I'm having trouble showing the following:

If $f_n$ is a sequence of measurable functions such that $f_n$ converges to $f$ almost everywhere, then $f$ is measurable.

I was thinking of using $\limsup$ since I know that $\limsup f_n$ is measurable. But now I'm not sure how to continue my argument.

  • 2
    $\begingroup$ Well, if $a_n$ converges to $a$, then $\limsup a_n = a$. So if $f_n$ converges to $f$ almost everywhere, then $\limsup f_n$ is equal to $f$ almost everywhere. Can you show that if $g$ is measurable, and $f=g$ a.e., then $f$ is measurable? $\endgroup$ Feb 26 '11 at 0:51
  • $\begingroup$ ah i see now. i was just missing the last piece. thanks $\endgroup$
    – jack
    Feb 26 '11 at 0:55
  • 9
    $\begingroup$ Note that this only holds if your measure is complete. $\endgroup$ Feb 26 '11 at 1:13

Based on your comment, it looks like you have a handle on the argument. A limsup of a sequence of measurable functions is an inf of a sequence of functions each of which is a sup of a sequence of measurable functions, so it reduces to showing that both infs and sups of sequences of measurable functions are measurable. Assuming you're working with a complete measure, it doesn't matter what happens on the set of measure zero where $(f_n)$ does not converge to $f$.

Just for fun, here's another way to think of this, assuming the functions are all defined on $[0,1]$ with Lebesgue measure. By Egoroff's theorem, off of a set $A$ of arbitrarily small measure, $f_n\to f$ uniformly. By Lusin's theorem, off of a set $B$ of arbitrarily small measure each $f_n$ is continuous (you can apply Lusin to each function with progressively smaller exceptional sets and take $B$ to be the union of these sets). Off of $A\cup B$, $f$ is a uniform limit of a sequence of continuous functions, hence continuous. As came up at another question, a function that is continuous off of sets of arbitrarily small measure is measurable.


It is not always true. See Exercise V chapter 3 of Bartle's Book.

  • 5
    $\begingroup$ Can you post the exercise here please . . . $\endgroup$
    – GA316
    Jan 4 '14 at 15:09

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.