14.9 The Riesz Representation Theorem

Proof. (1): $\dpb{g - f, I}{C_c(X; \real)}\ge 0$.

(2): By Urysohn’s lemma, there exists $g \in C_{c}(X; [0, 1])$ such that $g|_{K} = 1$. In which case,

so $C_{K} = \dpn{g, I}{C_c(X; \real)}$ is a desired constant.$\square$

Proof. (1): For any $U \in \topo$ and $f \in C_{c}(X; \real)$, denote $f \prec U$ if $f \in C_{c}(X; [0, 1])$ and $\supp{f}\subset U$. Let

and

then

1. Since $\emptyset \in \topo$ and $\mu_{0}(\emptyset) = 0$, $\mu^{*}(\emptyset) = 0$.

2. Let $E, F \subset X$ with $E \subset F$, then $\cn^{o}(E) \subset \cn^{o}(F)$, so $\mu^{*}(E) \le \mu^{*}(F)$.

3. Let $\seq{E_n}\subset 2^{X}$, $E = \bigcup_{n \in \natp}E_{n}$, $U \in \cn^{o}(E)$, and $\seq{U_n}\subset \topo$ such that $U_{n} \in \cn^{o}(E_{n})$ for each $n \in \natp$.

   Let $f \prec U$, then by compactness of $\supp{f}$, there exists $N \in \natp$ such that $\supp{f}\subset \bigcup_{n = 1}^{N} U_{n}$. By Proposition 4.20.6, there exists a partition of unity $\seqf{\phi_j}\subset C_{c}(X; [0, 1])$ on $K$ subordinate to $\seqf[N]{U_n}$. In which case,

   \[\dpb{f, I}{C_c(X; \real)}= \sum_{n = 1}^{N} \dpb{\phi_n f, I}{C_c(X; \real)}\le \sum_{n = 1}^{N} \mu_{0}(U_{n}) \le \sum_{n \in \natp}\mu_{0}(U_{n})\]

   Since this holds for all $f \prec U$,

   \[\mu^{*}(E) \le \mu_{0}(U) \le \sum_{n \in \natp}\mu_{0}(U_{n})\]

   Let $\eps > 0$, then there exists $\seq{U_n}\subset \topo$ such that $U_{n} \supset E_{n}$ and $\mu_{0}(U_{n}) \le \mu^{*}(E_{n}) + \eps/2^{n}$ for each $n \in \natp$. In which case,

   \[\mu^{*}(E) \le \sum_{n \in \natp}\mu_{0}(U_{n}) \le \eps + \sum_{n \in \natp}\mu^{*}(E_{n})\]

   As this holds for all $\eps > 0$, $\mu^{*}(E) \le \sum_{n \in \natp}\mu^{*}(E_{n})$.

Therefore $\mu^{*}: 2^{E} \to [0, \infty]$ is an outer measure.

To see that all Borel sets are $\mu^{*}$-measurable, let $E \subset X$ and $U \in \topo$. First suppose that $E$ is open. Let $f \prec E \cap U$, then for any $g \prec E \setminus \supp{f}$, $\supp{f}\cap \supp{g}= \emptyset$ and $f + g \prec E$, so

Since this holds for all $g \prec E \setminus \supp{f}$,

As $E \setminus \supp{f}\supset E \setminus U$,

Finally, the above holds for all $f \prec E \cap U$,

Now suppose that $E$ is arbitrary. Let $V \in \cn^{o}(E)$, then

As this holds for all such $V$,

By Carathéodory’s Extension Theorem, there exists a Borel measure $\mu: \cb_{X} \to [0, \infty]$ such that for all $U \in \topo$, $\mu(U) = \mu_{0}(U)$.

(2): Let $K \subset X$ be compact and $U \in \cn^{o}(K)$. By Urysohn’s lemma, there exists $f \prec U$ with $f \ge \one_{K}$. In which case,

On the other hand, let $f \in C_{c}(X; [0, 1])$ with $f \ge \one_{K}$. For any $r \in (0, 1)$, let $g \prec \bracs{f > r}$, then $r^{-1}f \ge g$ and

As this holds for all $g \prec \bracs{f > r}$,

Since the above holds for all $r \in (0, 1)$,

(3): Let $f \in C_{c}(X; \real)$. Using linearity, assume without loss of generality that $f \in C_{c}(X; [0, 1])$. Let $N \in \natp$. For each $1 \le n \le N$, let

For any $x \in X$, there exists $1 \le n \le N$ such that $f(x) \in [(n-1)/N, n/N]$. In which case,

• For each $1 \le j < n$, $f_{j}(x) = 1/N$.

• $f_{n}(x) = f(x) - \frac{n - 1}{N}$.

• For each $n < j \le N$, $f_{j}(x) = 0$.

so $\sum_{j = 1}^{N}f_{j}(x) = \frac{n-1}{N}+ f(x) - \frac{n - 1}{N}= f(x)$.

Let $K_{0} = \supp{f}$, and for each $1 \le n \le N$, let $K_{n} = \bracs{f \ge n/N}$.

Since $\supp{f_n}\subset \bracs{f \ge (n-1)/N}$, for any $U \supset \bracs{f_n \ge (n - 1)/N}$, $f_{n} \prec U$, so

On the other hand, $f_{n} \ge \frac{1}{N}\one_{K_n}$,

so

Therefore

As this holds for all $N \in \natp$, $\int f d\mu = \dpb{f, I}{C_c(X; \real)}$.

(4):

1. For any $K \subset X$ compact, by Urysohn’s lemma, there exists $f \in C_{c}(X; [0, 1])$ with $f \ge \one_{K}$. In which case, $\mu(K) \le \int f d\mu = \dpb{f, I}{C_c(X; \real)}$.

2. By definition of $\mu^{*}$, $\mu$ is outer regular.

3. For any $U \in \topo$,

   \[\mu(U) = \sup_{f \prec U}\dpb{f, I}{C_c(X; \real)}\le \sup_{f \prec U}\mu(\supp{f}) \le \sup_{\substack{K \subset U \\ \text{compact}}}\mu(K) \le \mu(U)\]

(U): By Proposition 14.6.2, $\nu$ also satisfies (2). Thus for any $E \in \cb_{X}$, by (R2) and (R3’),

so $\nu$ is uniquely determined by $I$.$\square$