This question already has an answer here:

I want to find counterexamples for this:

  • $f_n\to f$ in measure implies $f_n\to f$ almost uniformly.
  • $f_n\to f$ in measure implies $f_n\to f$ almost everywhere.

I proved that almost uniformly implies almost everywhere, so it suffices to show a counterexample for the second point. I mean, to show a sequence such that $f_n\to f$ in measure but $f_n$ does not converge to $f$ almost everywhere.

I know that the sequence of functions of the form $f_n=\chi_{A_n}$, where $\mu(A_n)\to 0$, converges to $0$ in measure. So I wanted to find $A_n$ such that $f_n$ doesn't converge to zero a.e.

Any hint?

Thank you.


marked as duplicate by Nate Eldredge, kjetil b halvorsen, Claude Leibovici, user147263, apnorton May 2 '15 at 17:14

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


Consider the Lebesgue measure on $(0,1]$ and $$ \begin{align*} A_1 &:= \bigg( \frac{1}{2},1 \bigg] \\ A_2 &:= \bigg( 0, \frac{1}{2} \bigg] \\ A_3 &:= \bigg( \frac{3}{4}, 1 \bigg] \\ A_4 &:= \bigg( \frac{1}{2},\frac{3}{4} \bigg] \\ \vdots & \end{align*}$$

  • $\begingroup$ isn't it true that $\{x:\chi_{A_n}(x)\to 0\}=(0,1)$? $\endgroup$ – Sandor Apr 27 '15 at 6:07
  • 1
    $\begingroup$ @Sandor: No. In fact, $\{x:\chi_{A_n}(x)\to 0\}=\emptyset$. For every $x$, there are infinitely many $n$ such that $x \in A_n$ (draw yourself some pictures to see how the sets $A_n$ march back and forth across the unit interval, infinitely many times). $\endgroup$ – Nate Eldredge Apr 27 '15 at 6:12
  • $\begingroup$ Oh, you're right. I see it now. Thank you. $\endgroup$ – Sandor Apr 27 '15 at 7:40

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