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 $\displaystyle\lim_{n\rightarrow\infty}\mu(E_n)=0$.
Measure theory limit question
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3 Answers
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
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)
Hints:
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".
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 :)