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Question: $E$ is measurable, $m(E)< \infty$, and $f(x)=m[(E+x)\bigcap E]$ for all > $x \in \mathbb{R}$. Prove $\lim_{x \rightarrow \infty} f(x)=0$.

First, since measure is translation invariant, I'm assuming that $(E+x)\bigcap E=E$. But then I had this thought: if $E=\{1,2,3\}$ and $x=1$, then $E+x = \{2,3,4\}$. So the intersection is just a single point. This will have measure zero.

My question is, I'm not sure if this is the right train of thinking. And, if it is, I'm not sure how to make this rigorous.

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Yes. But if $m$ is the Lebesgue measure then $E = \{1, \dots , n \}$ also has measure zero. – Rudy the Reindeer Oct 11 '12 at 7:34
Assuming you meant to write $f(x) = m [(E + x) \cap E]$ then by your argument you'd have $f(x) = m [(E + x) \cap E] = m[E]$ which is an expression independent of $x$. Hence unless $E$ has measure zero to begin with, you wouldn't necessarily have $\lim_{x \to \infty} f(x) = 0$. Or am I missing anything? – Rudy the Reindeer Oct 11 '12 at 7:37
I did forget a brakcet. Furthermore, it just has finite measure, not necessarily finitely many elements. – emka Oct 11 '12 at 7:39
You know that Lebesgue measure is translation invariant, so do you also know that it's continuous from below? If so, one approach would be to begin by writing $E$ as the union of $E\cap[-n,n]$ for all $n$. – Kevin Carlson Oct 11 '12 at 7:39
Measure is translation invariant - yes. I do know that measure is continuous for both ascending and descending sequences of sets. The latter requires an extra condition of having the first set being finite. I don't follow the last part of your hint dealing with $E \bigcap [-n,n]$. – emka Oct 11 '12 at 7:44
up vote 3 down vote accepted

Here's another direction to take my hint.

Setting as above $E_n=E\cap[-n,n]$ we have $E=\cup_{n=1}^\infty E_n$ and $\{E_n\}$ is an increasing sequence, so $m(E)=\lim_{n\to\infty}m(E_n)$. In particular for any $\varepsilon$ we've got $n^*$ so that $$m(E\setminus E_{n^*})=m(E)-m(E_{n^*})<\varepsilon$$

Now $(E_{n^*}+2n^*) \cap E$ is contained in $E\setminus E_{n^*}$ and thus has measure less than $\varepsilon,$ while $(E\setminus E_{n^*} +2n^*)\cap E$ is contained in $E\setminus E_{n^*}+2n^*,$ which also has measure at most $\varepsilon$ by translation invariance.

So the whole set $$(E+2n^*)\cap E=[(E_{n^*}+2n^*) \cap E ]\cup [(E\setminus E_{n^*} +2n^*) \cap E]$$ has measure no more than $2\varepsilon$.

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Yes, I was also looking for this way.. – Berci Oct 11 '12 at 9:19
Yes, our solutions are similar, and I'm unconvinced yours isn't better. I guess I feel a bit most comfortable when I get my hands right on the $\varepsilon$. – Kevin Carlson Oct 11 '12 at 9:29
I don't think I understand your very last inequality. – emka Oct 11 '12 at 16:52
Sorry about that, there were remnants of a notation I abandoned in the midst of writing the answer. Is it readable now how I broke $(E+2n^*)\cap E$ into two disjoint measurable pieces of measure less than $\varepsilon$? – Kevin Carlson Oct 11 '12 at 19:14
@KevinCarlson No worries. I understand that a lot of symbols in \LaTeX becomes unbearable. – emka Oct 11 '12 at 19:19

Well, it seems you are a bit confused about which object lives where..

By translation invariance, we indeed have $m(E)=m(E+x)$, but not $E=E+x$ as you wrote. Also, $\{1,2,3\}\cap\{2,3,4\}$ has two common elements, not just one:)

The hint in one of the comments to consider $E_n:=E\cap[-n,n]$ is a great idea, because in case $x>2n$, $E_n$ and $E_n+x$ will be disjoint, and $$(E+x)\cap E = \bigcup_n \left( (E_n+x)\cap E_n \right) $$ so you can use continuity from below of $m$.

Denote $f_n(x):=m((E_n+x)\cap E_n)$. By the above argument, if $x>2n$, then $f_n(x)=0$. We are looking for $$\lim_{x\to\infty} m((E+x)\cap E) = \lim_{x\to\infty}\lim_{n\to\infty} f_n(x)$$ And exchange the limits.. ($f_n$ is nonnegative and bounded: $f_n(x)\le m(E)<\infty$)

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