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!



The Notion of Strong Measurability for Separable Banach Spaces
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.


