Measurable functions with values in Banach spaces

Solution 1:

Yes, it's rather trivial. You might note the following fact: if $f$ is measurable, $\mu$ is complete, and $f = g$ $\mu$-a.e., then $g$ is measurable. This is valid for functions from any complete measure space to any other measurable space. (Proving it is a good exercise.) Essentially, when you have a complete measure, you can change a function arbitrarily on a null set without breaking its measurability.

Now suppose $f_n \to f$ $\mu$-a.e., with each $f_n$ measurable. Let $A$ be the set of $x$ such that $f_n(x) \to f(x)$. Note $A$ is a measurable set since $A^C$ is $\mu$-null and hence measurable (by completeness of $\mu$). Let $$g_n = \begin{cases}f_n(x), & x \in A \\ 0, & x \notin A\end{cases}$$ Then clearly $g_n$ is measurable. Now $g_n(x)$ converges for all $x$, so the limit $g(x) = \lim_{n \to \infty} g_n(x)$ is measurable, as you know. But $g(x) = f(x)$ on $A$, so $f = g$ a.e. By the fact above, $f$ is measurable.

If your underlying measure space is not complete, certainly things can go wrong. For a trivial example, let $B$ be any set which is $\mu$-null but not measurable. If $f_n$ is the zero function for all $n$, we have $f_n \to 1_B$ a.e., but $1_B$ is not a measurable function.