Measure Theory/Riesz' representation theorem

Theorem (Riesz' representation theorem)

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

1. $\mathrm{\Lambda }f=\underset{X}{\int }fd\mu$ for all $f\in C_c(X)$
2. $\mu (K)<\mathrm{\infty }$ for all compact $K\in \mathrm{\Sigma }$
3. If $E\in \mathrm{\Sigma }$, and $\mu (E)<\mathrm{\infty }$ then $\mu (E)=\inf \{\mu (V)|E\subset V,V\text{ open}\}$
4. If $E\in \mathrm{\Sigma }$, and $\mu (E)<\mathrm{\infty }$ then $\mu (E)=\sup \{\mu (K)|E\supset K,K\text{ compact}\}$
5. The measure space $(X,\mathrm{\Sigma },\mu )$ 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 $ϵ>0$ be given.

By (3), there exists open $V\subset X$ with $V\supset K$ such that ${\mu }_1(V)<{\mu }_1(K)+ϵ$

Urysohn's lemma implies that there exists $f\in C_c(X)$ such that $K\prec f\prec V$

(1) implies that ${\mu }_2(K)\le \underset{X}{\int }fd{\mu }_2=\mathrm{\Lambda }f=\underset{X}{\int }fd{\mu }_1$. But ${\mu }_1(V)<{\mu }_1(K)+ϵ$, that is ${\mu }_2(K)<{\mu }_1(K)+ϵ$. We can similarly show that ${\mu }_1(K)<{\mu }_2(K)+ϵ$. Thus, ${\mu }_2(K)={\mu }_1(K)$

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

If $V_1\subset V_2$ are open , then $\mu (V_1)\le \mu (V_2)$

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

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

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

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

Let $E\subset X$ with $\mu (E)=0$

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

Step 1

Suppose $\{E_i{\}}_{i=1}^{\mathrm{\infty }}$ is a sequence of subsets of $X$. Then, $\mu (\bigcup E_i)\le \sum _{i=1}^{\mathrm{\infty }}\mu (E_i)$

Proof

Let $V_1,V_2$ be open subsets of $X$. We wish to show that $\mu (V_1\cup V_2)\le \mu (V_1)+\mu (V_2)$.

Given $ϵ>0$, let $g\in C_c(X)$ be such that $g\prec (V_1\cup V_2)$ (so that $\text{supp }g=K\subset (V_1\cup V_2)$) and $\mu (V_1\cup V_2)-ϵ\le \mathrm{\Lambda }g$. This is possible because $\mu (V_1\cup V_2)=\sup \{\mathrm{\Lambda }g|g\prec V_1\cup V_2\}$.

Now by Urysohn's lemma we can find $h_i$, $i=1,2,\dots$ 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 $\mathrm{\Lambda }$ is a linear functional $\mathrm{\Lambda }f\le \mathrm{\Lambda }g$ for all $f\le g$,

$\mathrm{\Lambda }g=\mathrm{\Lambda }{h}_{1}g+\mathrm{\Lambda }{h}_{2}g\le \mu (V_1)+\mu (V_2).$

Thus, $\mu (V_1\cup V_2)\le \mu (V_1)+\mu (V_2)+ϵ$ for every $ϵ>0$, i.e. $\mu (V_1\cup V_2)\le \mu (V_1)+\mu (V_2).$

If $\{E_i{\}}_{i=1}^{\mathrm{\infty }}$ is a sequence of members of $\mathrm{\Sigma }$, there exist open $V_i$ such that given $ϵ>0$,

$\mu (V_i)<\mu (E_i)+\frac{ϵ}{2^i}$. Define $E=\bigcup E_i\subset \bigcup V_i=V$, $V$ is open. Let $f\prec V$. Then $\mu (V)\ge \mu (E)$ but $\mu (V)<\mathrm{\Lambda }f$

Thus, $\mu (V)\le \sum \mu (V_i)\le \sum \mu (E_i)+ϵ.$

Step 2

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

Proof

It suffices to show that $\mu (K)<\mathrm{\infty }$ 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, $\alpha g\le f$ on $V_{\alpha }$

Since $\mathrm{\Lambda }g\ge \mu (K)$, we have $\mu (K)\le \frac{1}{\alpha }\mathrm{\Lambda }f$

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

Let $ϵ>0$ be

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

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

$\mu (K)>\mathrm{\Lambda }f+ϵ$.

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

Step 3

Every open set $V$ satisfies

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

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

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 $\mathrm{\Lambda }f>\alpha$

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

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

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

Step 4

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

Proof

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

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

Suppose, $K_1,K_2$ are compact and disjoint then $K_1,K_2\in {\mathrm{\Sigma }}_F$; $K_1\cup K_2\in {\mathrm{\Sigma }}_F$

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 $\mathrm{\Lambda }g\le \mu (K_1\cup K_2)+ϵ$

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

Thus, $\mu (K_1)+\mu (K_2)<\mathrm{\Lambda }g\le \mu (K_1+K_2)+ϵ$

Assume $\mu (E)<\mathrm{\infty }$. Given $E_i\in {\mathrm{\Sigma }}_F$, there exists compact $H_i\subset E_i$ such that $\mu (H_i)>\mu (E_i)-\frac{ϵ}{2^i}$

Let $K_N=H_1\cup H_2\cup \dots H_N$. Obviously, $K_N$ is compact.

Thus, $\mu (E)\ge \mu (K_N)=\sum _{i=1}^N\mu (H_i)\ge \sum _{i=1}^N\mu (E_i)-ϵ$

and hence, $\mu (E)\ge \sum _{i=1}^{\mathrm{\infty }}\mu (E_i)$ By step 1, we have $\mu (E)\le \sum _{i=1}^{\mathrm{\infty }}\mu (E_i)$.

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

Step 5

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

Proof

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

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

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

Thus, $\mu (V\setminus K)<ϵ$

Step 6

${\mathrm{\Sigma }}_F$ is a field of subsets of $X$

Proof

Let $A,B\in {\mathrm{\Sigma }}_F$ and let $ϵ>0$ 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

$\mu (V_1\setminus K_1),\mu (V_2\setminus K_2)<ϵ$

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, $\mu (A\setminus B)\le \mu (V_1\setminus K_1)\cup \mu (K_1\setminus V_2)\cup \mu (V_2\setminus K_2)=\mu (K_1\setminus V_2)+2ϵ$

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

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

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

Step 7

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

Proof

Let $C$ be closed

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

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

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

Now let $A=\bigcup _{i=1}^{\mathrm{\infty }}{A}_i$ where $A_i\in \mathrm{\Sigma }$

We know that $A_i\cap K\in {\mathrm{\Sigma }}_F$ for every compact $K$

Let $B_1=A_1\cap K$, $B_n=A_n\cap K\setminus (B_1\cup B_2\dots \cup B_{n-1})$. $A\cap K=\bigcup _{i=1}^{\mathrm{\infty }}$, but $\mu (B_i)<\mathrm{\infty }$ and hence, $A\cap K\in \mathrm{\Sigma }$

Step 8

${\mathrm{\Sigma }}_F=\{E\in \mathrm{\Sigma }:\mu (E)<\mathrm{\infty }\}$

Proof

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

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

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

$\mu (E\cap K)<\mu (H)+ϵ$

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

As $ϵ>0$ is arbitrary, we are done.

Step 9

For $f\in C_c(X)$, $\mathrm{\Lambda }f=\underset{X}{\int }fd\mu$

Proof

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

It is obvious from the definition of $\mu$ that $\mathrm{\Lambda }f\ge \underset{X}{\int }fd\mu$

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 $a,b\in ℝ$. Let $ϵ>0$. Let $𝒫=\{a=y_0<y_1<\dots <y_n=b\}$ be an $ϵ$-fine partition of $[a,b]$

Let $E_i=\{x\in X|y_{i-1}<f(x)<y_i\}\cap K$. As $f$ is continuous, $K$ is compact, $E_i$ is measurable for every $i$, and hence, $K=\bigcup _{i=1}^n{E}_i$

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

Hence, we can find open sets $V_i\supset E_i$ such that $\mu (V_i)<\mu (E_i)+\frac{ϵ}{n}$

$f(x)<y_i+ϵ$ for all $x\in V_i$

We know that if compact $K\subset V_1\cup V_2\cup \dots \cup V_n$ with $V_i$ open then there exists $h_i\in C_c(X)$ wiht $h_i\prec V_i$ and $\sum _{i=1}^n{h}_i=1$ on $K$

Hence, there exist functions $h_i\prec V_i$ such that $\sum _{i=1}^n{h}_i=1$ on $K$.

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

By step 2, we have $\mu (K)\le \sum _{i=1}^n\mathrm{\Lambda }{h}_i$

$h_{i}f<(y_i+ϵ)h_i$ on each $V_i$

Thus, $\mathrm{\Lambda }f=\sum _{i=1}^n\mathrm{\Lambda }{h}_{i}f\le \sum _{i=1}^n\mathrm{\Lambda }{h}_i(y_i+ϵ)=(\sum _{i=1}^n(y_i+ϵ)\mathrm{\Lambda }{h}_i+\mathrm{\Lambda }\sum _{i=1}^n|a|h_i)-\mathrm{\Lambda }\sum _{i=1}^n|a|h_i$

$=\sum _{i=1}^n(y_i+ϵ+|a|)\mathrm{\Lambda }{h}_i-\mathrm{\Lambda }\sum _{i=1}^n|a|h_i$

$\le \sum _{i=1}^n(y_i+ϵ+|a|)(\mu (E_i)+\frac{ϵ}{n})-\mathrm{\Lambda }\sum _{i=1}^n|a|h_i$

$=\sum _{i=1}^n{y}_i\mu (E_i)+\sum _{i=1}^nϵ(\mu (E_i)+\frac{ϵ}{n}+\frac{|a|}{n})\le \sum _{i=1}^n(y_i-ϵ)\mu (E_i)+\sum _{i=1}^nϵ(2\mu (E_i)+\frac{ϵ}{n}+\frac{|a|}{n})$

$\le \sum _{i=1}^{n}f(x_i)\mu (E_i)+\sum _{i=1}^nϵ(2\mu (E_i)+\frac{ϵ}{n}+\frac{|a|}{n})$

As $ϵ>0$ is arbitrary, we have $\mathrm{\Lambda }f\le \underset{X}{\int }fd\mu$ which completes the proof.

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