14.6 Radon Measures
Definition 14.6.1 (Radon Measure). Let $X$ be a LCH space and $\mu: \cb_{X} \to [0, \infty]$ be a Borel measure, then $\mu$ is a Radon measure if:
For any $K \subset X$ compact, $\mu(K) < \infty$.
$\mu$ is outer regular on all Borel sets.
$\mu$ is inner regular on all open sets.
Proposition 14.6.2. Let $X$ be a LCH space and $\mu: \cb_{X} \to [0, \infty]$ be a Radon measure, then
For any $U \subset X$ open,
\[\mu(U) = \sup\bracs{\int f d\mu \bigg | f \in C_c(X; [0, 1]), \supp{f} \subset U}\]For any $K \subset X$ compact,
\[\mu(K) = \inf\bracs{\int f d\mu \bigg | f \in C_c(X; [0, 1]), f \ge \one_K}\]
Proof. (1): Let $U \subset X$ be open. By Urysohn’s lemma, for any $K \subset U$ compact, there exists $f_{K} \in C_{c}(X; [0, 1])$ such that $f_{K}|_{K} = 1$ and $\supp{f_K}\subset U$. In which case,
By (R3’),
(2): Let $K \subset X$ be compact and $U \in \cn^{o}(K)$. By Urysohn’s lemma, there exists $f_{U} \in C_{c}(X; [0, 1])$ such that $f_{U}|_{K} = 1$ and $\supp{f_U}\subset U$. In which case,
By (R2),
Proposition 14.6.3. Let $X$ be a LCH space and $\mu: \cb_{X} \to [0, \infty]$ be a Radon measure, then
$\mu$ is inner regular on all its $\sigma$-finite sets.
If $X$ is $\sigma$-compact or $\mu$ is $\sigma$-finite, then $\mu$ is regular.
Proof. (1): Let $E \in \cb_{X}$ with $\mu(E) < \infty$ and $\eps > 0$, then there exists
$U \in \cn^{o}(E)$ with $\mu(U \setminus E) < \eps/2$.
$K \subset U$ compact with $\mu(K) > \mu(U) - \eps/2$.
$V \in \cn^{o}(U \setminus E)$ with $\mu(V) < \eps/2$.
In which case, $K \setminus V \subset E$ is compact with
so $\mu$ is inner regular on $E$.
Now suppose that $E$ is $\sigma$-finite, then there exists $\seq{E_n}\subset \cb_{X}$ with $\mu(E_{n}) < \infty$ for each $n \in \natp$ and $E = \bigsqcup_{n \in \natp}E_{n} = E$. Let $N \in \natp$ and $\seqf{K_n}\subset 2^{X}$ compact with $K_{n} \subset E_{n}$ for each $n \in \natp$, then $\bigcup_{n = 1}^{N} K_{n} \subset E$ is compact. Hence
As the above holds for all $N \in \natp$,
Proposition 14.6.4. Let $X$ be a LCH space, $\mu: \cb_{X} \to [0, \infty]$ be a $\sigma$-finite Radon measure, and $E \in \cb_{X}$, then
For every $\eps > 0$, there exists $U \in \cn^{o}(E)$ and $F \subset E$ closed such that $\mu(U \setminus F) < \eps$.
There exists a $F_{\sigma}$ set $A$ and a $G_{\delta}$ set $B$ such that $A \subset E \subset B$ and $\mu(B \setminus A) = 0$.
Proof. (1): Let $\seq{E_n}\subset \cb_{X}$ such that $\mu(E_{n}) < \infty$. By outer regularity, there exists $\seq{U_n}$ open such that $U_{n} \in \cn^{o}(E_{n})$ and $\mu(U_{n}) < \mu(E_{n}) + \eps/2^{n}$ for all $n \in \natp$. In which case,
so $U = \bigcup_{n \in \natp}U_{n}$ is the desired open set.
Applying the above result to $E^{c}$, there exists $V \in \cn^{o}(E^{c})$ such that $\mu(V \setminus E^{c}) < \eps$. Let $F = V^{c}$, then $F \subset E$ is closed and
Proposition 14.6.5. Let $X$ be a LCH space and $\mu: \cb_{X} \to [0, \infty]$ be a Borel measure such that:
For any $U \subset X$ open, $U$ is $\sigma$-compact.
For any $K \subset X$ compact, $\mu(K) < \infty$.
then $\mu$ is a regular measure on $X$.
Proof. By assumption (b), $f \mapsto \int f d\mu$ is a positive linear functional on $C_{c}(X; \real)$. By the Riesz Representation Theorem, there exists a Radon measure $\nu: \cb_{X} \to [0, \infty]$ such that for any $f \in C_{c}(X; \real)$, $\int f d\mu = \int f d\mu$.
Let $U \subset X$ be open, then by Proposition 14.6.2,
By assumption (a), there exists $\seq{K_n}\subset 2^{X}$ compact such that $K_{n} \upto U$. By Urysohn’s lemma, there exists $\seq{f_n}\subset C_{c}(X; [0, 1])$ such that $\one_{K_n}\le f_{n} \le \one_{U}$ for all $n \in \natp$, and $f_{n} \upto f$ pointwise. Using the Monotone Convergence Theorem,
Therefore $\mu(U) = \nu(U)$.
Now let $E \in \cb_{X}$ be arbitrary and $\eps > 0$. By Proposition 14.6.4, there exists $U \in \cn^{o}(E)$ and $V \subset E$ closed such that $\nu(U \setminus E), \nu(E \setminus V) < \eps$. In which case, since $U \setminus V$ is open,
so $\mu(U) = \mu(U \setminus E) + \mu(E) \le \mu(E) + \eps$. Therefore
and $\mu = \nu$.
Since $X$ is $\sigma$-compact, $\mu$ is $\sigma$-finite, so $\mu$ is regular by Proposition 14.6.3.$\square$
Proposition 14.6.6. Let $X$ be a LCH space, $\mu: \cb_{X} \to [0, \infty]$ be a Radon measure, $E$ be a normed space, and $p \in [1, \infty)$, then $C_{c}(X; E)$ is dense in $L^{p}(X; E)$.
Proof. By Proposition 12.1.8, $\Sigma(X, \cm; E) \cap L^{p}(X; E)$ is dense in $L^{p}(X; E)$. Using linearity, it is sufficient to approximate indicator functions of Borel sets with finite measure.
Let $A \in \cb_{X}$ and $\eps > 0$. By Proposition 14.6.3, there exists $U \in \cn^{o}(A)$ and $K \subset A$ compact such that $\mu(U \setminus A), \mu(A \setminus K) < \eps/2$. By Urysohn’s lemma, there exists $f \in C_{c}(X; [0, 1])$ such that $f|_{K} = 1$ and $\supp{f}\subset U$. In which case, for any $y \in E$,