I consider quite a standard example of a function that is weakly measurable but is not strongly measurable. Unfortunatelly, I don't fully understand it.
Let $X=l_2([0,1])$. Then $X$ equipped with a standard inner product is a nonseparable Hilbert space. Since every Hilbert space has an orthonormal basis, let $\{e_t\,:\, t\in[0,1]\}$ denote an orthonormal basis of $X$. Define a map $f:[0,1]\to X$ by
$$f(t):=e_t$$
for all $t\in [0,1]$. Assume the the measure $\mu$ we consider is the Lebesgue measure on $[0,1]$. First, I'd like to prove that $f$ is weakly-$\mu$ measurable. Pick $x^*\in X^*$. Then, invoking Riesz Representation Theorem we deduce that there is unique $x\in X$ such that
$$\langle x^*, f(t)\rangle_{X^*\times X}=\langle x, f(t) \rangle_{X}.$$
Hence, in our situation, we obtain that
$$\langle x^*, f(t)\rangle_{X^*\times X}=\langle x, f(t) \rangle_{X}=\langle x, e_t \rangle_{X},$$
for all $t\in [0,1]$. How to deduce that a function $g(t)=\langle x^*, f(t)\rangle_{X^*\times X}=0$ $\mu$ a.e. on $[0,1]$??? The only that comes to my mind is that $x=\sum_{t\in[0,1]}\langle x, e_t\rangle e_t$, so
$$\langle x^*, f(t)\rangle_{X^*\times X}=\langle x, f(t) \rangle_{X}=\langle x, e_t \rangle_{X}=\langle \sum_{t\in[0,1]}\langle x, e_t\rangle e_t, e_t \rangle_{X}=\sum_{t\in[0,1]}\langle x, e_t\rangle.$$
Anyway, having that $\langle x^*, f(\cdot)\rangle$ is $0$ $\mu$ a.e. will give its measurablity and in consequance the weak measurablity of $f$.
How to prove that $f$ is not strongly measurable? I try arguing by contradiction. Assume that $f$ is strongle measurble. Then, by Pettis's theorem, we deduce that there exists a subset $E\subset[0,1]$ such that $\mu(E)=0$ and $f([0,1]\setminus E)$ is (norm) separable. If $\mu(E)= 0$, we deduce that $E$ is at most countable and so $[0,1]\setminus E$ is uncountable. Hot to obtain a contradiction now?
Best Answer
The proof of the Pettis theorem shows in fact that $f:[0,1]\to \ell^2([0,1]$ is strongly measurable iff it is the a.e. uniform limit of a sequence of countably-valued $\mu$-measurable functions. If $E\subseteq [0,1]$, then $f([0,1]\smallsetminus E)$ is separable iff $[0,1]\smallsetminus E$ is countable, in which case, $\mu(E)\neq 0.$