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During one of my take home exams I was asked to prove in a finite measure space, convergence in measure is the same as converging almost everywhere. I did not think seriously and wrote a proof receiving full mark. But now I realize this statement cannot be right, because there are easy counter-example. So I am wondering where is wrong in my "proof" below. I remember I examined it several times and thought it was correct.

If $f_{n}$ does not converge to $f$ almost everywhere, there exist some measurable subset $L$ with positive measure such that $\lim f_{n}\not=f$ on $L$. $L$ is measurable because if we let $$L_{n}=(\bigcup^{\infty}_{k=n}\{x||f(x)-f_{k}(x)|>0)$$ Then $L=\bigcap_{n=1}^{\infty} L_{n}$. In particular this implies $\mu(L_{n})\ge\mu(L)>T>0$ for all $n$ for some real number $T$.

Now we have $$L_{n}=(\bigcup^{\infty}_{i=1}(\bigcup^{\infty}_{k=n}\{x||f(x)-f_{k}(x)|\ge\frac{1}{i})$$ By our assumption we should have $$\lim_{n\rightarrow \infty}\mu(\bigcup^{\infty}_{k=n}\{x||f(x)-f_{k}(x)|\ge \frac{1}{i}\})=0,\forall i\in \mathbb{N}$$

Hence in particular for any $\epsilon<\frac{T}{20}$ there is a sequence $N_{j}\rightarrow \infty$ such that $$\mu(\bigcup^{\infty}_{k=N_{j}}\{x||f(x)-f_{k}(x)|\ge \frac{1}{i}\})<\frac{\epsilon}{i^{2}2^{j}}$$ Then $$\mu(L_{N_{1}})< \sum^{\infty}_{i=1}\sum^{\infty}_{j=1}\frac{\epsilon}{i^{2}2^{j}}=\frac{\pi^{2}\epsilon}{6}<\frac{\pi^{2}}{120}T<T$$ This contradicts with our conclusion $\mu(L_{n})>T$ above. Hence $f_{n}\rightarrow f$ almost everywhere.

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You have to consider the possibility that the limit does not exist at some points, too. – Mariano Suárez-Alvarez Dec 27 '12 at 23:31
Thanks. Let me think about it. Sorry for the ugly and wrong proof. – Bombyx mori Dec 27 '12 at 23:32
up vote 3 down vote accepted

Your third displayed equation is not true. $(f_n)$ converges to $f$ in measure if for every $\epsilon>0$, we have $$\lim\limits_{n\rightarrow\infty}\mu\bigl(\{x:|f(x)-f_n(x)|\ge\epsilon\}\bigr)=0.$$

This does not imply that $$ \lim\limits_{n\rightarrow\infty} \mu\Bigl(\bigcup\limits_{n=k}^\infty\{x:|f(x)-f_n(x)|\ge\epsilon\}\Bigr)=0.$$

The sets over which the $f_n$ are far away from $f$ may differ. For example, consider the sequence of functions over the unit interval: $$\chi_{[0,1]}, \chi_{[0,1/2]}, \chi_{[1/2,1]}, \chi_{[0,1/4]}, \chi_{[1/4,1/2]}, \dots$$ This sequence converges in measure to the zero function; but, here we have $\mu\Bigl(\bigcup\limits_{n=k}^\infty\{x:|f_n(x) |\ge\epsilon\}\Bigr)=1$ for any $0<\epsilon\le 1$.

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Very well. Thanks! I was wondering why for a long time. – Bombyx mori Dec 28 '12 at 0:27

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