Show $\forall t \in \mathbb{R}$ we have : $F(t) = \int_B \mathbb{1}_A (x-t) d\lambda(x)$

Question

Let $\lambda$ Lebesgue measure on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$.
Let $A, B \in \mathbb{R}$ Borel set with $\lambda(A) < +\infty$ and $\lambda(B) < +\infty$. For $t \in \mathbb{R}, A+t =\{x+t , x \in A\}$ the translate of $A$. Let $F : \mathbb{R} \to \mathbb{R}$ with $F(t) = \lambda((A+t) \cap B)$.

Show $\forall t \in \mathbb{R}$ we have : $F(t) = \int_B \mathbb{1}_A (x-t) d\lambda(x)$.

Answer 1

Well, for every point $t$ you have $$\int_B 1_A(x-t) d \lambda(x) = \int_B 1_{A+t}(x) d \lambda(x) = \int 1_B (x) 1_{A+t}(x) d \lambda(x) = \int 1_{B \cap (A+t)}(x) d \lambda(x) = \lambda(B \cap (A+t))$$