Lebesgue Integral of Lebesgue Measure Function

Question

I'm curious about how to prove the following claim:

Let $A$ and $B$ be subsets of $\mathbb{R}$ both with finite Lebesgue measure. Let $f(x)=m(A\cap (B+x))$, then $$\int_{-\infty}^\infty f(x)=m(A)m(B)$$ where $m$ denotes Lebesgue measure.

Answer 1

That follows from Fubini's theorem:

$$\begin{align} \int_\mathbb{R} f &=\int_\mathbb{R}\Big(\int_\mathbb{R}\mathbb{1}_A(y)\mathbb{1}_{B+x}(y)\,dy\Big)\,dx=\int_\mathbb{R}\Big(\int_{\mathbb{R}}\mathbb{1}_{A}(y)\mathbb{1}_B(y-x)_\,dy\Big)\,dx\\ &=\int_\mathbb{R}\mathbb{1}_A(y)\Big(\int_{\mathbb{R}}\mathbb{1}_B(y-x)_\,dx\Big)\,dy=\int_\mathbb{R}\mathbb{1}_A(y)\,m(y-B)\,dy=m(A)m(B) \end{align}$$ where the last equality follows from translation invariance and symmetry of the Lebesgue measure, i.e. $\mu(V+h)=m(V)$ for all $h$, and $m(-U)=m(U)$ for all measurable set $U$.