I heard a lot of times during my classes that if a Banach space $X$ is separable then if a function is $f\colon \Omega\to X$ is strongly measurable, then it is weakly measurable and vice versa. Is that true? – I have some doubts.
The first statement is very easy. Take $x^*\in X^*$ and let $f_n\colon \Omega\to X$ be e sequance of simple functions such that
$$\|f_n(x)-f(x)\|\to 0$$
for a.e. $x\in \Omega$. Then, easy computation shows that $\langle x^*, f_n()\rangle_{X^*\times X}$ is also a sequqnce of simple functions. Due to continuity of dual pairing we obtain weak measurablity. So, no separability was needed in here.
Suppose now, that $f$ is weakly measurable. If I find $E\subset \Omega$ such that $\mu(E)=0$ and $f(\Omega\setminus E)$ is (norm) separable, then it's done (due to Pettis theorem). If for instance function $f$ is surjective, then we pick $E=\emptyset$ and $f(\Omega\setminus E)=X$ – done. But what happens if $f$ is not surjective?
Best Answer
Any subset of a separable metric space is separable. You can take $E=\emptyset$; $f(\Omega)$ is separable because $X$ is. You don't need $f(\Omega\setminus E)=X$ to apply Pettis' Theorem.