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](https://www.jstor.org/stable/24892914). 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?



Measurability of superposition operator with non-separable Banach space
[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.


