# When can we replace countably valued simple functions by finitely valued simple functions

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 $$\mu(\Omega)$$?

## 1 Answer

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.