Let $(\Omega, \Sigma)$ be a measurable space and $X$ a Banach space. Let $f: \Omega \rightarrow X$.
- $f$ is called measurable if every the preimage of every Borel set in $X$ is an element of $\Sigma$.
- $f$ is called strongly measurable if $f$ is the pointwise limit of a sequence of simple functions.
It is known that strongly measurable and measurable are equivalent when $X$ is separable. For this reason, the notion of strong measurability is only relevant when dealing with Bochner integration in full generality. What is an example of a function $f$ taking values in a non-separable Banach space $X$ such that $f$ is measurable but not strongly measurable?
Let me try to show that there is no counterexample when $(\Omega, \Sigma) = ([0, 1], \Sigma)$ where $\Sigma$ is the class of Lebesgue measurable subsets of $[0, 1]$.
Claim: Suppose $X$ is a metric space and $f:[0, 1] \rightarrow X$ is measurable - This means preimages of Borel sets are Lebsgue measurable. Then, there is Lebesgue null set $N \subset [0, 1]$ such that the image of $f \upharpoonright [0, 1] \backslash N$ is separable.
Proof: Suppose not. Let $\{A_n : n \geq 1\}$ be a maximal family of pairwise disjoint positive measure subsets of $[0, 1]$ such that the image of $f \upharpoonright A_n$ is separable. Let $B = [0, 1] \backslash \bigcup_{n \geq 1} A_n$. If $B$ is null, we are done. So assume otherwise. WLOG, assume $B = [0, 1]$. Let $m$ be a Borel measure on $X$ defined by $m(E) = \mu(f^{-1}(E))$. Note that whenever $E$ is separable, $m(E) = 0$. Let $\mathcal{F}$ be the family of all open balls in $X$ which are $m$-null. Then, we claim that $U = \bigcup \mathcal{F}$ is $m$-null. Let us assume this and finish the proof. Let $Y = X \backslash U$. Then every open ball in $Y$ has positive $m$-measure. If $Y$ is not separable, then we can find uncountably many pairwise disjoint open balls in $Y$ each with positive $m$-measure which is impossible. Now, towards a contradiction, suppose $m(U) > 0$. Then the family $\mathcal{N} = \{f^{-1}[V]: V \in \mathcal{F}\}$, consists of pairwise disjoint Lebesgue null subsets of $[0, 1]$ whose union $W$ is not null and the union of every subfamily of $\mathcal{N}$ is Lebesgue measurable. Let $T \subset [0, 1]$ be a set of reals of size continuum which does not contain any perfect set (so you can take $T$ to be a Bernstein set). Let $h: W \rightarrow T$ be a function such that the set of preimages of points in $T$ is the family $\mathcal{N}$. Then $h$ is Lebesgue measurable. By Lusin's theorem, the restriction of $h$ to a positive measure compact subset $K \subseteq W$ is continuous. But $h[K]$ is an uncountable compact subset of $T$ which is impossible.