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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.

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  • $\begingroup$ @Ramiro Yes but metric completeness is not the same as measure completeness ? $\endgroup$
    – Alphie
    Apr 5 '21 at 19:26
  • $\begingroup$ @Ramiro Ok so there is an issue right? Or did I misunderstood to definition of weak measurability? $\endgroup$
    – Alphie
    Apr 5 '21 at 19:34
  • $\begingroup$ Actually, there is an issue with the definition of strong measurability. $\endgroup$
    – Ramiro
    Apr 5 '21 at 19:49
  • $\begingroup$ I have posted a detailed answer to your question. Please, let me know if you have any question regarding my answer. If my answer provides relevant / helpful information regarding your question, please, upvote it. If my answer actualy answers your question, accept it too, please. To upvote, click the triangle pointing upward above the number (of votes) in front of the question. To accept the answer, click on the check mark beside the answer to toggle it from greyed out to filled in. $\endgroup$
    – Ramiro
    Apr 5 '21 at 22:41
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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 $[f] \in L^p$.

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  • $\begingroup$ 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... $\endgroup$
    – Alphie
    Apr 5 '21 at 21:14
  • $\begingroup$ 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. $\endgroup$
    – Alphie
    Apr 5 '21 at 21:24
  • $\begingroup$ @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.. $\endgroup$
    – Ramiro
    Apr 5 '21 at 22:20
  • $\begingroup$ 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? $\endgroup$
    – Alphie
    Apr 5 '21 at 23:49
  • $\begingroup$ @Alphie , Yes. Exactly. $\endgroup$
    – Ramiro
    Apr 5 '21 at 23:54

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