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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 $mathbb{R}^2$ ?

In order to see that
$$
S=(B_1 times N_2) cup (N_1 times B_2) cup (N_1 times N_2)=S_1cup S_2cup S_3.$$
is a null-set of $mathbb R^2$, it suffices to show that if $A$ is a Borel-measurable set and $N$ a null-set, then $Atimes N$ is a null-set. This would prove that $S_1$ is a null-set; for $S_2$ it work similarly with switched coordinates and for $S_3$ pick a Borel set $A$ of measure $0$ containing $N_1$.

Since $N$ is a null-set, there exists a Borel-measurable set $N'$ containing $N$ and whose Lebesgue measure is $0$. Therefore, $Atimes Nsubset Atimes N'$, and the latter is a Borel subset of $mathbb R^2$ of measure zero.