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Assuming $\left\lbrace f_n\right\rbrace$ is measurable for all n, I was able to show using the fact that sequences of real numbers converge iff they are Cauchy that:

$\left\lbrace x: \lim f_n(x) \text{ exists} \right\rbrace =\bigcap_{k=1}^{\infty} \bigcup_{N=1}^{\infty} \bigcap_{m=N}^{\infty} \bigcap_{n=N}^{\infty} \left\lbrace x: f_n(x)-f_m(x) < \frac{1}{k} \right\rbrace \cap \left\lbrace x: f_m(x)-f_n(x) < \frac{1}{k} \right\rbrace$

Since $f_n$ is measurble for all n, $\left\lbrace x: f_n(x)-f_m(x) < \frac{1}{k} \right\rbrace $ should be a measurable set for all $n,m,$ and $k$, and then since there's a bunch of countable intersections and unions of all these sets, I should have that $\left\lbrace x: \lim f_n(x) exists \right\rbrace$ is measurable as well. What I don't understand is why $f_n$ being measurable immediately implies that $\left\lbrace x: f_n(x)-f_m(x) < \frac{1}{k}\right\rbrace $ is measurable. We just started learning about measurable functions and I know that $f:X \rightarrow Y$ is $M,N-$measurable if $f^{-1}(E) \in M$ for all $E \in N$ where $M$ is the sigma-algebra on $X$ and $N$ the sigma algebra on $Y$, but I don't see how to apply that definition here.

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  • $\begingroup$ The difference of two measurable functions is measurable; hence $f_n-f_m$ is measurable. This, in turn, implies that $$\{x; f_n(x)-f_m(x)< \frac{1}{k} \} = (f_n-f_m)^{-1}((-\infty,1/k))$$ is measurable. $\endgroup$ – saz Oct 18 '15 at 19:16
  • $\begingroup$ With notations $f_n=f$, $f_m=-g$ you may use this answer. $\endgroup$ – A.Γ. Oct 18 '15 at 19:16
  • $\begingroup$ Thanks guys, I appreciate the help! $\endgroup$ – Brenton Oct 18 '15 at 22:49
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An easier way to do this so that you won't have to take unions/intersections over both $m$ and $n$:

Define $f(x) = \limsup_{n \to \infty} f_{n}(x)$. Now, if $f_{n}(x)$ converges at $x$, it must converge to $f(x)$ (recall that convergence implies that your limit is equal to your limsup and liminf). Define $$ A_{m,n} = \{x: |f_{n}(x) - f(x)| < \frac{1}{m}\} $$ And notice that $$ \{x: f_{n}(x) \text{ converges}\} = \bigcap_{m \in \mathbb{N}} \bigcup_{N \in \mathbb{N}}\bigcap_{n \geq N} A_{m,n} $$ Showing that $f$ and consequetly $|f_{n} - f|$ are measurable isn't too hard, and if you're doing this problem, you've probably already proved this in your class/textbook.

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For sake of completness I shall give a sketch for another proof.

$1)$ Show that the maps $x \mapsto \lim \inf f_n(x)$ and $x \mapsto \lim \sup f_n(x)$ are measurables.

$2)$ Show that the set if $f, g$ are measurable functions, then the set $\{x : f(x) = g(x)\}$ is measurable. To do this, use that since $\overline{\mathbb{R}}$ is Hausdorff, then the diagonal $\Delta := \{(x,x) : x \in \mathbb{R}\}$ is closed. Further, use thtat $f\times g(x) := (f(x),g(x))$.

$3)$ Recall that a $\lim f_n(x)$ exists if, and only if, $\lim \sup f_n(x) = \lim \inf f_n(x).$

What is the advantage of such approach? It is because you can change the counterdomain to any second countable Hausdorff space with Borel sigma algebra.

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