> [!definition] > > Let $X$ be a LCH space, $U \subset X$ be open and $f \in C_c(X, [0, 1])$. $f \prec U$ if $\supp{f} \subset U$. > [!theoremb] Theorem > > Let $X$ be a [[Locally Compact Hausdorff Space|LCH]] space, $I$ be a [[Positive Linear Functional|positive linear functional]] on $C_c(X)$, then there exists a unique [[Radon Measure|Radon measure]] $\mu$ on $X$ such that $I(f) = \int f d\mu$ for all $f \in C_c(X)$. Moreover, > $ > \mu(U) = \sup\bracs{I(f): f \in C_c(X), f \prec U} > $ > for every $U$ [[Open Set|open]], and > $ > \mu(K) = \inf\bracs{I(f): f \in C_c(X), f \ge \one_K} > $ > for every $K$ [[Compactness|compact]]. > [!theoremb] Theorem > > Let $X$ be a LCH space, and $\mu \in M(X)$ be a [[Space of Radon Measures|complex Radon measure]]. For each $f \in C_0(X)$, let $I_\mu f = \int f d\mu$, then the mapping $M(X) \to C_0(X)^*$ defined by $\mu \mapsto I_\mu$ is an isometric isomorphism. # Uniqueness > [!theorem] > > Let $\mu$ be a Radon measure on $X$ such that $I(f) = \int f d\mu$ for all $f \in C_c(X)$, then > $ > \mu(U) = \sup\bracs{I(f): f \in C_c(X), f \prec U} > $ > for all $U$ open. Thus a Radon measure satisfying the integral condition is unique. > > *Proof*. Let $K \subset U$, then by [[Urysohn's Lemma]], there exists $f \prec U$ such that $f|_K = 1$. By the inner regularity of $\mu$, $\mu(U) \le \sup\bracs{I(f): f \in C_c(X), f \prec U}$. # Existence Let $ \mu(U) = \sup\bracs{I(f): f \in C_c(X), f \prec U} $ for all $U$, open, and $\mu^*(E)$ by $ \mu^*(E) = \inf \bracs{\mu(U): U \supset E, U \text{ open}} $ Since $\mu(U) \le \mu(V)$ whenever $U \subset V$, $\mu^*(U) = \mu(U)$ for all $U$ open. > [!theorem] > > Let $\seq{U_j}$ be a family of open sets, then $\mu(U) \le \sum_{n \in \nat}\mu(U_n)$. > > *Proof*. Let $f \prec U$ and $K = \supp{f}$, then since $K$ is compact, there exists $n \in \nat$ such that $K \subset \bigcup_{j = 1}^n U_j$. Let $\seqf{g_j} \subset C_c(X)$ be a [[Partition of Unity|partition of unity]] subordinate to $\seqf{U_j}$. By writing $f = \sum_{j = 1}^n fg_j$, then $fg_j \prec U_j$, and > $ > I(f) = \sum_{j = 1}^n I(fg_j) \le \sum_{j = 1}^n\mu(U_j) \le \sum_{n \in \nat}\mu(U_j) > $ > therefore $\mu(U) \le \sum_{n \in \nat}\mu(U_n)$. > [!theorem] > > Let $E \subset X$, then > $ > \mu^*(E) = \inf\bracs{\sum_{n \in \nat}\mu(U_j): U_j \text{ open}, E \subset \bigcup_{n \in \nat}U_j} > $ > so $\mu^*$ is an [[Outer Measure|outer measure]]. > > *Proof*. Let $\nu(E)$ denote the expression on the right. Since single covers are also countable covers, $\nu(E) \le \mu^*(E)$. On the other hand, for any countable cover $U = \bigcup_{n \in \nat}U_j$, $\mu(U) \le \sum_{n \in \nat}\mu(U_j)$, so $\mu^*(E) \le \nu(E)$. > [!theorem] > > Let $U \subset X$ be open, then $U$ is $\mu^*$-measurable. > > *Proof*. Let $E$ be an open set such that $\mu^*(E) < \infty$, then $E \cap V$ is also open. Let $\eps > 0$ and $f \prec E \cap U$ such that $f \in C_c(X)$ and $I(f) > \mu(E \cap U) - \eps$. Since $E \setminus \supp{f}$ is open, there exists $g \prec E \setminus \supp{f}$ such that $g \in C_c(X)$ and $I(g) > \mu(E \setminus \supp{f}) - \eps$. As $f + g \prec E$, > $ > \begin{align*} > \mu^*(E) = \mu(E) &\ge I(f) + I(g) > \mu(E \cap U) + \mu(E \setminus \supp{f}) - 2\eps \\ > &\ge \mu^*(E \cap U) + \mu(E \setminus U) - 2\eps > \end{align*} > $ > as $\eps$ is arbitrary, $\mu^*(E) \ge \mu^*(E \cap U) + \mu^*(E \setminus U)$. > > Now suppose that $E$ is arbitrary with $\mu^*(E) < \infty$. Let $\eps > 0$ and $V \supset E$ be an open neighbourhood such that $\mu(V) < \mu^*(E) + \eps$. Therefore > $ > \begin{align*} > \mu^*(E) + \eps > \mu(V) &\ge \mu^*(V \cap U) + \mu^*(V \setminus U) \\ > &\ge \mu^*(E \cap U) + \mu^*(E \setminus U) > \end{align*} > $ > since $\eps$ is arbitrary, $\mu^*(E) \ge \mu^*(E \cap U) + \mu^*(E \setminus U)$. > [!theorem] > > $\mu = \mu^*|_{\cb(X)}$ is a [[Borel Measure|Borel measure]] on $X$. > > *Proof*. By [[Carathéodory's Extension Theorem]]. # Regularity > [!theorem] > > For any $K \subset X$ compact, > $ > \mu(K) = \inf\bracs{I(f): f \in C_c(X), f \ge \one_K} > $ > *Proof*. Let $f \in C_c(X)$ such that $f \ge \one_K$. Let $U_\eps = \bracs{x: f(x) > 1 - \eps}$, then $U_\eps \supset K$ is open. Let $g \prec U_\eps$, then $(1 - \eps)^{-1}f - g \ge 0$, so $I(g) \le (1 - \eps)^{-1}I(f)$. Therefore $\mu(K) \le \mu(U_\eps) \le (1 - \eps)^{-1}I(f)$. Since $\eps > 0$ is arbitrary, $\mu(K) \le I(f)$. > > The other direction follows from the outer regularity of $\mu$. > [!theorem] > > $\mu$ is a Radon measure. > > *Proof*. Let $U$ be open and $f \prec U$, let $g \in C_c(X)$ with $g \ge \one_{\supp{f}}$, then $g - f \ge 0$. Therefore $I(f) \le I(g)$. Since this holds for all $g \ge \one_{\supp{f}}$, $I(f) \le \mu(\supp{f})$. As $\supp{f}$ is compact, this implies that $\mu$ is inner regular on open sets. # Representation > [!definition] > > > ![[layercake.png|500]] > > Let $f \in C_c(X, [0, 1])$ and $N \in \nat$. For each $0 \le j \le N$, define $K_j = \bracs{x \in X: f(x) \ge j/N}$, then each $K_j$ is compact with $K_0 = \supp{f}$. Let > $ > f_j = \begin{cases} > 0 &x\not\in K_{j - 1} \\ > f(x) - \frac{j - 1}{N} &x \in K_{j - 1} \setminus K_j \\ > \frac{1}{n} &x \in K_j > \end{cases} > $ > such that $f_j \in C_c(X, [0, 1/N])$, $\frac{1}{N}\one_{K_j} \le f_j \le \frac{1}{N}\one_{K_{j - 1}}$, and $f = \sum_{j = 1}^nf_j$ is the **layercake decomposition** of $f$. > [!theorem] > > For any $f \in C_c(X, [0, 1])$, $I(f) = \int f d\mu$. Since $C_c(X)$ is in the linear span of $C_c(X, [0, 1])$, this extends to $C_c(X)$ by linearity. > > *Proof*. Let $N \in \nat$, and $\bracs{f_j}_1^N$ be the layercake decomposition of $f$. Since $\frac{1}{N}\one_{K_j} \le f_j \le \frac{1}{N}\one_{K_{j - 1}}$, > $ > \frac{1}{N}\mu(K_j) \le \int f_j \le \frac{1}{N}\mu(K_{j - 1}) > $ > From here, if $U \supset K_{j - 1}$ is open, then $Nf_j \prec U$ and $I(f_j) \le \frac{1}{N}\mu(K_{j - 1})$. As $Nf_j \ge \one_{K_j}$, $I(f_j) \ge \frac{1}{N}\mu(K_j)$, so > $ > \frac{1}{N}\mu(K_j) \le I(f_j) \le \frac{1}{N}\mu(K_{j - 1}) > $ > By summing up across the indices, we get that > $ > \frac{1}{N}\sum_{j = 1}^N \mu(K_j) \le I(f),\int f \le \frac{1}{N}\sum_{j = 0}^{N - 1}\mu(K_j) > $ > so > $ > \abs{I(f) - \int f} \le \frac{\mu(K_0) - \mu(K_N)}{N} \le \frac{\mu(\supp{f})}{N} > $ > as $N$ is arbitrary, $I(f) = \int f d\mu$. # Complex > [!theorem] > > Let $X$ be a LCH space, and $\mu \in M(X)$ be a [[Space of Radon Measures|complex Radon measure]]. For each $f \in C_0(X)$, let $I_\mu f = \int f d\mu$, then the mapping $M(X) \to C_0(X)^*$ defined by $\mu \mapsto I_\mu$ is an isometric isomorphism. > > *Proof*. Firstly, by taking the real and imaginary parts, then applying the [[Jordan Decomposition Theorem|Jordan decomposition]], every functional on $C_0(X)$ can be expressed as a complex Radon measure. > > On the other hand, let $\mu \in M(X)$, then > $ > \abs{\int f d\mu} \le \int \abs{f}d\abs{\mu} \le \norm{f}_u \norm{\mu} > $ > so $I_\mu \in C_0(X)^*$ and $\norm{I_\mu} \le \norm{\mu}$. On the other hand, let $h = \frac{d\mu}{d\abs{\mu}}$, then $\abs{h} = 1$, and by [[Lusin's Theorem]], there exists $f \in C_c(X)$ with $\norm{f}_u = 1$ and $\norm{f} = \abs h$ except on a set $E$ with $\abs{\mu}(E) < \eps/2$. From here, > $ > \begin{align*} > \norm{\mu} &= \int \abs{h}^2 d\abs \mu = \int \ol h d\mu \\ > &\le \abs{\int f d\mu} + \abs{\int (f - \ol h)d\mu} \\ > &\le \norm{I_\mu} + 2\abs{\mu}(E) \\ > &\le \norm{I_\mu} + \eps > \end{align*} > $