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Assume $A\subset\mathbb{R}^p,B\subset\mathbb{R}^q$, then $A\times B=\left\{(x,y)\big|x\in A\land y\in B\right\}\subset\mathbb{R}^{p+q}$

Then is it true that $m^*(A\times B)=m^*(A)m^*(B)$? Here $$m^*(E)=\inf\left\{\sum_{i=1}^\infty|I_i|\big|E\subset\bigcup_{i=1}^\infty I_i\right\}$$ is the Lebesgue outer measure.

I think it is true that $m^*(A\times B)\leq m^*(A)m^*(B)$, but cannot prove the equality.

Thus, I would like to ask if it is true or is true under certain conditions, or examples for being not true.

Jose Avilez
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Shore
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    https://math.stackexchange.com/questions/566048/on-the-lebesgue-measure-of-a-cartesian-product – Snaw Jan 25 '22 at 03:57
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    It is true if $A$ and $B$ are both Lebesgue measurable. To prove it, note that when $A$ and $B$ are Borel sets, the result is immediate by construction of Lebesgue measure as a product measure. Then it is simple to extend it to Lebesgue measurable sets since any Lebesgue measurable set is a union of a Borel set and a subset of a Borel set of measure 0. – Mason Jan 25 '22 at 04:20
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    @Mason May I further ask for an example indicating this is not true when one of them is not measurable?? – Shore Jan 25 '22 at 05:53
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    @snaw Hi thanks, I've read the post and trying to follow the prove. But if I may ask further for a example that m∗(A×B)<m∗(A)m∗(B) if at least one of them is not measurable?? – Shore Jan 25 '22 at 06:49
  • Your second inequality is proven in this post, so if you can also show $m^* (A \times B) \geq m^* (A) \cdot m^* (B)$ then your first inequality also holds. – ムータンーオ Jun 22 '24 at 10:27

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