$\{f_n \}_{n=1}^{\infty}$ : 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.


1 Answer 1


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.

  • $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.