# The relationship between measurability and weak measurability

For a Banach-valued random mapping $$f:\Omega\rightarrow X$$, there are three kind of measurability: strong measurability (can be approximated by sequence of simple functions, measurability (the preimage of each measurable set is measurable) and weak measurability (the composition with any element of the dual space is a real-valued measurable function in the usual sense); see Section 1.2.1 in the book of Neerven https://fa.ewi.tudelft.nl/~neerven/publications/notes/ISEM.pdf. The well-known Pettis theorem (see Theorem 1.5 in Section 1.2.1 of the above reference book) tells us that the strong measurability implies both measurability and weak measurability ($$f$$ is strongly measurable if it is measurable (or weakly measurable) and separably valued). But what is the relationship between measurability and weak measurability? Seems there is no book discusses this issue (If $$X$$ is separable, these three kind of measurability are equivalent due to the Pettis theorem, but what about for the general (non-separable) Banach space $$X$$?).

• If $f:\Omega\to X$ is measurable (in the preimage sense), and $g\in X^*$, then for any Borel-measurable $B\subset\mathbb R$, we have that $g^{-1}(B)$ is $\text{Borel}(X)$-measurable, so $(g\circ f)^{-1}(B)=f^{-1}(g^{-1}(B))$ is $\Omega$-measurable. So we have weak measurability. Am I missing something? Commented Jul 3, 2022 at 9:16
• @MaximilianJanisch I think your proof is right. This proof is also what I think before. But I haven’t find any reference that writing this property explicitly, which makes me confused. Thanks a lot for your kind reply! Commented Jul 3, 2022 at 12:10
• It seems what you call "measurable" is a useless notion (when $X$ is not separable). For example, we cannot prove the sum of two measurable functions is measurable. With both "strongly measurable" and "weakly measurable", we can prove that. See mathoverflow.net/a/313792/454 and mathoverflow.net/q/294728/454 Commented Jul 3, 2022 at 13:29
• @GeraldEdgar Okay, I see. So the strong and weak measurability should be the right notions in the general non-separable spaces. Thanks for your comment. Commented Jul 3, 2022 at 15:13
• @GeraldEdgar So if we denote by $L_s$, $L$ and $L_w$ the spaces of integrable (i.e., with finite $L^1$ norm) strongly measurable, measurable and weakly measurable functions, respectively. It is well known that $L_s$ is a Banach space. $L$ is even not a linear space since the sum of two measurable functions may not be measurable. Is $L_w$ also a Banach space? First it is easy to check that $L_w$ is a normed linear space, but is it complete? Commented Jul 6, 2022 at 16:37