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$\{f_n \}_{n=1}^{\infty}$ : sequence of Lebesgue-measurable functions $(f_n : \mathbb{R} \to \overline{\mathbb{R}})$

$\displaystyle\lim_{n\to \infty} f_n (x)=f(x)$ almost everywhere.

Then, prove that $f$ is Lebesgue-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.

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1 Answer 1

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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.

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  • $\begingroup$ @Kavi_Rama_Murthy Isn't $g_n(x)$ defined by $g_n(x)=f_n(x) $ when $x\notin N$ and $0$ when $x \in N$? $\endgroup$
    – daㅤ
    Jun 1, 2021 at 6:20
  • $\begingroup$ @EPA Yes, of course. Utter carelessness on my part. $\endgroup$ Jun 1, 2021 at 6:22

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