The set of convergence of a sequence of measurable functions is measurable

Solution 1:

I preassume that $E^{*}:=\{x\in E: (F_n(x))_n\text{ is a convergent sequence}\}$ and mention that $\mathbb N$ denotes the set of positive integers in this answer.

Then for $x\in E$:

$$x\in E^{*}\iff\forall k\in\mathbb{N}\exists n\in\mathbb{N}\forall r,s\in\mathbb{N}\left[r,s\geq n\implies\left|F_{r}\left(x\right)-F_{s}\left(x\right)\right|<\frac{1}{k}\right]$$ So if $\mathbb N_n$ denotes $\{n,n+1,\dots\}$ then:

$$E^{*}=\bigcap_{k\in\mathbb{N}}\bigcup_{n\in\mathbb{N}}\bigcap_{\left(r,s\right)\in\mathbb{N_n}^{2}}\left\{ x\in E:\left|F_{r}\left(x\right)-F_{s}\left(x\right)\right|<\frac{1}{k}\right\}$$

The sets $\left\{ x\in E:\left|F_{r}\left(x\right)-F_{s}\left(x\right)\right|<\frac{1}{k}\right\} $ are measurable and we are dealing with countable unions and intersections.

So $E^{*}$ is measurable.