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If I have a subset of $\mathbb R$ with Lebesgue outer measure $0$ and I take the Cartesian product with an arbitrary subset of $\mathbb R$, does the resulting set also have Lebesgue outer measure $0$ in $\mathbb R^2$?

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The title of your question is missing some words, no? –  Mariano Suárez-Alvarez Jan 18 '12 at 4:07
    
please see this –  leo Jan 18 '12 at 4:26

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It is enough to show that it is true when the arbitrary subset is all of $\mathbb R$. Since $\mathbb R$ is a countable union of bounded intervals, it is enough to show it is true when the arbitrary subset is a bounded interval. So I recommend proving that it is true for $[0,1]$ first.

If $U$ is an open subset of $\mathbb R$, you can figure out if you don't already know what the measure of $U\times [0,1]$ is.

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What do you mean in that last sentence by "you can figure out"? –  Eric Jan 18 '12 at 4:52
    
I mean that there is a nice expression for the measure of $U\times [0,1]$ in $\mathbb R^2$ in terms of the measure of $U$ in $\mathbb R$, and if you do not already know what this expression is, rather than immediately telling you what it is and how to show it, I recommend that you try to determine what it is and show it (i.e., "figure out" what it is). –  Jonas Meyer Jan 18 '12 at 5:03
    
Isn't it just the measure of $U$? Since you just add on an interval of length one to each of the product boxes? –  Eric Jan 18 '12 at 5:20
    
Right, $|I_1\times I_2|=|I_1||I_2|$, and $U$ is a countable union of disjoint intervals. Can you see how to apply this to show that $|E|=0\implies |E\times [0,1]|=0$? –  Jonas Meyer Jan 18 '12 at 5:23
    
I mean, isn't it just a direct application of the formula you just posted? –  Eric Jan 18 '12 at 5:37

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