# Convergence in measure, almost everywhere and almost uniformly. [duplicate]

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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, apnortonMay 2 '15 at 17:14

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## 1 Answer

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*}

• isn't it true that $\{x:\chi_{A_n}(x)\to 0\}=(0,1)$? – Sandor Apr 27 '15 at 6:07
• @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). – Nate Eldredge Apr 27 '15 at 6:12
• Oh, you're right. I see it now. Thank you. – Sandor Apr 27 '15 at 7:40