TITLE: If $\{f_n\}$ is a measurable sequence of functions, then $\{x : \lim f_n(x) $ exists $\}$ is measurable
QUESTION [10 upvotes]: Was hoping someone could help me out on this problem (self-study not hw).
If $\{f_n\}$ is a sequence of measurable functions on $X$, then $\{x : \lim f_n(x) $ exists $\}$ is a measurable function.
Thank you

REPLY [8 votes]: Another proof not using $\limsup$ and $\liminf$ (a bit lengthy, though). 
We know that, given a fixed $x \in X$, the sequence $f_n(x)$ converges iff it is a Cauchy sequence. We will construct sets using this idea.
Let $E_{n,m,r} = \{ x: |f_n(x) - f_m(x)| \leq \frac{1}{r}\}$. This set is a preimage of $|f_n(x) - f_m(x)|$, which is a measurable function due to the fact that absolute value is a continuous function, and difference of measurable functions is measurable.
Now, we can define $T_{n_0, r} = \{ x: |f_n(x) - f_m(x)| \leq \frac{1}{r}, \forall n,m \geq n_0 \}$. This set is a countable intersection of $E_{n,m,r}$, which are measurable.
Then $S_{r} = \{ x: \exists n_0 \in \mathbb{N}: \forall n,m \geq n_0,   |f_n(x) - f_m(x)| \leq \frac{1}{r} \}$. This set is a countable union of $T_{n_0, r}$.
Finally, the desired set is just $\bigcap_\limits{r=1}^{\infty} S_r$, which of course is measurable.