20.1 Radon Measures

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),

$\square$

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 $N \in \natp$,

$\square$

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

$\square$

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 20.1.2,

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 20.1.4, 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 20.1.3.$\square$

Proof. Let $A \in \cb_{X}$ and $\eps > 0$. By outer regularity of $\mu$, there exists $U \in \cn_{X \times Y}(A \times Y)$ such that $\mu(U \setminus (A \times Y)) < \eps$. By the Tube Lemma, there exists $V \in \cn_{X}(A)$ such that $V \times Y \subset U$. In which case,

so $\nu$ is outer regular on $A$.

On the other hand, by Proposition 20.1.3, there exists $K \subset A \times Y$ compact such that $\mu((A \times Y) \setminus K) < \eps$. By Proposition 5.16.2, $\pi_{1}(K) \subset A$ is also compact. Thus

so $\nu$ is inner regular on $A$.$\square$

Proof. By Proposition 14.1.9, $\Sigma(X, \cm; E) \cap L^{p}(X; E)$ 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 20.1.3, 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 $y \in E$,

$\square$

Proof [Theorem 7.10, Fol99]. First assume that $f$ is bounded.

(1, bounded): If $f$ is bounded, then $f \in L^{1}(X; E)$. By Proposition 20.1.7, there exists $\seq{\phi_n}\subset C_{c}(X)$ such that $\phi_{n} \to f$ in $L^{1}(\mu)$. Since $\phi_{n} \to f$ in $L^{1}(\mu)$, $\phi_{n} \to f$ in measure by Proposition 14.1.12. By taking a subsequence using 21.6.4, assume without loss of generality that $\phi_{n} \to f$ almost everywhere.

By Egoroff’s Theorem, there exists $A \subset \bracs{f \ne 0}$ such that $\phi_{n} \to f$ uniformly and $\mu(\bracs{f \ne 0}\setminus A) < \eps/3$. By Proposition 20.1.3, there exists $K \subset A$ compact such that $\mu(A \setminus K) < \eps/3$.

Let $\phi_{0} = \lim_{n \to \infty}\phi_{n}$ on $K$, then $\phi_{0} \in C(K; E)$ by Proposition 7.2.2, $\phi_{0} = f$ almost everywhere on $K$, and $\mu(\bracs{f \ne 0}\setminus K) < \eps$.

(2, bounded): By outer regularity, there exists $U \in \cn^{o}(\bracs{f \ne 0})$ such that $\mu(U \setminus \bracs{f \ne 0}) < \eps/3$.

By the Tietze Extension Theorem, there exists $\phi \in C_{c}(X; \complex)$ such that $\phi|_{K} = \phi_{0}$ and $\supp{\phi}\subset U$. In which case,

(3): Let

then $\psi$ is continuous with $\bracs{\phi = f}= \bracs{\psi \circ \phi = f}$ and $\norm{\psi \circ \phi}_{u} \le \norm{f}_{u}$.

Now assume that $f$ is arbitrary.

(1, unbounded): Since $\mu\bracs{f \ne 0}< \infty$, there exists $\alpha > 0$ such that $\mu\bracs{|f| > \alpha}< \eps/2$. Let $g \in L^{\infty}(X; E)$ such that $g = f$ on $\bracs{|f| \le \alpha}$. By (1) applied to $g$, there exists $A \subset \bracs{|f| \le \alpha}$ such that $f|_{A}$ is continuous and $\mu(\bracs{|f| \le \alpha}\setminus A) < \eps/2$. In which case, $\mu(\bracs{f \ne 0}\setminus A) < \eps$, as desired.

(2, unbounded): By (1) applied to $g$, there exists $\phi \in C_{c}(X; \complex)$ such that $\mu\bracs{\phi \ne g}< \eps/2$, and $\mu(\bracs{\phi \ne f}) < \eps$.$\square$

Proof [Proposition 7.12, Fol99]. Assume without loss of generality that $\int f d\mu < \infty$. By Proposition 5.22.2, $f$ is Borel measurable, so $f \ge f_{\alpha}$ for all $\alpha \in A$ implies that

Let $\phi \in \Sigma^{+}(X, \cm)$ with $0 \le \phi < f$ and $\beta < \int \phi d\mu$. Let $\seqf{a_j}\subset (0, \infty)$ and $\seqf{E_j}\subset \cb_{X}$ such that $\phi = \sum_{j = 1}^{n} a_{j} \one_{E_j}$. Since $\int \phi d\mu < \infty$, for each $1 \le j \le n$, $\mu(E_{j}) < \infty$ by Markov’s Inequality. By Proposition 20.1.3, there exists compact sets $\seqf{K_j}\subset 2^{X}$ such that $K_{j} \subset E_{j}$ for each $1 \le j \le n$ and $\int \sum_{j = 1}^{n} a_{j} \one_{K_j}d\mu > \beta$.

Let $\psi = \sum_{j = 1}^{n} a_{j} \one_{K_j}$ and $K = \bigcup_{j = 1}^{n} K_{j}$, then $K$ is compact by Proposition 5.16.2 and $-\psi$ is lower semicontinuous by Proposition 5.22.2.

For any $x_{0} \in K$, since $f_{\alpha} \upto f$ and $\psi \le \phi < f$, there exists $\alpha(x_{0}) \in A$ such that $f_{\alpha(x_0)}(x_{0}) > \psi(x_{0})$. By Proposition 5.22.2, $f_{\alpha(x_0)}- \psi$ is lower semicontinuous, so

is an open cover of $K$. Let $\bracs{x_j}_{1}^{N} \subset K$ such that $K \subset \bigcup_{j = 1}^{N} \bracsn{f_{\alpha(x_j)} > \psi}$. Since $f_{\alpha} \upto f$, there exists $\alpha_{0} \in A$ such that

so

As such an $\alpha_{0} \in A$ exists for all $\beta < \int \phi d\mu$,

Since the above holds for all $\phi \in \Sigma^{+}(X, \cm)$ with $0 \le \phi < f$,

by Lemma 22.2.3.$\square$