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Let $ (X,\Sigma,\mu) $ be a measure space and $ B $ a Banach space. According to my understanding, a function $ f: X \to B $ is said to be strongly $ \mu $-measurable if and only if it is the almost-everywhere pointwise limit of a sequence $ (s_{n}: X \to B)_{n \in \mathbb{N}} $ of integrable simple functions (ISF’s).

Note: An integrable simple function from $ X $ to $ B $ has the form $ \displaystyle \sum_{k = 1}^{n} \chi_{E_{k}} \cdot b_{k} $, where $ b_{1},\ldots,b_{n} \in B $ and $ E_{1},\ldots,E_{n} \in \Sigma $ have finite $ \mu $-measure.

If $ B $ is a separable Banach space, I have seen several authors say that a function $ f: X \to B $ is strongly $ \mu $-measurable if and only if it is the everywhere pointwise limit of a sequence $ (s_{n}: X \to B)_{n \in \mathbb{N}} $ of ISF’s.

I simply do not see how, in the separable case, one can replace the notion of ‘almost-everywhere’ by the stronger notion of ‘everywhere’ without affecting the definition of strong $ \mu $-measurability.

I would appreciate any help. Thank you!

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  • $\begingroup$ When there is an uncountable $\ A\subseteq X\ $ of measure 0 (as for instance for $\ X:=[0;1]\ $ with the classical Lebesgue measure), and if Banach space $\ B\ $ has uncountable dimension (meaning here that there is an uncountable set of vectors which are not convergent countable linear combinations of any fixed countable set of vectors) then the equivalence fails. Thus Per H. Enflo's example strongly suggests that there is a counter-example even for a separated Banach space. There is still the nonuniqueness question (Schauder base assumes uniqueness). Specialists should know this better. $\endgroup$ – Włodzimierz Holsztyński Oct 9 '14 at 6:32
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So if $\Sigma$ is a complete measurable space, the notions should be equivalent, I think. Let $f$ be strongly measurable in the almost everywhere sense and let $(s_n)$ be the sequence of simple integrable functions converging almost everywhere to $f$. Let $Z$ be the set where the convergence does not take place. Then we will define simple functions $\tilde s_n$ which converge everywhere to $f$.

Let $(x_n)$ be a dense sequence in $B$. For $i\le n$, define a subset $A_{n,i}=\{x\in B\colon d(x,x_j)>d(x,x_i)\forall j<i;\ d(x,x_j)\ge d(x,x_i)\forall i<j\le n\}$. That is $A_{n,i}$ is the set of points in $B$ nearer to $x_i$ than any other element of $\{x_1,\ldots,x_n\}$, breaking ties in a sensible way.

Now define $$ \tilde s_n(\omega)=\begin{cases} s_n(\omega)&\text{if $\omega\in Z^c$;}\\ x_i&\text{if $\omega\in Z$ and $f(\omega)\in A_{n,i}$} \end{cases} $$ This is a simple function. It remains measurable (since $Z$ is a set of measure 0 and so any subset is measurable) and the sequence of functions converges to $f$ everywhere.

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    $\begingroup$ Yes, in complete measure space, you can redefine your approximations on a set of measure zero. $\endgroup$ – Gerald Edgar Oct 9 '14 at 11:53
  • $\begingroup$ Thanks, Anthony! Very innovative construction of the required sequence of integrable simple functions. $\endgroup$ – Transcendental Oct 9 '14 at 14:20

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