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.

  • $\begingroup$ @Ramiro Yes but metric completeness is not the same as measure completeness ? $\endgroup$
    – Alphie
    Apr 5, 2021 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, 2021 at 19:34
  • $\begingroup$ Actually, there is an issue with the definition of strong measurability. $\endgroup$
    – Ramiro
    Apr 5, 2021 at 19:49
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    – Ramiro
    Apr 5, 2021 at 22:41

1 Answer 1


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 $[f] \in L^p$.

  • $\begingroup$ 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, 2021 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, 2021 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, 2021 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, 2021 at 23:49
  • $\begingroup$ @Alphie , Yes. Exactly. $\endgroup$
    – Ramiro
    Apr 5, 2021 at 23:54

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