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How can I show that the cartesian product of a nonmeasurable set in $\mathbb R$ and a nonmeasurable set or a measurable set with nonzero measure in $\mathbb R$ is nonmeasurable? I have only learned some basic measure theory (i.e. the definitions and some very basic theorems like measure continuity). Can anyone give me a hint?

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By "product" do you mean living in the product space, or do you mean you take the pointwise product of pairs of elements in the sets? (Assuming these sets are subsets of the real numbers) –  Alex Zorn Mar 14 '13 at 3:57
    
well i'm talking about the cartesian product, so pairwise product i guess. –  Aden Dong Mar 14 '13 at 4:01
    
Ok. Sorry if I was unclear, by "pairwise product" I meant "multiplication". When I said "product space", I meant "cartesian product". –  Alex Zorn Mar 14 '13 at 4:03
    
ahh ok ok. sorry i misundertood. well perhaps i should edit my question because i'm talking about the cartesian product. –  Aden Dong Mar 14 '13 at 4:06

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Assuming you mean Cartesian product and that $(X_i, \mathcal{M_i})$ are measurable spaces for $i=1, 2$.

Suppose that $A_1\notin \mathcal{M_1}$ and $A_{2}\notin \mathcal{M_2}$ but their product $A_1\times A_{2}\in\mathcal{M_1}\bigotimes\mathcal{M2}$ where $\mathcal{M_1}\bigotimes\mathcal{M2}$ is the collection generated by subsets $S_1\times S_2$ where $S_i\in\mathcal{M_i}$ for $i=1, 2.$

Then the product measure would be: \begin{equation} (\mu_1\times\mu_2)(A_{1}\times A_2)=\mu_1(A_1)\mu_2(A_2). \end{equation}

But for the RHS to make sense that is, the measures are well defined, then it means that $A_1$ and $A_2$ must have been measurable. This is a contradiction.

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Note also even if just $A_1$ or just $A_2$ was measurable the above argument would lead to a contradiction. –  Nirav Mar 14 '13 at 4:38
    
hmm isn't the product measure only defined on the the cartesian product of $M_1$ and $M_2$? so it wouldn't make sense in the first place to say $(\mu_1 \times \mu_2)(A_1 \times A_2)$? –  Aden Dong Mar 15 '13 at 0:22
    
...wherein lies the contradiction. If $A_1\times A_2\in \mathcal{M}_1\bigotimes\mathcal{M}_2$ that is it is measurable then it must be because it's a union or intersection or complement of sets in $\{S\times T\ \vert\ S\in\mathcal{M}_1, T\in\mathcal{M}_2\}$ but because the $A_i$'s are not measurable then this is not possible so the product measure is not defined on them. Also note that $\mathcal{M}_1\bigotimes\mathcal{M}_2$ is not really a Cartesian product of two sets. It is a collection sets whose members are cartesian products of sets from $\mathcal{M}_1$ and $\mathcal{M}_2$ –  Nirav Mar 15 '13 at 8:06

To get the measurable subsets of $\mathbb{R} \times \mathbb{R}$, you do the following steps:

  1. Take all cartesian products $A \times B$, where $A$ and $B$ are measurable subsets of $\mathbb{R}$.

  2. Take countable unions of sets of the type listed in $1$. Now you have a $\sigma-$algebra, call it $\mathcal{F}$. You should verify that, if $E$ is in $\mathcal{F}$, then the projections $\pi_1$ and $\pi_2$ of $E$ onto the coordinates should be measurable subsets of $\mathbb{R}$.

  3. Now, you "complete" $\mathcal{F}$. ie, you consider all sets $S$ such that: For all $\varepsilon > 0$ there exists $E_1 \subset S \subset E_2$, with $E_1,E_2 \in \mathcal{F}$, such that $\mu(E_1) - \mu(E_2) < \varepsilon$ ($\mu$ is our measure).

Now you should have shown that, for instance, a product of two non-measurable sets $A \times B$ cannot be in $\mathcal{F}$. So you just have to show that it cannot be "boxed between" two sets in $\mathcal{F}$. To do this, notice that the property described in $3$ holds for non-measurable sets in $\mathbb{R}$. In other words, since $A$ is nonmeasurable we cannot find measurable sets $E_1 \subset A \subset E_2$ with $\mu(E_2) - \mu(E_1)$ arbitrarily small. This fact should help you prove that you cannot "box" $A \times B$ in between two sets in $\mathcal{F}$.

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Taking countable unions of rectangles of measurable sets doesn't give you a $\sigma$-algebra. –  Thomas Mar 14 '13 at 9:24
    
Hm, yeah I think you're right. Dang. –  Alex Zorn Mar 14 '13 at 18:52

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