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Prove that if $(f_n)$ is a sequence of nonnegative, measurable functions on $[a,b]$ such that $lim_{n\to\infty}\int_a^b f_n(x)dx=0$, then $(f_n)$ converges to $0$ in measure. Show by example that we cannot replace the conclusion with the assertion that $(f_n)$ converges to $0$ almost everywhere.

I don't really know how to go about proving this. I know that convergence almost everywhere implies convergence in measure and that if a sequence converges in measure then there exists a subsequence that converges almost everywhere but I haven't done a lot with convergence in measure.

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  • $\begingroup$ For an counterexample take an example with a 'running maximum', that is $f_{n,k}(x) = 1_{[k/,(k+1)/n]}(x)$. $\endgroup$ – p4sch Dec 4 '18 at 22:03
  • $\begingroup$ So I understand the example that shows this but without using Markovs inequality where should I start in proving the theorem? $\endgroup$ – ICanMakeYouHateME Dec 5 '18 at 3:28
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Use Markov's inequality namely, $$ \mu(|f_n|>\varepsilon)\leq \frac{1}{\varepsilon}\int_a^b f_n(x)\, dx. $$

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  • $\begingroup$ So I don't have Markov's inequality at my disposal. Some theorems I have are The Dominated Convergence Theorem, Egorov's Theorem, Almost everywhere implies in measure, and Riesz's theorem which I stated above. $\endgroup$ – ICanMakeYouHateME Dec 4 '18 at 22:04
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    $\begingroup$ And you have outed yourself as a statistician :). +1 $\endgroup$ – qbert Dec 4 '18 at 22:04
  • $\begingroup$ @ICanMakeYouHateME It is not too hard to prove the inequality from scratch in your particular case. $\endgroup$ – Foobaz John Dec 4 '18 at 22:13

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