Strongly measurable functions are weakly measurable? From  Pettis measurability theorem, if $f:X\to B$ is a function on a measure space$(X,\Sigma,\mu)$  taking values in a Banach space $B$, then $f$ strongly measurable should imply $f$ weakly measurable.
$f$ strongly measurable means that there exist a sequence of simple functions $(f_n)$ converging almost everwhere to $f$. If $g\in B^*$, with $B^*$ denoting the continuous dual of $B$, then $(g\circ f_n)$ is a sequence of measurable functions converging almost everywhere to $g\circ f$.
But in general the almost everywhere limit of a sequence measurable functions is not measurable unless the measure space is complete.
Am I missing something? Thanks a lot for your help.
 A: Actually, there is a small "issue" with the definition of strong measurability for non-complete measure spaces.
To see it take the Banach space to be $\Bbb R$. Let $(\Omega, \Sigma, \mu)$ to be a measure space that is not complete.
Then using the definition of strong measurability as "$f$ strongly measurable means that there exists a sequence of simple functions $(f_n)$ converging almost everwhere to $f$" will result in having non-measurable functions that will be "strong measurable".
However, given any such strong measurable function $h$ that is non-measurable, there is a measurable function $k$ such that $k=h$ a.e. (and of course, $k$ is also strong measurable).
Remark  In fact, strong measurability is not defined for individual functions $f$ but for the class $[f]$ (class of functions $h$ such that $h=f$ a.e.).
It is similar to what happens with $L^p$. We say that a function is in $L^p$, but to be precise the elements of $L^p$ are equivalence classes of functions by relation $=$ a.e.. For a function $f$, we may write $f \in L^p$, to actually mean $[f] \in L^p$.
