Showing $\lim_{n \to \infty} m(E_n) = 0$, assuming $f > 0$ a.e. and $\lim_{n \to \infty} \int_{E_n}f \,dm =0$ Let $f:[0,1] \to \mathbb{R}$ be Lebesgue measurable with $f &gt; 0$ a.e. Suppose that $\{E_n\}$ is a sequence of measurable sets in $[0,1]$ with the property that $\displaystyle \lim_{n \to \infty}\int_{E_n} f \,dm = 0$. Prove that $\displaystyle \lim_{n \to \infty} m(E_n) = 0$.


This question is from an old analysis qual I am studying. So far I have tried a proof by contradiction: if $\displaystyle \lim_{n \to \infty} m(E_n) \neq 0$, then there is an $\epsilon &gt; 0$ and a subsequence $\{n_k\}$ so that $m(E_{n_k}) \ge \epsilon$ for all $k$. I am trying to somehow use this subsequence and show $\displaystyle \lim_{k \to \infty}\int_{E_{n_k}} f \,dm \neq 0$, which would give me a contradiction.

Another fact I know from my measure theory course is that, for meausurable $E \subseteq [0,1]$, the map $\displaystyle \nu(E) = \int_E f \,dm$ defines a measure on the Lebesgue measurable subsets of $[0,1]$. Will this fact be useful to me?
Here is a direct proof: Let $k,n  \in \mathbb{N}$. Then

$$\int_{E_n} f \, dm \geq \int_{E_n \cap \left[f&gt;\frac{1}{k}\right]} f \, dm \geq \frac{1}{k} \cdot m \left( E_n \cap \left[f&gt; \frac{1}{k} \right] \right) \geq 0$$

Since $\int_{E_n} f \, dm \to 0$ as $n \to \infty$ we obtain

$$m \left( E_n \cap \left[f&gt; \frac{1}{k} \right] \right) \to 0 \qquad (n \to \infty)$$

for all $k \in \mathbb{N}$. Thus

$$m(E_n) \leq m \left( E_n \cap \left[f&gt; \frac{1}{k} \right] \right) + m \left( \left[f &gt; \frac{1}{k} \right]^c \right) \to m \left( \left[f &gt; \frac{1}{k} \right]^c \right) \qquad (n \to \infty)$$

We have

$$m \left( \left[f &gt; \frac{1}{k} \right]^c \right) \to 0 \qquad (k \to \infty)$$

since $f&gt;0$ a.s. and therefore conclude $m(E_n) \to 0$ as $n \to \infty$.
EDIT: Wrong!

If $\lim_{n\to\infty} m(E_n)&gt;0$, then(*) you can find $E_{n_k}$ such that $m(E_{n_k})&gt;\epsilon$ and $\bigcap E_{n_k}=E$ has positive measure, say $m(E)&gt;\epsilon/2$.

Then $\int_{E_n}f(x)dm(x)\geq\int_E f(x)dm(x)&gt;0$, so $\lim \int_{E_n}fdm\neq 0$.



(*)Wlog assume that $m(E_n)&gt;\epsilon$ and consider $F=\bigcup E_n$. We can find a finite number $k$ such that $m(F\setminus (E_1\cup\ldots\cup E_k))&lt;\epsilon/4$.

If for every $\delta&gt;0$ there exists $h&gt;0$ such that $m(E_j\cap E_l)&lt;\delta$ for $j=1,\ldots, k$ and every $l&gt;h$, then we can fix $\delta&lt;\epsilon/4k$ and we get that
$$m((E_1\cup\ldots\cup E_k)\cap E_l)\leq \epsilon/4$$
so
$$m(E_l\cap(F\setminus (E_1\cup\ldots\cup E_k)))\geq 3\epsilon/4$$
but that's impossible, because $m(F\setminus (E_1\cup\ldots\cup E_k))\leq \epsilon/4$.