The Borel $\sigma$-algebra of $\mathbb{R}^n$, $\mathcal{B}_{\mathbb{R}^n}$, is defined as the smallest $\sigma$-algebra of $\mathbb{R}^n$ containing the open sets of $\mathbb{R}^n$ for its usual topology.
The Lebesgue $\sigma$-algebra of $\mathbb{R}^n$, $\mathcal{L}_{\mathbb{R}^n}$, is characterized as the set of all subsets $A$ of $\mathbb{R}^n$ that can be written as $A = B \cup N$, where $B$ is a Borel set and $N$ is a null-set (with respect to the Borel-Lebesgue measure). It is the completion of $\mathcal{B}_{\mathbb{R}^n}$ with respect to the Borel-Lebesgue measure : $\mathcal{L}_{\mathbb{R}^n} = \widehat{\mathcal{B}_{\mathbb{R}^n}}$.
I know that $\mathcal{B}_{\mathbb{R}^2} = \mathcal{B}_{\mathbb{R}} \otimes \mathcal{B}_{\mathbb{R}}$.
I want to show the following (clearly) equivalent assertions :
$\mathcal{L}_{\mathbb{R}} \otimes \mathcal{L}_{\mathbb{R}} \subset \mathcal{L}_{\mathbb{R}^2}$
$\widehat{\mathcal{B}_{\mathbb{R}}} \otimes \widehat{\mathcal{B}_{\mathbb{R}}} \subset \widehat{\mathcal{B}_{\mathbb{R}^2}}$
$\widehat{\mathcal{B}_{\mathbb{R}}} \otimes \widehat{\mathcal{B}_{\mathbb{R}}} \subset \widehat{\mathcal{B}_{\mathbb{R}} \otimes \mathcal{B}_{\mathbb{R}}}$ .
I think that what I need to show is that if $B_1$, $B_2$ are Borel sets of $\mathbb{R}$ and $N_1$, $N_2$ null sets of $\mathbb{R}$, then $(B_1 \times N_2) \cup (N_1 \times B_2) \cup (N_1 \times N_2)$ is a null-set of $\mathbb{R}^2$.
Thanks.
Edit :
@G. Sassatelli : Thank you for pointing out this already existing topic. Nevertheless, they don't explain there why $(B_1 \times N_2) \cup (N_1 \times B_2) \cup (N_1 \times N_2)$ is a null-set of $\mathbb{R}^2$. So here is my new question :
Let $B_1$, $B_2$ be Borel sets of $\mathbb{R}$ and $N_1$, $N_2$ be null sets of $\mathbb{R}$.
Why is $(B_1 \times N_2) \cup (N_1 \times B_2) \cup (N_1 \times N_2)$ a null-set of $\mathbb{R}^2$ ?