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!