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) $$\Lambda f=\displaystyle\int_Xfd\mu$$ 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 $$\mu_2(K)\leq\displaystyle\int_Xfd\mu_2=\Lambda f=\int_Xfd\mu_1$$. 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, $$h_ig\prec V_i$$ and $$h_1g+h_2g=g$$ on $$K$$.

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

$$\Lambda g=\Lambda h_1g+\Lambda h_2g\leq\mu(V_1)+\mu(V_2).$$

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 $$f\in C_c(X)$$, $$\Lambda f=\displaystyle\int_Xf d\mu$$

Proof

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

It is obvious from the definition of $$\mu$$ that $$\Lambda f\geq \displaystyle\int_Xfd\mu$$

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, $$K=\displaystyle\bigcup_{i=1}^nE_i$$

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

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 $$\displaystyle\sum_{i=1}^nh_i=1$$ on $$K$$

Hence, there exist functions $$h_i\prec V_i$$ such that $$\displaystyle\sum_{i=1}^nh_i=1$$ on $$K$$.

Thus, $$\displaystyle\sum_{i=1}^nh_i(x)f(x)=f(x)$$ for all $$x\in X$$

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

$$h_if<(y_i+\epsilon)h_i$$ on each $$V_i$$

Thus, $$\Lambda f=\displaystyle\sum_{i=1}^n\Lambda h_if\leq\sum_{i=1}^n\Lambda h_i(y_i+\epsilon)=\left(\sum_{i=1}^n(y_i+\epsilon)\Lambda h_i+\Lambda\sum_{i=1}^n|a|h_i\right)-\Lambda\sum_{i=1}^n|a|h_i$$

$$=\displaystyle\sum_{i=1}^n(y_i+\epsilon+|a|)\Lambda h_i-\Lambda\sum_{i=1}^n|a|h_i$$

$$\leq\displaystyle\sum_{i=1}^n(y_i+\epsilon+|a|)\left(\mu(E_i)+\frac{\epsilon}{n}\right)-\Lambda\sum_{i=1}^n|a|h_i$$

$$=\displaystyle\sum_{i=1}^ny_i\mu(E_i)+\sum_{i=1}^n\epsilon\left(\mu(E_i)+\frac{\epsilon}{n}+\frac{|a|}{n}\right)\leq\sum_{i=1}^n(y_i-\epsilon)\mu(E_i)+\sum_{i=1}^n\epsilon\left(2\mu(E_i)+\frac{\epsilon}{n}+\frac{|a|}{n}\right)$$

$$\leq\displaystyle\sum_{i=1}^nf(x_i)\mu(E_i)+\sum_{i=1}^n\epsilon\left(2\mu(E_i)+\frac{\epsilon}{n}+\frac{|a|}{n}\right)$$

As $$\epsilon>0$$ is arbitrary, we have $$\Lambda f\leq \displaystyle\int_Xfd\mu$$ which completes the proof.