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 $\Bbb R^{m+n}$?
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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 $\mathbb R^{m+n}$.
answered Jul 19, 2021 at 8:19
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$\begingroup$ Thanks. But at the last step, how to show that it is not a Borel set in $\Bbb R^{m+n}$? $\endgroup$– EricJul 19, 2021 at 8:29
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1$\begingroup$ 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 $\endgroup$ Jul 19, 2021 at 8:38
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$\begingroup$ 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. $\endgroup$– EricJul 19, 2021 at 8:45
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1$\begingroup$ 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 $\endgroup$ Jul 19, 2021 at 8:50