I am reading a theorem about measurability of vector-valued functions in a note on functional analysis:
Theorem 3.6.1. If $X$ is a separable, metrizable locally convex space, $(\Omega, \Sigma, \mu)$ is a $\sigma$-finite measure space, and $f : \Omega \to X$ is weakly measurable, then $f$ is strongly measurable.
The proof begins
We can restrict ourselves to the case of finite measures.
Could somebody explain how the general case (when $(\Omega,\Sigma,\mu)$ is "$\sigma$-finite") can be deduced from the "finite" case?
For completeness, here are some definitions. Let $( \Omega, \Sigma, \mu )$ be a measure space, let $X$ be a Hausdorff locally convex space, and let $f : \Omega \to X$. We say that
- $f$ is measurable if $f^{-1} (G) \in \Sigma$ for every open set $G \subset X$.
- $f$ is weakly measurable if $\varphi \circ f$ is measurable for every $\varphi \in X^*$.
- $f$ is strongly measurable if there exist simple functions $f_n : \Omega \to X$ such that $f(\omega) = \lim_{n \to \infty} f_n(\omega)$ for $\mu$-almost every $\omega \in \Omega$. (A simple function is a measurable function which takes only finitely many values.)
An image of proof, and surrounding definitions, can be seen here.
A general strategy in scenarios where you are trying to extend a proposition like this from a finite measure space to a $\sigma$-finite one is to break the $\sigma$-finite space in question into countably many finite-measure disjoint measurable pieces and apply the theorem to each chunk individually, and then try to stitch things back up again.
To do this here, assume the hypotheses of the question. Then, write $\Omega=\cup_{n\in \mathbb{N}}U_n$, where the $U_n$ are disjoint and measurable (why can we do this?). Restricting f, we obtain functions $f_n:U_n\rightarrow X$ defined as $f_n=f|_{U_n}$, which you should convince yourself are weakly measurable on $U_n$. Then, applying the proposition in the finite case to each $f_n$, we get a sequence of simple functions $\phi^n_m$ converging a.e. (in $U_n$) to $f_n$.
To obtain a sequence of simple functions defined on $\Omega$ which converges to $f$, we define $\psi_k:\Omega\rightarrow X$ by defining it piecewise on each $U_n$ in the following way: For $k\in\mathbb{N}$ let $\psi_k(x)=\phi^n_k(x)$ for $x\in U_n$, $n\le k$ and $\psi_k(x)=0$ elsewhere. Convince yourself that $\psi_k$ is simple (hint: the fact that it assumes finitely many values is straightforward, that it is measurable follows from our having specified that $U_n$ should be measurable).