Let $f_n(x): \mathbb{R}\to\mathbb{R}$ be a sequence of measurable functions.
Show that the set $$\{x: (f_n(x))_{n=1}^{\infty} \ \text{converges to a real number}\}$$
is measurable.
My attempt:
$\begin{align}
\{x: (f_n(x))_{n=1}^{\infty} \, \text{converges to a real number}\}&=\{x: (f_n(x))_{n=1}^{\infty} \, \text{is Cauchy }\}\\
&=\underbrace{\{x:\forall\epsilon>0, \exists N \ \text{s.t}\ \forall n,m>N, |f_n-f_m|<\epsilon\}}_A
\end{align}$
Since $f_n(x)$ is measurable for all $n$, $|f_n-f_m|$ is also measurable.
So I'm trying to write the set $A$ as a combination of unions and intercepts of sets like $\{x: |f_n-f_m|<k\}$. But I was so far not successful.
I saw that there is a proof considering limsup and liminf in here and here
But I would like to know whether it is possible to do it by considering the Cauchy property too..
I appreciate your help
Best Answer
Note that for all $n,m,k$ the set
$$A_{n,m,k} = \{x \in \Bbb R: |f_n(x)-f_m(x)| < \frac{1}{k}\}$$
is measurable for measurable $f_n, f_m$. And a sequence is convergent iff it's Cauchy, so your set can eb written as
$$\bigcap_k \bigcup_N \bigcap_{m,n \ge N} A_{n,m,k}$$
which is clearly measurable too.