Let $\{f_n : \mathbb{R} \to \overline{\mathbb{R}} \}_{n=1}^{\infty}$ be a sequence of measurable functions, and suppose
$\displaystyle\lim_{n\to \infty} f_n (x)=f(x)$ for a.e. $x$.
Then, show that $f$ is measurable.
Since $\displaystyle \lim_{n\to \infty} f_n (x)=f(x)$ almost everywhere, there exists Lebesgue-measurable set $N$ s.t. $m(N)=0$ and $f_n(x) \to f(x)$ on $N^c$.
On $N^c$, $f(x)=\displaystyle \limsup_{n\to \infty} f_n(x)=\liminf_{n\to \infty} f_n(x)$ $\cdots (\ast)$
It seems that I can use the fact that "If $f_n$ is Lebesgue measurable, then $\displaystyle\limsup_{n\to \infty} f_n, \displaystyle\liminf_{n\to \infty} f_n$ are also Lebesgue-measurable".
However, $(\ast)$ holds on only $N^c$.
I'm stacked.
I would like you to give me some ideas.
Best Answer
Let $g_n(x)=f_n(x)$ when $x \notin N$ and $0$ when $x \in N$. Then $g_n$ is measurable so $g \equiv \lim \sup g_n$ is measurable. Check that $f=g$ almost everywhere.