Let $A,B \subset \mathbb R$ measurable subsets and $h:\mathbb R \to\overline{\mathbb R}$ defined s $h(x)=m((A-x) \cap B)$. Show that $h$ is measurable and that it satisfies $\int_{\mathbb R}h(x)dx=m(A)m(B)$
This is what I could do:
If we consider $f:\mathbb R^2 \to \mathbb R$ defined as $f(x,y)=\chi_{A-x}(y)\chi_B(y)$, then we have $f(x,y)=\chi_{(A-x) \cap B}(y)$. By Fubini-Tonelli, $$\int_{\mathbb R^2} f(x,y)=\int_{\mathbb R}(\int_{\mathbb R}\chi_{(A-x) \cap B}(y)dy)dx$$$$=\int_{\mathbb R} m((A-x)\cap B)dx$$$$=\int_{\mathbb R}h(x)dx$$
So $h(x)=\int_{\mathbb R} \chi_{(A-x) \cap B}(y)dy$, then $h$ is a measurable function.
I don't know what to do in order to prove that $$\int_{\mathbb R}h(x)dx=m(A)m(B)$$
I would appreciate any suggestions. Thanks in advance.
It should be \begin{align} \int_{\mathbb R^2} f(x,y)&=\int_{\mathbb R}\left(\int_{\mathbb R}\chi_{(A-x) \cap B}(y)dy\right)dx \\ &=\int_{\mathbb R}\left(\int_{B}\chi_{(A-x) }(y)dy\right)dx \\ &=\int_{\mathbb R}\left(\int_{B}\chi_{A }(y+x)dy\right)dx \\ &=\int_{B}\left(\int_{\mathbb R}\chi_{A }(y+x)dx\right)dy \\ &=\int_{B}m(A)dy \\ &=m(A)m(B) \end{align}