# Measurability of superposition operator with non-separable Banach space

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?

• 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. Oct 6, 2020 at 19:24
• ... and can be checked directly : if $w$ is Bochner-measurable function, it is an almost everywhere limit of a sequence of simple functions, 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. Oct 6, 2020 at 19:43
• @MikaeldelaSalle thank you! Please put as answer if you wish
– MMML
Oct 6, 2020 at 20:12

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.