> [!definition] > > Let $(X, \cm)$ be a [[Sigma Algebra|measurable space]], $E$ be a [[Normed Vector Space|normed space]] over $F \in \bracs{\real, \complex}$, and $f: X \to E$ be a [[Function|function]], then $f$ is **finitely-valued/simple** if $f(X) \subset E$ is finite, and **separably-valued** if $f(X) \subset E$ is [[Separable Topological Space|separable]]. > > Let $\mu: \cm \to \real^+$ be a [[Complete Measure|complete measure]], then $f$ is **$\mu$-almost finitely/countably/separably-valued** if there exists $A \in \cm$ [[Null Set|co-null]] such that $f(X) \subset E$ is finite/at most countable/separable. > [!definitionb] Definition > > Let $f: X \to E$ be a function, then the following are equivalent: > 1. There exists a [[Sequence|sequence]] $\seq{f_n}$ of finitely-valued $(\cm, \cb(E))$-measurable functions such that $f_n \to f$ strongly [[Pointwise Convergence|pointwise]] [[Almost Everywhere|a.e]] and $\norm{f_n(x)}_E \le \norm{f(x)}_E$ for a.e. $x \in E$. > 2. There exists a [[Sequence|sequence]] $\seq{f_n}$ of finitely-valued $(\cm, \cb(E))$-measurable functions such that $f_n \to f$ strongly [[Pointwise Convergence|pointwise]] [[Almost Everywhere|a.e]]. > 3. $f$ is [[Weakly Measurable Function|weakly measurable]] and $\mu$-almost separably-valued. > 4. $f$ is $(\cm, \cb(E))$-measurable and $\mu$-almost separably-valued. > > If the above holds, then $f$ is **strongly measurable**. In particular, if $E$ is a separable Banach space, then $f$ is strongly measurable if and only if $f$ is $(\cm, \cb(E))$-measurable. > > *Proof*. $(1) \Rightarrow (2)$ directly. > > $(2) \Rightarrow (3)$: First note that if $g: X \to E$ is countably-valued, then $\phi \circ g$ is also $(\cm, \cb(F))$-measurable. Let $\phi \in E^*$, then $\phi \circ f_n \to \phi \circ f$ a.e., and as $\mu$ is complete, $\phi \circ f$ is also $(\cm, \cb(F))$-[[Measurable Function|measurable]] as an a.e. limit of measurable functions. > > Let $A \in \cm$ be co-null such that $f_n \to f$ strongly pointwise and $f_n(A)$ is finite for all $n \in \nat$, then $\bigcup_{n \in \nat}f_n(A)$ is [[Dense|dense]] in $f(A)$ and $f$ is $\mu$-almost separably-valued. > > $(3) \Leftrightarrow (4)$: Let $A \in \cm$ be co-null such that $f(A) \subset E$ is separable. Let $E' \subset E$ be the smallest closed subspace containing $f(A)$, then $E'$ is a separable Banach space. In which case, $\cb(E')$ coincides with the $\sigma$-algebra generated by the weak topology. By setting $f|_{X \setminus A} = 0$, assume without loss of generality that $f(X) \subset E'$. Thus $f$ is weakly measurable if and only if $f$ is $(\cm, \cb(E'))$-measurable. > > Now, for each $V \subset E'$ open with respect to $E'$, there exists $U \subset E$ open such that $V = U \cap E'$. As $E' \subset E$ is closed, $V \in \cb(E)$. Therefore $\cb(E') \subset \bracs{B \cap E': B \in \cb(E)}$. On the other hand, since $E' \in \cb(E)$, $\cb(E') \supset \bracs{B \cap E': B \in \cb(E)}$. Since $f(X) \subset E$, $f$ is $(\cm, \cb(E'))$-measurable if and only if $(\cm, \cb(E))$-measurable. > > $(4) \Rightarrow (1)$: Let $A \in \cm$ be co-null such that $f(A) \subset E$ is separable. By taking the smallest closed subspace of $E$ containing $A$, assume without loss of generality that $E$ is also separable. > > Let $\seq{y_n} \subset E$ be a dense subset such that $y_1 = 0$. For each $N \in \nat$, let > $ > A_N(x) = \bracs{1 \le n \le N: \norm{y_n}_E \le \norm{f(x)}_E} > $ > then $A_N(x) \ne \emptyset$ for all $x \in E$. Define > $ > I_N(x) = \min\bracs{n \in A_N(x): \norm{y_n - f(x)}_E = \min_{k \in A_N(x)}\norm{y_k - f(x)}_E} > $ > be the smallest $n \in \nat$ such that $\norm{y_n - f(x)}_E$ is minimised among $\bracs{y_n: n \in A_N(x)}$. > > Let $L > \max_{1 \le k \le N}\norm{y_k - f(x)}_E$, $1 \le k \le N$, and define > $ > D_k(x) = \begin{cases} > \norm{y_k - f(x)}_E &k \in A_N(x) \\ > L &k \not\in A_N(x) > \end{cases} > $ > then since $\bracs{x \in X: k \in A_N(x)} \in \cm$ and $f$ is $(\cm, \cb(E))$-measurable, $D_k: X \to \real$ is $(\cm, \cb(E))$-measurable. This allows writing > $ > \min_{k \in A_N(x)}\norm{y_k - f(x)}_E = \min_{1 \le k \le N} D_k(x) > $ > By continuity of $\min$, > $ > M_{N, n} = \bracs{x \in X: \norm{y_n - f(x)}_E = \min_{k \in A_N(x)}\norm{y_k - f(x)}_E} \in \cm > $ > This allows expressing $\bracs{x \in X: I_N(x) = 1} = M_{N, 1}$ and > $ > \bracs{x \in X: I_N(x) = k} = M_{N, k} \setminus \bigcup_{j = 1}^{k - 1}M_{N, j} > $ > so $I_N: X \to \nat$ is a finitely-valued $(\cm, \cb(E))$-measurable function. From here, let $f_N(x) = y_{I_N(x)}$, then $f_N: X \to E$ is a finitely-valued $(\cm, \cb(E))$-measurable function. > > For each $x \in A$ and $\eps > 0$, there exists $N \in \nat$ such that $\norm{f(x) - y_N}_E < \eps$ and $\norm{y_N} \le \norm{f(x)}$. In which case, for each $n \ge N$, $\norm{f(x) - y_N}_E < \eps$. Therefore $f_n \to f$ $\mu$-a.e. > [!theorem] > > Let $\seq{f_n}$ be strongly measurable functions such that $f_n \to f$ a.e., then $f$ is also strongly measurable. > > *Proof*. Since $E$ is a metric space, $f$ is $(\cm, \cb(E))$-measurable. For each $n \in \nat$, let $A_n \in \cm$ be co-null such that $f_n(A_n) \subset E$ is separable, then $\bigcap_{n \in \nat}A_n$ is co-null, and $f(\bigcap_{n \in \nat}A_n) \subset \bigcup_{n \in \nat}f_n(A_n) \subset E$ is also separable. > [!theorem] > > The set all strongly measurable functions $X \to E$ forms a $F$-[[Vector Space|vector space]]. > > *Proof*. Let $f, g: X \to E$ be strongly measurable, then there exists finitely-valued $(\cm, \cb(E))$-measurable functions $\seq{f_n}$ and $\seq{g_n}$ such that $f_n \to f$ and $g_n \to g$ a.e. as $n \to \infty$. For any $\lambda \in F$, $\lambda f_n + g_n \to \lambda f + g$ a.e. with $\lambda f_n + g_n$ also being finitely-valued.