Measure Theory/Riesz' representation theorem

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Theorem (Riesz' representation theorem)[edit]

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[edit]

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 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)

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 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=\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 \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[edit]

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}

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 \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[edit]

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[edit]

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[edit]

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[edit]

\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[edit]

\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[edit]

\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[edit]

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 \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.