Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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)

share|improve this question
    
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

2 Answers 2

Hint: Write $\int_{-1}^1 m(A \cap B_t)\, dt$ as a double integral.

share|improve this answer

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 $m(A \cap B_t) \geq {1 \over 32} > {1 \over 1000}$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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