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, $\displaystyle \mu \left(\bigcup E_{i}\right)\leq \sum _{i=1}^{\infty }\mu (E_{i})$

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 $V_{\alpha }$

Since $\Lambda g\geq \mu (K)$, we have $\mu (K)\leq {\frac {1}{\alpha }}\Lambda f$

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 obvious 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 ${\mathcal {P}}=\{a=y_{0}<y_{1}<\ldots <y_{n}=b\}$ be an $\epsilon$-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, $K=\displaystyle \bigcup _{i=1}^{n}E_{i}$

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

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

$f(x)<y_{i}+\epsilon$ 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.