> [!definition] > > Let $(X, \cm)$ be a [[Sigma Algebra|measurable space]] and $E$ be a [[Banach Space|Banach space]] over $F \in \bracs{\real, \complex}$. A **vector measure** on $\cm$ is a mapping $\mu: \cm \to E$ such that > 1. $\mu(\emptyset) = 0$. > 2. For every $\seq{A_j} \subset \cm$, $\mu\paren{\bigcup_{n \in \nat}A_n} = \sum_{n \in \nat}\mu(A_j)$, where the sum converges unconditionally[^1]. > [!theorem] > > Let $(X, \cm, \mu)$ be a vector measure space, then > $ > \sup_{A \in \cm}\norm{\mu(A)}_E < \infty > $ > *Proof*. For each $A \in \cm$, $\mu$ induces a [[Bounded Linear Map|bounded linear map]] > $ > \mu_A: E^* \to F \quad x^* \mapsto \angles{\mu(A), x^*}_E > $ > For any $x^* \in E^*$, $x^* \circ \mu$ is a [[Complex Measure|complex measure]]. Hence $\sup_{A \in \cm}\abs{\angles{\mu_A, x^*}_E} < \infty$ for every $x^* \in E^*$. By the [[Uniform Boundedness Principle]], > $ > \sup_{A \in \cm}\norm{\mu(A)}_E = \sup_{A \in \cm}\norm{\mu_A}_{E^{**}} < \infty > $ > [!definition] > > Let $(X, \cm, \mu)$ be a vector measure space. For every $A \in \cm$, let > $ > \abs{\mu_f}(A) = \sup\bracs{\sum_{j = 1}^n\norm{\mu(A_j)}_E: \seqf{A_j} \subset \cm, A = \bigsqcup_{j = 1}^n A_j} > $ > and > $ > \abs{\mu_i}(A) = \sup\bracs{\sum_{i \in I}\norm{\mu(A_i)}_E: \seqi{A} \subset \cm, A = \bigsqcup_{i \in I}A_i} > $ > then $\abs{\mu_f} = \abs{\mu_i}$, and $\abs{\mu_f} = \abs{\mu_i} = \abs{\mu}: \cm \to [0, \infty]$ is a [[Measure Space|measure]] on $(X, \cm)$, known as the **total variation** of $\mu$. > > *Proof of equality*. Since all finite partitions are partitions, $\abs{\mu_f} \le \abs{\mu_i}$. On the other hand, let $\seqi{A} \subset \cm$ such that $A = \bigsqcup_{i \in I}A_i$, then for any $J \subset I$ finite, > $ > \abs{\mu_f}(A) \ge \norm{\mu\paren{A \setminus \bigsqcup_{j \in J}A_j}}_E + \sum_{j \in J}\norm{\mu(A_j)}_E \ge \sum_{j \in J}\norm{\mu(A_j)}_E > $ > As the above holds for all $J \subset I$ finite, > $ > \abs{\mu_f}(A) \ge \sum_{i \in I}\norm{\mu(A_i)}_{E} > $ > Thus $\abs{\mu_f} \ge \abs{\mu_i}$. > > *Proof of measure.* Since $\mu(\emptyset) = 0$, $\abs{\mu}(\emptyset) = 0$. Let $\seq{A_j} \subset \cm$ and $A \in \cm$ such that $A = \bigsqcup_{j \in \nat}A_j$. Firstly, if $\seqi{B} \subset \cm$ such that $A = \bigsqcup_{i \in I}B_i$, then each $A_j = \bigsqcup_{i \in I}B_i \cap A_j$. Thus > $ > \sum_{n \in \nat}\abs{\mu}(A_j) \ge \sum_{n \in \nat}\sum_{i \in I}\norm{\mu(B_i \cap A_j)}_E \ge \sum_{i \in I}\norm{\mu(A_i)}_E > $ > As the above holds for every partition of $A$, $\sum_{n \in \nat}\abs{\mu}(A_j) \ge \abs{\mu}(A)$. > > On the other hand, if $\bracs{A_{i, j} \in \cm: i \in I_j, j \in \nat}$ is a family of measurable sets such that $\bigsqcup_{i \in I_j}A_{i, j} = A_j$ for all $j \in \nat$, then $A = \bigsqcup_{j \in \nat}\bigsqcup_{i \in I_j}A_{i, j}$, and > $ > \abs{\mu}(A) \ge \sum_{j \in \nat}\sum_{i \in I_j}\norm{\mu(A_{i, j})}_{E} > $ > If $\abs{\mu}(A_j) < \infty$ for all $j \in \nat$, then for every $\eps > 0$ and $j \in \nat$, there exists $\bracs{A_{i,j}: i \in I_j}$ such that $\bigsqcup_{i \in I_j}A_{i,j} = A_j$ and $\sum_{i \in I_j}\norm{\mu(A_{i, j})}_E \ge \abs{\mu}(A_j) - \eps/2^j$. Thus > $ > \abs{\mu}(A) \ge \sum_{j \in \nat}\sum_{i \in I_j}\norm{\mu(A_{i, j})}_{E} \ge \sum_{j \in \nat}{\abs\mu(A_j)} - \eps > $ > Otherwise, there exists $j \in \nat$ such that for every $\alpha > 0$, there exists $\seqi{B}$ such that $\bigsqcup_{i \in I}B_i = A_j$ and $\sum_{i \in I}\norm{\mu(B_i)}_{E} \ge \alpha$. In which case, > $ > \abs{\mu}(A) \ge \sum_{i \in I}\norm{\mu(B_i)}_E \ge \alpha > $ > As the above holds for all $\eps > 0$ or $\alpha > 0$, $\abs \mu(A) \ge \sum_{j \in \nat}\abs \mu(A_j)$. # Approximation with Delta Masses > [!definition] > > Let $X$ be a $\sigma$[[Sigma Compact|-compact]] [[Metric Space|metric space]] and $\seq{E_n} \subset X$, $E$ be a Banach space, and $\mu: \cb(X) \to E$ be a finite vector measure, then there exists a sequence[^2] $\seq{\mu_k}$ of the form > $ > \mu_k = \sum_{j = 1}^{n_k}\alpha_{k, j}\delta_{x_{k, j}} \quad \alpha_{k, j} \in E, x_{k, j} \in X, n_k \in \nat > $ > such that $\mu_k \Rightarrow \mu$ weakly. > > *Proof*. Let $E_0 = \emptyset$, $\seq{E_n}$ be an exhaustion of $X$ with [[Compactness|compact]] sets, and $U_n = E_{n} \setminus E_{n - 1}$ for all $n \in \nat$. > > Since each $U_n$ is [[Totally Bounded|totally bounded]], there exists $N_n \in \nat$ and $C_n = \bracs{x_{n, j} \in E_n: 1 \le j \le N_n}$ such that $\bigcup_{x \in C_n}B(x, 1/n) \supset U_n$. From here, let $A_{n, j} = B(x_{n, j}, 1/n) \cap U_n$, and $B_{n, j} = A_{n, j} \setminus \bigcup_{k = 1}^{j - 1}A_{n, j}$, then $\bigsqcup_{j = 1}^{N_j}B_{n, j} = U_n$ and $d(y, x_{n, j}) < 1/n$ for all $y \in B_{n, j}$. > > This allows defining the approximation as > $ > \mu_N = \sum_{n = 1}^N \sum_{j = 1}^{N_n}\mu(A_{n, j})\delta_{x_{n, j}} > $ > > Now, let $\phi \in BC(X)$ and $\eps > 0$, then since $\mu$ is finite, there exists: > 1. $k \in \nat$ such that $\abs{\mu}(X \setminus E_k) < \eps/(3\norm{\phi}_u)$. > 2. $N_0 \ge k$ such that for all $x, y \in E_k$, $d(x, y) < 1/N_0$ implies that $\abs{\phi(x) - \phi(y)} < \eps/(3\norm{\mu}_{\text{var}})$. > > Let $N \ge N_0$, then for each $1 \le n \le k$, > $ > \begin{align*} > \abs{\int_{U_n}\phi d\mu - \int_{U_n}\phi d\mu_N} &= \abs{\int_{U_n}\phi d\mu - \sum_{j = 1}^{N_n}\mu(A_{n, j})\phi(x_{n, j})} \\ > &\le \sum_{j = 1}^{N_n}\int_{A_{n, j}}\abs{\phi(y) - \phi(x_{n, j})}d\mu(y) \\ > &\le \sum_{j = 1}^{N_n}\frac{\eps\abs{\mu}(A_{n, j})}{3\norm{\mu}_{\text{var}}} \le \frac{\eps \abs{\mu}(U_n)}{3\norm{\mu}_{\text{var}}} > \end{align*} > $ > thus > $ > \begin{align*} > \abs{\int_{E_k} \phi d\mu - \int_{E_k} \phi d\mu_N} &\le \sum_{j = 1}^{k}\abs{\int_{U_j}\phi d\mu - \int_{U_j}\phi d\mu_N} \\ > &\le \sum_{j = 1}^k \frac{\eps \abs{\mu}(U_j)}{3\norm{\mu}_{\text{var}}} = \frac{\eps \abs{\mu}(E_k)}{3\norm{\mu}_{\text{var}}} \le \eps/3 > \end{align*} > $ > > On the other hand, for any $1 \le n \le N < \infty$, > $ > \abs{\mu_N}(X \setminus E_k) = \sum_{n = k + 1}^N\sum_{j = 1}^{N_n}\norm{\mu(A_{n,j})}_E \le \sum_{n = k + 1}^N\abs{\mu}(U_n) \le \abs{\mu}(X \setminus E_k) > $ > thus by $(1)$, > $ > \begin{align*} > \abs{\int_{X \setminus E_k} \phi d\mu - \int_{X \setminus E_k} \phi d\mu_N} &\le \norm{\phi}_u \braks{\abs{\mu}(X \setminus E_k) + \abs{\mu_N}(X \setminus X_k)} \\ > &= 2\norm{\phi}_u\norm{\mu}_{\text{var}} < \eps/2 > \end{align*} > $ > > so $\abs{\int \phi d\mu - \int \phi d\mu_N} < \eps$ for all $N \ge N_0$. Therefore $\mu_N \Rightarrow \mu$ as $N \to \infty$. [^1]: In arbitrary Banach spaces, this apparently is not equivalent to absolute convergence. [^2]: Approximation using delta masses holds in arbitrary topological spaces. The assumptions allows sequential approximations.