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.

• @Ramiro Yes but metric completeness is not the same as measure completeness ? Apr 5 '21 at 19:26
• @Ramiro Ok so there is an issue right? Or did I misunderstood to definition of weak measurability? Apr 5 '21 at 19:34
• Actually, there is an issue with the definition of strong measurability. Apr 5 '21 at 19:49

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 $$[f] \in L^p$$.

• Thanks a lot. So the right definition of weak measurability here would be $f=f'$ a.e. for some weakly measurable function $f'$? On the Wikipedia page there were no mention of complete measure space... Apr 5 '21 at 21:14
• I also had a basic question relating to measurability here : math.stackexchange.com/q/4087241/522332. Do you you think you could have a look at it? Thanks for your time. Apr 5 '21 at 21:24
• @Alphie , the "issue" is not in the definition of weak measurability itself, but in the fact that strong measurability is actually defined for the class of function module $=$ a.e.. Apr 5 '21 at 22:20
• Right but the definition of weak measurability is $g\circ f$ measurable for all $g\in B^*$, and for this definition to make sense we need $f$ to be a function (not a class of functions). If I define $\tilde{f}(x)$ to be equal to $\lim_{n\to\infty} f_n(x)$ outside a null set and $0$ otherwise, then I obtain a $\Sigma-\mathcal{B}(B)$ measurable function (also weakly measurable) with $\tilde{f}=f$ a.e. no? Apr 5 '21 at 23:49
• @Alphie , Yes. Exactly. Apr 5 '21 at 23:54