Show that $B=\{x\in A : (f_n(x)) \mbox{ converges} \}$ is measurable.

Question

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.

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The hint is to use Cauchy criterion of convergence, but I can't figure out how that would work.

Answer 1

$f_n(x)$ 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 ...

Answer 2

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.