Measure Theory/Riesz' representation theorem

Theorem (Riesz' representation theorem)

Let X be a locally compact Hausdorff space and let Λ be a positive linear functional on C_c(X). Then, there exists a σ-field Σ containing all Borel sets of X and a unique measure μ 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,Σ,μ) 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 μ₁,μ₂ are measures that satisfy (1) through (5)

It suffices to show that μ₁(K) = μ₂(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, μ₂(K) = μ₁(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 μ is obvious for all subsets of X

Let $E\subset X$ with μ(E) = 0

It is obvious that $E\in\Sigma_F$ which implies that $E\in\Sigma$. Hence, we have that {X,Σ,μ} 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 V₁,V₂ be open subsets of X. We wish to show that $\mu(V_1\cup V_2)\leq \mu(V_1)+\mu(V_2)$

Let $g\in C_c(X)$ such that $g\prec (V_1\cup V_2)$, then $\displaystyle\text{supp }g=K\subset (V_1\cup V_2)$. By Urysohn's lemma we can find h_i, $i=1,2,\ldots$ such that $h_i\prec V_i$ and h₁ + h₂ = 1 on K with $h_i\in C_c(X)$

Thus $h_ig\prec V_i$ and h₁g + h₂g = g on K

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

$\Lambda g=\displaystyle\int_Xh_1gd\mu+\int_Xh_2gd\mu\leq\mu(V_1)+\mu(V_2)$

By definition, $\mu(V_1\cup V_2)=\inf\{\Lambda g|g\prec V_1\cup V_2\}$

If $\{E_i\}_{i=1}^{\infty}$ is a sequence of members of Σ, 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 μ(V) < Λf

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 < α < 1 and $K\prec f$, define V_α = {x:f(x) > α} Then V_α 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_α

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 ε > 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 α > 0 such that α < μ(V). It suffices to show that there exists compact K such that μ(K) > α.

By definition of μ, there exists $f\in C_c(X)$ such that $f\prec V$ and Λf > α

Let K = suppf. 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, μ(K) > α

Step 4

Suppose μ(E_i) is a sequence of pairwise disjoint sets in Σ_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 μ(E) is finite then, $E\in\Sigma_F$ and hence, μ is countably additive on Σ_F

Suppose, K₁,K₂ 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 V₁, V₂ 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, suppfg = 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 ε > 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

Σ_F is a field of subsets of X

Proof

Let $A,B\in\Sigma_F$ and let $\epsilon>0$ be given.

There exist compact K₁,K₂ and open V₁,V₂ 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

Σ is a σ-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, Σ 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 obviuos from the definition of μ that $\Lambda f\geq \displaystyle\int_Xfd\mu$

Let K = suppf. 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 ε > 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.