$\|f\|_{L^1}=\lambda(A)\cdot \lambda (B)$

Question

Let $A,B\subseteq \mathbb R^d$ be two Borel-measurable sets with a finite Lebesgue measure and let $f=\chi_A * \chi_B$ ("*" is the convolution). Show $\|f\|_{L^1}=\lambda(A)\cdot \lambda (B)$
I already showed "$\leq$" with Young inequality. How can I show the equality?

Answer 1

The function $f$ is non-negative and $$ \lVert f\rVert_1=\int_{\mathbb R}f(t)dt=\int_{\mathbb R}\int_{\mathbb R}\mathbf{1}_A(x)\mathbf{1}_B(t-x)dxdt, $$ then switching the integrals gives $$ \lVert f\rVert_1=\int_{\mathbb R}\mathbf{1}_A(x)\int_{\mathbb R}\mathbf{1}_B(t-x)dtdx $$ and for each fixed $x$, $\int_{\mathbb R}\mathbf{1}_B(t-x)dt=\lambda(x+B)=\lambda(B)$, where $x+B=\{x+b,b\in B\}$.