Measurability of superposition operator with non-separable Banach space

Question

Let $f\colon I \times X \to \mathbb{R}$ be a map where $I \subset \mathbb{R}$ is an interval, $X$ is a Banach space (possibly non-separable) and we have $$t \mapsto f(t,x) \text{ is measurable}$$ $$x \mapsto f(t,x) \text{ is continuous}.$$

My question is: given $w \in L^1(0,T;X)$, is $t \mapsto f(t,w(t))$ measurable or not?

When $X$ is a separable space, it is true. See this paper. I've seen also claims that this holds when $X$ is not separable, but all such claims have "proofs" cited to the above linked paper, which only handles the separable case. So is separability required or not?

Answer 1

[Copying here the content of the comments, for the question not to appear as unanswered]

If by $L^1(0,T;X)$ you mean (as it is standard) the space of Bochner-measurable functions, then by definition any $w \in L^1(0,T;X)$ takes values in a separable subspace of $X$, so the general case follows from the separable case.

This can be checked directly : if $w$ is a Bochner-measurable function, it is an almost everywhere limit of a sequence of simple functions (=measurable taking finitely many different values), so (by the second hypothesis) $t \mapsto f(t,w(t))$ is an almost every limit of a sequence of functions of the same form but for $w$ simple, and (by the first hypothesis) for simple functions $t\mapsto f(t,w(t))$ is measurable.