Product of Lebesgue measures $\Bbb R^m$ and $\Bbb R^n$ strictly larger than Borel sets in $\Bbb R^{m+n}$?

I know the completion of the product of Lebesgue measures of $$\Bbb R^m$$ and $$\Bbb R^n$$ (on Lebesgue measurable $$\sigma$$-algebras) equals to the Lebesgue measure in $$\Bbb R^{m+n}$$. However, if we do not make a completion, is the product $$\sigma$$- algebra of Lebesgue $$\sigma$$-algebras of $$\Bbb R^m$$ and $$\Bbb R^n$$ strictly larger than the Borel $$\sigma$$-algebra in $$\Bbb R^{m+n}$$?

If $$A$$ is a Lebesgue measurable set in $$\mathbb R^{m}$$ which is not a Borel set then $$A \times \mathbb R^{n}$$ belongs to the product of the Lebesgue $$\sigma$$ algebras but it is not a Borel set in $$\mathbb R^{m+n}$$.
• Thanks. But at the last step, how to show that it is not a Borel set in $\Bbb R^{m+n}$?
• The Borel $\sigma$ algebra of $\mathbb R^{m+n}$ is the product of the Borel $\sigma$ algebras of $\mathbb R^{m}$ and $\mathbb R^{n}$. If a set belongs to this product then all its sections are Borel sets. Take any $y \in \mathbb R^{n}$ and take the section of $A \times \mathbb R^{n}$ by $y$. You will get $A$. @Eric Jul 19, 2021 at 8:38
• Thanks. I came up with the same arguement as you just said. But since the fact that "the Borel $\sigma$-algebra of $\Bbb R^{m+n}$ is the product of the Borel $\sigma$-algebras of $\Bbb R^m$ and $\Bbb R^n$" can not be found in my books (I found in in this site), I was not sure there is another arguement.
• This fact is an easy consequence of the fact that any open set in $\mathbb R^{m+n}$ ois a countable union of sets of the form $U\times V$ where $U$ is open in $\mathbb R^{m}$ and $V$ is open in $\mathbb R^{n}$. @Eric Jul 19, 2021 at 8:50