# Cartesian products with a set of outer measure zero

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

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

we know that lebesgaue outer mearsure of an countable singelton set zero .so outer measure of (N*N)=o if {(a,b)} can be treated as a singelton set wehere 'a'and 'b'belongs 'N' wehere 'N' is natural numbers which are countabel.

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this does not really provide an answer to the question... The set in question doesn't have to be singleton or $\Bbb N$ they just have measure $0$ (ex: $\Bbb Q$)... – Surb Apr 14 '15 at 9:51