1
$\begingroup$

Suppose that $A \subset [0,1]$ and $B \subset [0,1]$ are measurable sets, each of Lebesgue measure 1/2. Prove that there exists an $x \in [-1,1]$ such that $m((A+x)\cap B) \geq 1/10.$ Note: $A+x$ is the translation of $A$ by $x$ which preserves the Lebesgue Measure of $A$. (Hint: Use Fubini’s theorem).

I have no clue on this problem and in particular, no idea how it is related to Fubini’s theorem which is about exchange of double integral.

$\endgroup$
  • $\begingroup$ Hint: Integrate the function $x\mapsto m((A+x) \cap B)$ over $(-1,1)$. $\endgroup$ – PhoemueX Mar 16 '15 at 6:37
3
$\begingroup$

It suffices to show that

$$\int_{-1}^1 m((A+x) \cap B) dx \ge \frac{1}{5} . $$

The left hand side is

$$\int_{-1}^1 m((A+x)\cap B) dx = \int_{-1}^1 \int_B \chi_{A+x}(y) dy dx = \int_{-1}^1 \int_B \chi_{A}(y-x) dy dx$$

By Fubini's theorem,

$$\int_{-1}^1 \int_B \chi_{A}(y-x) dy dx = \int_B\int_{-1}^1 \chi_A(y-x) dx dy = \int_B m(A) dx = m(A) m(B) = \frac{1}{4} > \frac{1}{5}$$

(Note that when $y\in B \subset [0,1]$, we have $\int_{-1}^1 \chi_A (y-x) dx = m(A)$ and thus the second equality)

$\endgroup$

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.