I have proved the following statement(s) and I would like to know if my proof is correct and/or if it could be improved somehow.
(a) Suppose $f_1,f_2,\dots$ is a sequence of functions from a set $X$ to $\mathbb{R}$.
Explain why $\{x\in X:\text{ the sequence }f_1(x), f_2(x),\dots\text{ has a limit in }\mathbb{R}\}=\bigcap_{n=1}^{\infty}\bigcup_{j=1}^{\infty}\bigcap_{k=j}^{\infty}(f_j-f_k)^{-1}((-\frac{1}{n},\frac{1}{n}))$
(b) Suppose $(X,\mathcal{S})$ is a measurable space and $f_1,f_2,\dots$ is sequence of $\mathcal{S}$-measurable functions from $X$ to $\mathbb{R}$. Prove that $\{x\in X:\text{ the sequence }f_1(x), f_2(x),\dots\text{ has a limit in }\mathbb{R}\}$ is an $\mathcal{S}$-measurable subset of $X$.
My proofs:
(a) Let $x\in\{x\in X:\text{ the sequence }f_1(x), f_2(x),\dots\text{ has a limit in }\mathbb{R}\}$: then $(f_n(x))_{n=1}^{\infty}$ is a convergent sequence of real numbers so it is also a Cauchy sequence of real numbers which implies* that for every $n\geq 1$ there exists $N\geq 1$ such that $|f_j(x)-f_k(x)|<\frac{1}{n}$ for all $j,k>N$ so if we set $j_{N}:=N+1$ we have that $|f_{j_N}(x)-f_k(x)|<\frac{1}{n}$ for all $k>j_N$ thus $-\frac{1}{n}<f_{j_N}(x)-f_k(x)<\frac{1}{n}\Leftrightarrow (f_{j_N}-f_k)(x)\in (-\frac{1}{n},\frac{1}{n})\Leftrightarrow x\in (f_{j_N}-f_k)^{-1}((-\frac{1}{n},\frac{1}{n}))$; since this is valid for every $n\geq 1$, for a certain $j\geq 1$ (which we have called $j_N$), and for all the $k$s equal or greater than this $j$ it follows that $x\in\bigcap_{n=1}^{\infty}\bigcup_{j=1}^{\infty}\bigcap_{k=j}^{\infty}(f_j-f_k)^{-1}((-\frac{1}{n},\frac{1}{n}))$.
Now, let $x\in\bigcap_{n=1}^{\infty}\bigcup_{j=1}^{\infty}\bigcap_{k=j}^{\infty}(f_j-f_k)^{-1}((-\frac{1}{n},\frac{1}{n}))$: then for every $n\geq 1$ there exists some $j\geq 1$ such that for all $k\geq j$ it is $x\in (f_j-f_k)^{-1}((-\frac{1}{n},\frac{1}{n}))$ (which is equivalent to saying that $|f_j(x)-f_k(x)|<\frac{1}{n}$) so if we fix $n^*\geq 1$ there is $j^*\geq 1$ such that for all $k\geq j^*,\ |f_j(x)-f_k(x)|<\frac{1}{2n}$ thus $|f_j(x)-f_k(x)|=|f_j(x)-f_{j^*}(x)+f_{j^*}(x)-f_k(x)|\leq |f_j(x)-f_{j^*}(x)|+|f_{j^*}(x)-f_k(x)|<\frac{1}{2n}+\frac{1}{2n}=\frac{1}{n}$ for all $n\geq 1$ and $j,k\geq j^*$. In summary we have found that for every $n\geq 1$ there exists $N\geq 1$ (the one we called $j^*$ in the previous sentence) such that $|f_j(x)-f_k(x)|<\frac{1}{n}$ for all $j,k\geq N$ so $(f_n(x))_{n=1}^{\infty}$ is a Cauchy sequence* of real numbers and hence convergent thus $x\in\{x\in X:\text{ the sequence }f_1(x), f_2(x),\dots\text{ has a limit in }\mathbb{R}\}$.
(b) By hypothesis all the $f_n$ are $\mathcal{S}$-measurable functions so their difference is also a $\mathcal{S}$-measurable function and since $(-\frac{1}{n},\frac{1}{n})$ is a Borel set for every $n\geq 1$ (being open) the preimages $(f_j-f_k)^{-1}((-\frac{1}{n},\frac{1}{n}))\in\mathcal{S}$ and being $\mathcal{S}$ a $\sigma$-algebra it is closed under intersections and unions so $\bigcap_{n=1}^{\infty}\bigcup_{j=1}^{\infty}\bigcap_{k=j}^{\infty}(f_j-f_k)^{-1}((-\frac{1}{n},\frac{1}{n}))\in\mathcal{S}$, as desired.
Note: in part (a) of the proof I have implicitly used the fact that if $(a_n)_{n=1}^{\infty}$ is a Cauchy sequence of real numbers then for every $n\geq 1$ there exists $j\geq 1$ such that $|a_j-a_k|<\frac{1}{n}$ for all $k\geq j$.
Proof. Let $n\geq 1$ and set $\varepsilon=\frac{1}{n}$: then being $(a_n)_{n=1}^{\infty}$ Cauchy we have that there exists $N\geq 1$ such that $|a_j-a_k|<\frac{1}{n}$ for all $j,k>N$ so if we set $j:=N+1$ we have $|a_j-a_k|<\frac{1}{n}$ for all $k\geq j$ and so for $n\geq 1$ we have thus found that there exists $j\geq 1$ such that $|a_j-a_k|<\frac{1}{n}$ for all $k\geq j$, and the claim is proved.
Best Answer
This looks fine to me, but for my tastes "explain" calls for something a little less involved:
$$ x \in \bigcap_{n=1}^\infty \bigcup_{j = 1}^\infty \bigcap_{k = j}^\infty \;(f_j - f_k)^{-1} \left( \left( -\frac 1n, \frac 1n \right) \right) $$
if and only if for each $n \in \mathbb N$, there exists $j \in \mathbb N$ such that for all $k \ge j$, we have that
$$ x \in (f_j - f_k)^{-1} \left( \left( \frac 1n, \frac 1n \right) \right) \iff \left| f_j(x) - f_k(x) \right | < \frac 1n . $$
This is true if and only if $\{f_m(x)\}_{m=1}^\infty$ has a limit in $\mathbb R$, so the two sets in (a) are the same.