I study Orlicz-Pettis theorem.And this is a lemma among the proof of it.
Define a measurable $f$ from a probability space $(\Omega,\Sigma,\mu)$ to a Banach space $X$ with norm $\Vert \cdot \Vert$, then $X$ has a Borel measure $B$ with respect to the norm topology.
It is plain to define a $\mu$-measurable function as usual : $f^{-1}(M)\in\Sigma$ whenever $M \in B$.
Further, we define $s:\Omega \rightarrow B$ as a simple function if there are disjoint members $E_1, E_2, \dots ,E_k$ in $\Omega$ and vectors $x_1,x_2,\dots,x_k$ in $X$ such that $f(\omega)=\Sigma_{i=1}^{k}\chi_{E_i}(\omega)x_i \forall\omega\in\Omega$ where $\chi(E)$ is an indicator function of the set $E\subseteq\Omega$.
Can I have the following similar result as in the function space from a Lebesgue measurable space to a real line? : $f$ is $\mu$-measurable if and only if there is a sequence of simple functions.
I think the only if part is true but I have no idea. For the if part I can't guest whether it is true or not. Any comments or contents? Thanks.