> [!theorem] > > Let $X$ be a [[Compactness|compact]] [[Hausdorff Space|Hausdorff space]], and $B(X)$ be the space of bounded [[Borel Measurable Function|Borel measurable functions]]. Identify the [[Space of Radon Measures|complex Radon measures]] $M(X)$ as a subspace of $B(X)^*$, and define the topology on $B(X)$ as the [[Weak Topology|weak topology]] generated by $M(\Sigma)$. > > If $\seq{f_n} \subset B(X)$[^1], then $f_n \to f \in B(X)$ if and only if $\seq{f_n}$ is uniformly bounded and $f_n \to f$ [[Pointwise Convergence|pointwise]]. > > $B(X)$ is a $C^*$-algebra under pointwise operations and the [[Uniform Convergence|uniform]] norm. > > *Proof*. If $\seq{f_n}$ is uniformly bounded and converges to $f$ pointwise, then $\angles{f_n, \mu} \to \angles{f, \mu}$ for all $\mu \in M(X)$ by the [[Dominated Convergence Theorem]]. > > On the other hand $\angles{f_n, \mu} \to \angles{f, \mu}$ for all $\mu \in M(X)$, then $f_n(x) = \angles{f_n, \delta_x} \to \angles{f, \delta_x} = f(x)$ for all $x \in X$, so $f_n \to f$ pointwise. Since $\angles{f_n, \mu}$ converges, it is bounded for all $\mu \in M(X)$. By the [[Uniform Boundedness Principle]], $\seq{f_n}$ is uniformly bounded. > [!theorem] > > The smallest closed algebra containing $C(X)$ is $B(X)$. > > *Proof*. Let $U \subset X$ be an open set, and $\seq{K_n} \subset U$ such that $K_n \subset K_{n + 1}$ and $\mu(U \setminus K_n) \to 0$ as $n \to \infty$. By [[Urysohn's Lemma]], there exists $f_n \in C(X)$ such that $\supp{f_n} \subset U$ and $f_n|_{K_n} = 1$. Thus $f_n \to \one_U$ weakly due to pointwise and uniform boundedness. > > Thus the closure of $C(X)$ contains indicators of all open sets. The space $\bracs{E \in \cb(X): \one_E \in \ol{C(X)}}$ is a $\sigma$-algebra due to being closed in countable disjoint unions, complements, and intersections (via multiplication). [^1]: This does not generalise to nets due to a potentially infinite initial segment.