5
$\begingroup$

Let $f:[0,1] \to \mathbb{R}$ be Lebesgue measurable with $f > 0$ 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 $\epsilon > 0$ 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?

$\endgroup$
6
$\begingroup$

Here is a direct proof: Let $k,n \in \mathbb{N}$. Then

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

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

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

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

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

We have

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

since $f>0$ a.s. and therefore conclude $m(E_n) \to 0$ as $n \to \infty$.

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

Hint: given $\epsilon > 0$, write $m(E_n) = m(E_n \cap \{f \ge \epsilon\}) + m(E_n \cap \{f < \epsilon\})$. Use Markov's inequality on the first term.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

EDIT: Wrong!

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

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


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

If for every $\delta>0$ there exists $h>0$ such that $m(E_j\cap E_l)<\delta$ for $j=1,\ldots, k$ and every $l>h$, then we can fix $\delta<\epsilon/4k$ 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 $m(F\setminus (E_1\cup\ldots\cup E_k))\leq \epsilon/4$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I don't think your first assertion is true: Let $E_1=[0,1/2]$, $E_2=[0,1/4]\cup[1/2,3/4]$, $E_3=[0,1/8]\cup[3/8,1/2]\cup[5/8,3/4]\cup[7/8,1]$, etc... The intersection of $n$ of these sets has measure $2^{-n}$. $\endgroup$ – David Mitra Jan 2 '13 at 19:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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