Show that $B=\{x\in A : (f_n(x)) \mbox{ converges} \}$ is measurable. Let $f_n$ be bounded and (Lebesgue) measurable on bounded and (Lebesgue) measurable set $A$, for $n=1,2,\ldots$. Show that $B=\{x\in A : (f_n(x)) \mbox{ converges} \}$ is (Lebesgue) measurable.

The hint is to use Cauchy criterion of convergence, but I can't figure out how that would work.
 A: $f_n(x)$ converges iff for every positive integer $M$ there is $N$ such that
for all $m,n > N$, $|f_n(x) - f_m(x)| < 1/M$.  Translate that into an intersection of a union of ... 
A: Another approach: 
Note that $\{x \in A : f_{n}(x) \text{ converges}\} = \{x \in A : -\infty < \liminf f_{n}(x) = \limsup f_{n}(x) < \infty \}$. 
Let $g_{0}^{*} := \liminf f_{n}$ and $g_{1}^{*} := \limsup f_{n}$. Further, let
$$g_{0}(x) := \left\{ \begin{array}{cl}
g_{0}^{*}(x) & \text{if } -\infty < g_{0}^{*}(x) < \infty, \\\\
0 & \text{if } \ \big|g_{0}^{*}(x)\big| = \infty,
\end{array}\right. $$
and
$$g_{1}(x) := \left\{ \begin{array}{cl}
g_{1}^{*}(x) & \text{if } -\infty < g_{1}^{*}(x) < \infty, \\\\
0 & \text{if } \ \big|g_{1}^{*}(x)\big| = \infty.
\end{array}\right.$$
It's easy to show that $g_{i}$ and $g_{i}^{*}$, $i = 1,2$ are measurable. Therefore $g := g_{0} - g_{1}$ is measurable. Thus,
$$\{ x \in A : f_{n}(x) \text{ converges}\} = g^{-1}[\{0\}] - \bigcup_{i=1}^{2}(g_{0}^{*})^{-1}[\{-\infty, \infty\}]$$
is measurable.
