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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

Calvin Khor
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Charith
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    Define $g_m(x)=\sup_{n \ge m}|f_n(x)-f_m(x)|$ for $x \in \mathbb{R}$ and $m \in \mathbb{N}$. Each $g_m$ is a measurable function. Because $\mathbb{R}$ is a complete metric space, for each $x \in \mathbb{R}$, ${f_n(x)}{n=1}^{\infty}$ is a convergent sequence iff ${g_m(x)}{m=1}^{\infty}$ converges to $0$. Notice that ${x \in \mathbb{R} \mid \lim_{m \to \infty}g_m(x)=0}=\bigcap_{k=1}^{\infty} \bigcup_{l=1}^{\infty} \bigcap_{m \ge l} {x \in \mathbb{R} \mid |g_m(x)| \le 1/k }$. – sharpe Aug 13 '20 at 22:30
  • @sharpe Thank you very much – Charith Aug 13 '20 at 22:42

2 Answers2

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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.

Henno Brandsma
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Let $A=\{x\mid\lim_{n}f_{n}(x)\in\mathbb{R}\mbox{ exists}\}$. Let $g=\limsup f_{n}$ and $h=\liminf_{n}f_{n}$. Recall that $g$ and $h$ are $[-\infty,\infty]$-valued measurable functions. (For, $g=\inf_{n}(\sup_{k\geq n}f_{k})$. Since supremum and infimum of a sequence of $[-\infty,\infty${]}-valued measurable functions is measurable, $g$ is measurable too.) Now $A=\{x\mid g=h\}\cap\{x\mid g(x)\in\mathbb{R}\}$, so $A$ is measurable.

Danny Pak-Keung Chan
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