# Measure Theory/Riesz' representation theorem

## Theorem (Riesz' representation theorem)

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

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

Proof

Recall the Urysohn's lemma:

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

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

short as ${\displaystyle K\prec f\prec V}$

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

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

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

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

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

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

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

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

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

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

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

monotonicity of ${\displaystyle \mu }$ is obvious for all subsets of ${\displaystyle X}$

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

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

### Step 1

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

Proof

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

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

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

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

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

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

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

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

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

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

### Step 2

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

Proof

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

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

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

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

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

Let ${\displaystyle \epsilon >0}$ be

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

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

${\displaystyle \mu (K)>\Lambda f+\epsilon }$.

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

### Step 3

Every open set ${\displaystyle V}$ satisfies

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

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

Proof

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

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

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

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

Thus, ${\displaystyle \mu (K)>\alpha }$

### Step 4

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

Proof

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

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

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

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

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

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

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

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

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

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

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

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

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

### Step 5

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

Proof

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

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

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

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

### Step 6

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

Proof

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

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

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

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

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

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

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

Now write ${\displaystyle A\cup B=(A\setminus B)\cup B}$

and ${\displaystyle A\cap B=A\setminus (A\setminus B)}$

### Step 7

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

Proof

Let ${\displaystyle C}$ be closed

Then, ${\displaystyle C\cap K}$ is compact for every ${\displaystyle K}$ compact

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

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

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

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

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

### Step 8

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

Proof

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

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

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

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

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

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

### Step 9

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

Proof

Without loss of generality, we may assume that ${\displaystyle f}$ is real valued.

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

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

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

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

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

${\displaystyle f(x) for all ${\displaystyle x\in V_{i}}$

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

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

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

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

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

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

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

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

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

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

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