2
$\begingroup$

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 $\mathbb R^2$?

  • 0
    please see [this](http://meta.math.stackexchange.com/q/3286/8271)2012-01-18

2 Answers 2

1

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.

  • 0
    @Eric: Well, I was implicitly assuming that $I_1$ and $I_2$ are intervals, because it is easier to prove in that case and it suffices for this application. If you want to prove (or have a reference for) the general fact that if $E_1,E_2\subseteq \mathbb R$ are measurable, then $|E_1\times E_2|=|E_1||E_2|$, then go ahead and use this. It is enough to handle the case of intervals, because of the reasons given in the first two sentences of my answer, and because a set of outer measure $0$ can be covered by a countable union of open intervals such that the sum of the lengths is arbitrarily small.2012-01-18
0

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.

  • 0
    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: $\Bbb Q$)...2015-04-14