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 do not see how it is related to Fubini’s theorem and need help on this.
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)