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?


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 $n \to \infty$.

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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.

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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 $m(F\setminus (E_1\cup\ldots\cup E_k))\leq \epsilon/4$.

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  • $\begingroup$ 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

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