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!

herethat 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