$f_{n}$ sequence of measurable functions on measurable set $E$, and $E_{0}$ the set of $x \in E$ that $f_{n}$ converges. Then $E_{0}$ is measurable.

Question

I need a hint on how to show $E_{0}$ is measurable. I've been trying to work with limsup and liminf of sets but am stuck. Thanks.

Answer 1

Note $x\in E_0$ $\iff$ for every positive integer $j$, there exists a positive integer $N$ such that $|f_m(x) - f_n(x)| < \frac{1}{j}$ for all $m, n \ge N$. So if $A_{m,n,j} := \{x\in E: |f_m(x) - f_n(x)| < \frac{1}{j}\}$, then

$$E_0 = \bigcap_{j = 1}^\infty \bigcup_{N = 1}^\infty \bigcap_{m = N}^\infty\bigcap_{n = N}^\infty A_{m,n,j}.$$

Show that $A_{m,n,j}$ is measurable for all $m$, $n$, and $j$, then use the fact that countable unions and intersections of measurable sets are measurable.

Answer 2

$\overline{f}(x) = \limsup_n f_n (x)$ and $\underline{f}(x) =\liminf f_n(x)$ are measurable functions and $$E_0 = \left\lbrace x \colon \left(\overline{f}-\underline{f} \right)(x) = 0\right\rbrace = \left(\overline{f}-\underline{f} \right)^{-1}(\{0\}) $$