# Approximation of measurable sets of [0,1]

Let $A,B$ be measurable subsets of $[0,1]$ where $m(A)=m(B)=1/4$. Prove that there exists a $t\in\mathbb{R}$ such that $m(A\cap B_{t})>1/1000$. $B_{t}$ denotes the translation of $B$ by $t$.

Perhaps some kind of set approximation works? as measurable sets can be approximated by "simple" sets (finite unions of intervals)

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The main issue is to obtain an upper bound of $m(A \cup B_t)$ with some distance from $m(A)+m(B)$. Now $A$ is approximately an interval $I$ (reads $A \subset I \cup E$, where $E$ is a set of small measure here), and $B$ is approximately an interval $J$. Then $A \cup B_t$ is approximately $I \cup J_t$, and by a suitable $t$ you can imagine that $I$ and $J_t$ can have significiant amount of overlap so that $m(I \cup J_t)$ is significantly smaller than $m(A)+m(B)$. –  Soarer Aug 7 '11 at 23:41
By interval I meant to say the union of finitely many open intervals. –  Soarer Aug 7 '11 at 23:42
Soarer:Are you assuming A,B are connected? I think if they were, the solution would be immediate. I think the approximation may work if you do it component-wise; take, e.g., A=[0,1/8) and B=(7/8,1] –  gary Aug 8 '11 at 0:04
Em, I did not. Is there a place that I used it implicitly? –  Soarer Aug 8 '11 at 2:10
Well, if A is disconnected, how would you approximate it by an interval? Do you mean a collection of intervals? If A were the union $A_:=[0,1/8)\cup (7/8,1]$ , how could you approximate it by an interval? –  gary Aug 8 '11 at 5:35

Hint: Write $\int_{-1}^1 m(A \cap B_t)\, dt$ as a double integral.
The old convolution trick works for this. (Possibly what Robert Israel had in mind in his answer). Let $\chi_A(x)$ and $\chi_{-B}(x)$ denote the characteristic functions of $A$ and $\{-x: x \in B\}$ respectively, extended to all of ${\mathbb R}$ by setting them equal to zero outside of $[0,1]$. Then $$\int_{\mathbb R} \chi_{-B} \ast \chi_A(t) \,dt = \int_{\mathbb R} \int_{\mathbb R} \chi_{-B}(t- x)\chi_A(x)\,dx\,dt$$ $$= \int_{\mathbb R} \int_{\mathbb R} \chi_{B}(x-t)\chi_A(x)\,dx\,dt$$ $$= \int_{\mathbb R} m(A \cap B_t)\,dt$$ But by Fubini's theorem the double integral is also $m(A)m(-B) = {1 \over 16}$. Since the $t$-integrand is nonzero only on $[-1,1]$, the integrand must be at least ${1 \over 32}$ for some $t$. Hence for that $t$, one has $m(A \cap B_t) \geq {1 \over 32} > {1 \over 1000}$.