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Suppose that $(\Omega,\mathcal{A},\mu)$ is a finite measure space and $X$ is a Banach space. Let $f:\Omega \to X$ be a function that is an a.e. pointwise limit of countable-valued functions $f_n:\Omega \to X$, i.e. $f_n(\Omega)$ is countable and $f_n^{-1}(\{x\})\in \mathcal{A}$ for all $x \in X$.

Is it then true that $f$ is an a.e. pointwise limit of finitely valued functions? What if we drop the finiteness of $\mu(\Omega)$?

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Define $d(f,g)=\int \frac {\|f-g\|} {1+\|f-g\|}d\mu$ for strongly measurable functions $f,g: \Omega \to X$. Then $f_n \to f$ in measure (in the sense $\|f_n-f\| \to 0$ in measure iff $d(f_n,f) \to 0$. We can first choose $n$ such that $d(f_n,f)<\epsilon $ and then choose a finitely valued function $g$ such that $d(f_n,g)<\epsilon $. This shows that there is a sequence of finitely valued measurable functions converging to $f$ in measure. There is a subsequence which converges almost everywhere.

If $\mu (E)=\infty$ for every non-empty set $E$ and $f:\Omega \to \mathbb R$ is a measurable functions which takes each of the values $1,2,...$ with positive (hence infinite!) measure then we cannot write this as the almost everywhere limit of a sequence of finite values measurable functions.

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