How can this theorem about weakly measurable functions on $\sigma$-finite measure spaces be deduced from the finite measure space case?

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.

• @Arthur Fischer: Thanks a lot for your editing.
– user9464
Nov 24 '15 at 14:56

1 Answer

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).

• Thank you for your answer. Would you elaborate why $\psi_k\to f$ $\mu$-a.e.?
– user9464
Nov 23 '15 at 16:23
• Hmm, maybe I could have added a hint about that too: For $n\in\mathbb{N}$ we have that $\psi_k=\phi_k^n$ on $U_n$ for all $k\geq n$. So there is a.e. convergence restricted to U_n. Put another way, the set of points in U_n without convergence is of measure 0. This is true for all $U_n$, and there are only countably many $U_n$. Nov 23 '15 at 16:59
• Ah, the point is $k\geq n$. I thought it was $k\leq n$. Fair enough. Thank you!
– user9464
Nov 23 '15 at 17:02
• Nice summary of the general strategy - you can see this technique applied in lots of proofs for $\sigma$-finite measure spaces. Nov 27 '15 at 6:48