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.

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

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)


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