_{1}\cup V_{2}\}$ .

Now by Urysohn's lemma we can find $h_{i}$ , $i=1,2,\ldots$  such that $h_{i}\prec V_{i}$  and $h_{1}+h_{2}=1$  on $K$  with $h_{i}\in C_{c}(X)$

Thus $h_{i}g\prec V_{i}$  and $h_{1}g+h_{2}g=g$  on $K$

As $\Lambda$  is a linear functional $\Lambda f\leq \Lambda g$  for all $f\leq g$

$\Lambda g=\Lambda h_{1}g+\Lambda h_{2}g\leq \mu (V_{1})+\mu (V_{2})$

Thus, $\mu (V_{1}\cup V_{2})\leq \mu (V_{1})+\mu (V_{2})+\epsilon$  for every $\epsilon >0$ , i.e. $\mu (V_{1}\cup V_{2})\leq \mu (V_{1})+\mu (V_{2}).$

If $\{E_{i}\}_{i=1}^{\infty }$  is a sequence of members of $\Sigma$ , there exist open $V_{i}$  such that given $\epsilon >0$

$\mu (V_{i})<\mu (E_{i})+{\frac {\epsilon }{2^{i}}}$ . Define $E=\bigcup E_{i}\subset \bigcup V_{i}=V$ , $V$  is open. Let $f\prec V$ . Then $\mu (V)\geq \mu (E)$  but $\mu (V)<\Lambda f$

Thus $\mu (V)\leq \sum \mu (V_{i})\leq \sum \mu (E_{i})+\epsilon$

### Step 2

If $K$  is compact, then $K\in \Sigma _{F}$  and $\mu (K)=\inf\{\Lambda f:K\prec f\}$

Proof

It suffices to show that $\mu (K)<\infty$  for every compact $K$

Let $0<\alpha <1$  and $K\prec f$ , define $V_{\alpha }=\{x:f(x)>\alpha \}$  Then $V_{\alpha }$  is open and $K\subset V_{\alpha }$

Then, by Urysohn's lemma, there exists $g\in C_{c}(X)$  such that $K\prec g\prec V_{\alpha }$ , and hence, $\alpha g\leq f$  on $V_{\alpha }$

By definition $\Lambda g\geq \mu (K)$  and $\mu (K)\leq {\frac {1}{\alpha }}\Lambda f$

As $\Lambda f<\infty$ , we have $\mu (K)<\infty$  and hence, $K\in \Sigma _{F}$

Let $\epsilon >0$  be

By definition, there exists open $V\supset K$  such that $\mu (K)>\mu (V)-\epsilon$

By Urysohn's lemma, there exists $f$  such that $K\prec f\prec V$ , which implies that $\mu (V)\geq \Lambda f$ , that is

$\mu (K)>\Lambda f+\epsilon$ .

Hence, $\mu (K)=\inf\{\Lambda f:K\prec f\}$

### Step 3

Every open set $V$  satisfies

$\mu (V)=\sup\{\mu (K):K\subset V,K{\text{ compact}}\}$

If $V$  is open and $V\subset X$ , $\mu (X)<\infty$ , then $V\in \Sigma _{F}$

Proof

Let $V$  be open. Let $\alpha >0$  such that $\alpha <\mu (V)$ . It suffices to show that there exists compact $K$  such that $\mu (K)>\alpha$ .

By definition of $\mu$ , there exists $f\in C_{c}(X)$  such that $f\prec V$  and $\Lambda f>\alpha$

Let $K={\text{supp}}f$ . Obviously $K\subset V$ .

Let $W$  be open such that $K\subset W$ , then $F\prec W$  and hence $\Lambda f\leq \mu (W)$ , further $\Lambda f\leq \mu (K)$

Thus, $\mu (K)>\alpha$

### Step 4

Suppose $\mu (E_{i})$  is a sequence of pairwise disjoint sets in $\Sigma _{F}$  and let $E=\displaystyle \bigcup _{i=1}^{\infty }E_{i}$ . Then, $\mu (E)=\displaystyle \sum _{i=1}^{\infty }\mu (E_{i})$

Proof

If $\mu (E)=\infty$ , by step 1, we are done.

If $\mu (E)$  is finite then, $E\in \Sigma _{F}$  and hence, $\mu$  is countably additive on $\Sigma _{F}$

Suppose, $K_{1},K_{2}$  are compact and disjoint then $K_{1},K_{2}\in \Sigma _{F}$ ; $K_{1}\cup K_{2}\in \Sigma _{F}$

Claim: $\mu (K_{1}\cup K_{2})=\mu (K_{1})+\mu (K_{2})$

As $X$  is a locally compact Hausdorff space, there exist disjoint open sets $V_{1}$ , $V_{2}$  with $K_{1}\subset V_{1}$ , $K_{2}\subset V_{2}$

Hence, by Urysohn's lemma, there exists $g\in C_{c}(X)$  such that $g\prec K_{1}\cup K_{2}$  and $\Lambda g\leq \mu (K_{1}\cup K_{2})+\epsilon$

Now, ${\text{supp}}fg=K$  and $K\prec fg$ , $K\prec (1-f)g$

Thus, $\mu (K_{1})+\mu (K_{2})<\Lambda g\leq \mu (K_{1}+K_{2})+\epsilon$

Assume $\mu (E)<\infty$ . Given $E_{i}\in \Sigma _{F}$ , there exists compact $H_{i}\subset E_{i}$  such that $\mu (H_{i})>\mu (E_{i})-{\frac {\epsilon }{2^{i}}}$

Let $K_{N}=H_{1}\cup H_{2}\cup \ldots H_{N}$ . Obviously, $K_{N}$  is compact.

Thus, $\mu (E)\geq \mu (K_{N})=\displaystyle \sum _{i=1}^{N}\mu (H_{i})\geq \sum _{i=1}^{N}\mu (E_{i})-\epsilon$

and hence, $\mu (E)\geq \displaystyle \sum _{i=1}^{\infty }\mu (E_{i})$  By step 1, we have $\mu (E)\leq \displaystyle \sum _{i=1}^{\infty }\mu (E_{i})$ .

Thus, $\mu (E)=\displaystyle \sum _{i=1}^{\infty }\mu (E_{i})$

### Step 5

If $E\in \Sigma _{F}$  and $\epsilon >0$  then there exists $K$  compact and $V$  open with $K\subset E\subset V$  and $\mu (V\setminus K)<\epsilon$

Proof

$(V\setminus K)$  is open. As $E\in \Sigma _{F}$ , there exist compact $K$  and open $V$  such that $K\subset E\subset V$  with

$\mu (V)-{\frac {\epsilon }{2}}<\mu (E)<\mu (K)+{\frac {\epsilon }{2}}$

Now, $\mu (V\setminus K)<\mu (V)<\mu (E)+{\frac {\epsilon }{2}}<\infty$  (by step 4)

Thus, $\mu (V\setminus K)<\epsilon$

### Step 6

$\Sigma _{F}$  is a field of subsets of $X$

Proof

Let $A,B\in \Sigma _{F}$  and let $\epsilon >0$  be given.

There exist compact $K_{1},K_{2}$  and open $V_{1},V_{2}$  such that $K_{1}\subset A\subset V_{1}$ , $K_{2}\subset A\subset V_{2}$  with

$\mu (V_{1}\setminus K_{1}),\mu (V_{2}\setminus K_{2})<\epsilon$

Write $A\setminus B=(V_{1}\setminus K_{1})\cup (K_{1}\setminus V_{2})\cup (V_{2}\setminus K_{2})$

As $K_{1}\setminus V_{2}$  is a closed subsetof $K$  , it is compact

Then, $\mu (A\setminus B)\leq \mu (V_{1}\setminus K_{1})\cup \mu (K_{1}\setminus V_{2})\cup \mu (V_{2}\setminus K_{2})=\mu (K_{1}\setminus V_{2})+2\epsilon$

thus, $\mu (A\setminus B$  is finite and hence, $A\setminus B\in \Sigma _{F}$

Now write $A\cup B=(A\setminus B)\cup B$

and $A\cap B=A\setminus (A\setminus B)$

### Step 7

$\Sigma$  is a $\sigma$ -field containing all Borel sets

Proof

Let $C$  be closed

Then, $C\cap K$  is compact for every $K$  compact

Therefore $C\cap K\in \Sigma _{F}$  and hence $C\in \Sigma$  (by definition) and hence, $\Sigma$  has all closed sets. In particular, $X\in \Sigma$

Let $A\in \Sigma$ . Then, $A^{c}\cap K\subset K$  and $A^{c}\cap K=K\setminus (K\setminus A)$  and hence, $A^{c}\in \Sigma$

Now let $A=\displaystyle \bigcup _{i=1}^{\infty }A_{i}$  where $A_{i}\in \Sigma$

We know that $A_{i}\cap K\in \Sigma _{F}$  for every compact $K$

Let $B_{1}=A_{1}\cap K$ , $B_{n}=A_{n}\cap K\setminus (B_{1}\cup B_{2}\ldots \cup B_{n-1})$ . $A\cap K=\displaystyle \bigcup _{i=1}^{\infty }$ , but $\mu (B_{i})<\infty$  and hence, $A\cap K\in \Sigma$

### Step 8

$\Sigma _{F}=\{E\in \Sigma :\mu (E)<\infty \}$

Proof

Let $E\subset \Sigma _{F}$ . Then $E\cap K\in \Sigma _{F}$  for every compact $K\subset X$

Now, let $E\in \Sigma$ , $\mu (E)<\infty$ . Given $\epsilon >0$ , there exists open $V$  such that $E\subset V$ , $\mu (V)<\infty$ , that is, $V\in \Sigma _{F}$ . Further, there exists compact $K\subset V$  such that $\mu (V\setminus K)<\infty$

$E\in \Sigma$  implies that $E\cap K\in \Sigma _{F}$ , that is, there exists compact $H\subset E\cap K$  such that

$\mu (E\cap K)<\mu (H)+\epsilon$

Therefore, $E\subset (E\cap K)\cup (V\setminus K)$  implies that $\mu (E)\leq \mu (H)+2\epsilon$

As $\epsilon >0$  is arbitrary, we are done.

### Step 9

For $f\in C_{c}(X)$ , $\Lambda f=\displaystyle \int _{X}fd\mu$

Proof

Without loss of generality, we may assume that $f$  is real valued.

It is obviuos from the definition of $\mu$  that $\Lambda f\geq \displaystyle \int _{X}fd\mu$

Let $K={\text{supp}}f$ . Hence, as $f$  is continuous, $f(K)$  is compact. and we can write $f(K)\subset [a,b]$  for some $a,b\in \mathbb {R}$ . Let $\epsilon >0$ . Let ${\mathcal {P}}=\{a=y_{0}  be an $\epsilon$ -fine partition of $[a,b]$

Let $E_{i}=\{x\in X|y_{i-1} . As $f$  is continuous, $K$  is compact, $E_{i}$  is measurable for every $i$ , and hence, $K=\displaystyle \bigcup _{i=1}^{n}E_{i}$

$\mu (E)=\inf\{\mu (V)|V\supset E,V{\text{ open}}\}$

Hence, we can find open sets $V_{i}\supset E_{i}$  such that $\mu (V_{i})<\mu (E_{i})+{\frac {\epsilon }{n}}$

$f(x)  for all $x\in V_{i}$

We know that if compact $K\subset V_{1}\cup V_{2}\cup \ldots \cup V_{n}$  with $V_{i}$  open then there exists $h_{i}\in C_{c}(X)$  wiht $h_{i}\prec V_{i}$  and $\displaystyle \sum _{i=1}^{n}h_{i}=1$  on $K$

Hence, there exist functions $h_{i}\prec V_{i}$  such that $\displaystyle \sum _{i=1}^{n}h_{i}=1$  on $K$ .

Thus, $\displaystyle \sum _{i=1}^{n}h_{i}(x)f(x)=f(x)$  for all $x\in X$

By step 2, we have $\displaystyle \mu (K)\leq \sum _{i=1}^{n}\Lambda h_{i}$

$h_{i}f<(y_{i}+\epsilon )h_{i}$  on each $V_{i}$

Thus, $\Lambda f=\displaystyle \sum _{i=1}^{n}\Lambda h_{i}f\leq \sum _{i=1}^{n}\Lambda h_{i}(y_{i}+\epsilon )=\left(\sum _{i=1}^{n}(y_{i}+\epsilon )\Lambda h_{i}+\Lambda \sum _{i=1}^{n}|a|h_{i}\right)-\Lambda \sum _{i=1}^{n}|a|h_{i}$

$=\displaystyle \sum _{i=1}^{n}(y_{i}+\epsilon +|a|)\Lambda h_{i}-\Lambda \sum _{i=1}^{n}|a|h_{i}$

$\leq \displaystyle \sum _{i=1}^{n}(y_{i}+\epsilon +|a|)\left(\mu (E_{i})+{\frac {\epsilon }{n}}\right)-\Lambda \sum _{i=1}^{n}|a|h_{i}$

$=\displaystyle \sum _{i=1}^{n}y_{i}\mu (E_{i})+\sum _{i=1}^{n}\epsilon \left(\mu (E_{i})+{\frac {\epsilon }{n}}+{\frac {|a|}{n}}\right)\leq \sum _{i=1}^{n}(y_{i}-\epsilon )\mu (E_{i})+\sum _{i=1}^{n}\epsilon \left(2\mu (E_{i})+{\frac {\epsilon }{n}}+{\frac {|a|}{n}}\right)$

$\leq \displaystyle \sum _{i=1}^{n}f(x_{i})\mu (E_{i})+\sum _{i=1}^{n}\epsilon \left(2\mu (E_{i})+{\frac {\epsilon }{n}}+{\frac {|a|}{n}}\right)$

As $\epsilon >0$  is arbitrary, we have $\Lambda f\leq \displaystyle \int _{X}fd\mu$  which completes the proof.