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Find a sequence of measurable functions defined on a measurable set $E$ such that the sequence converges everywhere on $E$, but the sequence does not converge almost uniformly on $E$.

I'm having troubles understanding this. I thought that a sequence of functions $\{f_n\}_{n=1}^\infty \to f$ converges almost uniformly if and only if $f_n\to f$ converges in measure. But that's wrong?

Any help would be welcome.

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In light of Egorov's theorem, your measure space will have to be infinite. A good sequence to keep in mind to check that the finite measure space assumption of a theorem is required is the "moving block" $f_n(x) = \chi_{[n,n+1]}(x)$. This converges pointwise to zero, but it fails to converge in a bunch of other senses, including almost uniformly. (To see that, note that it can't converge uniformly on any set whose complement has measure smaller than $1$).

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  • $\begingroup$ Do you know of anywhere I could get more information on this for further clarification? $\endgroup$ – Desperate Fluffy Apr 21 '15 at 17:44
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On $[0,\infty),$ let $f_n= \chi_{[0,n]}.$

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"The sequence $f_n$ converges almost uniformly on $E$" means for every $\delta > 0$ there is a set $B_\delta$ of measure $< \delta$ such that $f_n$ converges uniformly on $E \backslash B_\delta$.

For example, take $E = \mathbb R$ with Lebesgue measure, and $f_n$ the indicator function of the interval $[n,n+1)$.

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