Is $\{ x \in [0,1] : \textrm{the sequence }f_1(x),f_2(x),f_3(x),\ldots\textrm{converges}\}$ is measurable?

Question

If $f_n:[0,1] \to \Bbb R, n \in \Bbb N$ is measurable functions sequence, prove that set $\{ x \in [0,1] : \textrm{the sequence }f_1(x),f_2(x),f_3(x),\ldots\textrm{converges}\}$ is measurable.

Answer 1

I think that $\limsup f_n$ and $\liminf f_n$ are measurable and your set is the inverse image of $\{ 0 \}$ by the measurable function $g=\limsup f_n - \liminf f_n$.

Answer 2

Let $B:=\{x\in [0,1]\ |\ \{f_n(x)\}_n\ converges\}$. Then

$$B^c=\bigcup_{m≥1}\ \bigcap_{n≥1}\ \bigcup_{i≥n,j≥n} \ \{x\in [0,1]\ :\ |f_j(x)-f_i(x)|≥\frac{1}{m}\}$$

To prove this let $x\in B^c$ then the sequence $\{f_n(x)\}_n$ is not Cauchy , so there is $m\in \Bbb N$ such that for each $n\in \Bbb N$ we have some $i,j≥n$ with $|f_j(x)-f_i(x)|≥\frac{1}{m}$.

Conversely if $x$ is in set of right hand side then for some $m_0\in \Bbb N$ and for all $n\in \Bbb N$ we have some positive integers $i_n,j_n≥n$ such that $|f_{j_n}(x)-f_{i_n}(x)|≥\frac{1}{m_0}$ i.e. $\{f_n(x)\}_n$ does not satisfies Cauchy criterion for the positive number $\frac{1}{m_0}$. Hence such $x$ obviously is in $B^c$.

Now each set $\{x\in [0,1]\ :\ |f_j(x)-f_i(x)|≥\frac{1}{m}\}$ is measurable due to difference of two measurable functions is measurable, so that $B^c$ is measurable. Hence $B$ is also measurable.