# Convergence and Measure

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.

• For an counterexample take an example with a 'running maximum', that is $f_{n,k}(x) = 1_{[k/,(k+1)/n]}(x)$. – p4sch Dec 4 '18 at 22:03
• So I understand the example that shows this but without using Markovs inequality where should I start in proving the theorem? – ICanMakeYouHateME Dec 5 '18 at 3:28

## 1 Answer

Use Markov's inequality namely, $$\mu(|f_n|>\varepsilon)\leq \frac{1}{\varepsilon}\int_a^b f_n(x)\, dx.$$

• 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. – ICanMakeYouHateME Dec 4 '18 at 22:04
• And you have outed yourself as a statistician :). +1 – qbert Dec 4 '18 at 22:04
• @ICanMakeYouHateME It is not too hard to prove the inequality from scratch in your particular case. – Foobaz John Dec 4 '18 at 22:13