I'm trying to solve this exercise but I'm stuck. The exercise is the following:
Let $\mathcal{B}(\mathbb{R})$ and $\mathcal{B}(\mathbb{R}^2)$ be the borel $\sigma$ algebra on $\mathbb{R}$ and $\mathbb{R}^2$ respectively. We denote $\lambda$ the Lebesgue measure on $(\mathbb{R}, \mathcal{B}(\mathbb{R}))$ and by $\lambda_2$ the Lebesgue measure on $(\mathbb{R}^2, \mathcal{B}(\mathbb{R}^2))$.
I already proved that, for any nonnegative measurable function $f : \mathbb{R} \rightarrow \mathbb{R}^+$ it holds $\int_\mathbb{R} f(x-y) d \lambda(x) = \int_\mathbb{R} f(x) d \lambda(x)$ by the invariant in translation property of Lebesgue measure.
Now, given $f,g$ nonnegative functions and $h : (x,y) \in \mathbb{R}^2 \rightarrow h(x,y) = f(x-y)g(x)$, I have to prove that $\int_{\mathbb{R}^2} h d \lambda_2 = (\int_\mathbb{R} f d\lambda) (\int_\mathbb{R}g d\lambda) $ (I can assume h is measurable).
To prove the last point, I used the "standard procedure". In particular:
- According to Tonelli's theorem and the the invariance in traslation, since $\lambda_2 = \lambda \otimes \lambda$ and considering $D \in \mathcal{B}({\mathbb{R}^2}), A \in \mathcal{B(\mathbb{R})} \text{ and } B \in \mathcal{B(\mathbb{R})}$, I write the h function as $1_D$ such that:: $$ \int_{\mathbb{R}^2} 1_D (x,y) \lambda_2(x,y) = \int_\mathbb{R}\left(\int_\mathbb{R} 1_D(x,y)d\lambda(x)\right)d\lambda(y) $$
Now , $1_D$ is 1 if (x,y) is in D so if x is in A and y is in B. Hence we can rewrite as : $$ \int_{\mathbb{R}^2} 1_D (x,y) \lambda_2(x,y) = \int_\mathbb{R} \left(\int_\mathbb{R} 1_A(x) 1_B(y)d\lambda(x)\right)d\lambda(y) =\int_\mathbb{R}1_B(y)\left(\int_\mathbb{R}1_A(x)d\lambda(x)\right)d\lambda(y) = \int_\mathbb{R}1_B(y)\lambda(A)d\lambda(y) = \lambda(A)\int_\mathbb{R}1_Bd\lambda(y) = \lambda(A)\lambda(B) = \int_\mathbb{R} fd\lambda \int_\mathbb{R} gd\lambda $$
Now I don't know if this is the correct way and , if it is , I don't know how to continue the proof. Could you please give me some hints?