Strongly measurable implies Borel measurable in separable space Let $(M,\mu)$ be a measure space, $X$ be a Banach space, $f: M \to X$ be a function. $f$ is said to be strongly measurable if there is a sequence of simple functions $\{f_n\}\to f$ pointwisely a.e.. $f$ is said to be Borel measurable if $f^{-1}(U)$ is measurable for every open set $U$.
I have read in a functional analysis book that if $X$ is separable, strongly measurable and Borel measurable are equivalent, but regrettably the proof offered in the book is wrong. I have not been able to locate a proof of this myself, and I would be grateful if someone can provide me one or give a link containing a proof. Thank you in advance!
 A: The following proof is from "On the convergence of sample probability distributions" of Varadarajan. As you may not have access to jstor, i write it down here.
Remark that it is important that we lie in a probability space. So that it is a particuliar case of your question.
Let $f:M \to X$ measurable, suppose $\mu(M)=1$, let $\mu_f$ the "law" of $f$, ie: $\mu_f(A) := \mu(f^{-1}(A))$  and $n\geq 1$.
Since $X$ is separable, we can cover it with a countable set of sphere with radius $\frac{1}{2n}$. And we can chose a subsequence $(A_{nj})_j$ of mutually disjoint measurable sets such that $X=\cup_j A_{nj}$ and $\forall j$ , diam$(A_{nj})\leq 1/n$.
We now choose $k_n$ big enough such that $\mu_f (B_n) \leq 1/n^2$ where $B_n := \cup_{j>k_n} A_{nj}$.
We set $E_{nj} := f^{-1}(A_{nj})$ and $F_n := f^{-1}(B_n)$. We choose arbitrary points $t_n \in B_n$ and $t_{nj}\in A_{nj}$ for $1\leq j\leq k_n$. 
We set then $f_{n}(x) := t_{nj}$ if $x\in E_{nj}$ and $f_{n}(x):=t_n$ if $x\in F_n$.
With this construction, it follows that $\mu(F_n) \leq 1/n^2$ and for all $x\in M\cap (F_n)^c$, the distance in $X$ between $f(x)$ and $f_{n}(x)$ is less than $1/n$.
Finally, setting $F:= \limsup F_n$ and using Borel-Cantelli completes the proof.
