Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $(X, \cal{M},\mu)$ be a finite positive measure space and $f$ a $\mu$-a.e. strictly positive measurable function on $X$. If $E_n\in\mathcal{M}$, for $n=1,2,\ldots $ and $\displaystyle \lim_{n\rightarrow\infty} \int_{E_n}f d\mu=0$, prove that $\displaystyle\lim_{n\rightarrow\infty}\mu(E_n)=0$.

share|cite|improve this question
Where did you get stuck? – Egbert May 25 '12 at 22:28
I need help getting started on it – john May 25 '12 at 22:28
Have you tried proving the contrapositive? – Antonio Vargas May 25 '12 at 22:45
I'd just like to point out a good recent example of how to ask a question the good way. We feel much better answering questions to people who show in their question that they have thought about it. – Egbert May 26 '12 at 0:04
up vote 1 down vote accepted

Since $f$ is almost everywhere strictly positive, the increasing sequence of sets $$A_n=\{x\in X:f(x)>1/n\}$$ has the property that $$\lim_{n\to\infty} \mu(A_n)=\mu(X).$$ Now $\int_E f ~d\mu<r$ implies that $\mu(E\cap A_n)\cdot1/n<r$ and hence $$\mu(E\cap A_n)<rn$$ for all $n$ and $r>0$.

So let $\epsilon>0$. Choose $n$ so that $\mu(X\backslash A_n)<\epsilon/2$. For $N$ large enough, $\int_{E_N}f~d\mu<\epsilon/(2n)$ and hence $$\mu(E_N\cap A_n)<n~ \epsilon/(2n)=\epsilon/2.$$ Now $$\mu(E_N)\leq\mu(E_N\cap A_n)+\mu(X\backslash A_n)<\epsilon/2+\epsilon/2=\epsilon.$$

share|cite|improve this answer


  • If $f$ is strictly positive almost everywhere then for "most of $X$" (which you can make arbitrarily large) there is a positive value it is greater than.

  • So unless $E_n$ gets "small enough" permanently in the tail, you can show $\int_{E_n}f d\mu$ most be greater than a particular positive value infinitely often, by comparing the overlap between $E_n$ and "most of $X$".

  • You can formalise this, as with other limits, taking care with the relationship between "most" and "small enough".

share|cite|improve this answer

Since the measure space is assumed to be finite, we assume it to be a probability space. It is useful to first divide $X$ into pairwise disjoint measurable subsets $A_0:=\{x\in X:f(x)\geq 1\}$ $$ A_n:=\{x\in X:\tfrac{1}{n+1}\leq f(x)<\tfrac{1}{n}\}. $$ Note that since $f$ is assumed to be positive $\mu$-a.e., we have that $\sum_{n=0}^\infty\mu(A_n)=1$. Using the sets $A_n$, we are able to get the integrals $\int_{E_n}f\,\mathrm{d}\mu$ into the story: $$ \lim_{k\to\infty}\mu(E_k)=\lim_{k\to\infty}\sum_{n\in\mathbb{N}}\mu(E_k\cap A_n)\leq\lim_{k\to\infty}\sum_{n\in\mathbb{N}}\int_{E_k}\mathbf{1}_{A_n}(n+1)f\,\mathrm{d}\mu $$ Now, wouldn't it be good if we can take the limit inside? Try to prove if that is possible. And then see if you can use the hypothesis. Hopefully I've got you going now :)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.