How to show product of two nonmeasurable sets is nonmeasurable? 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?
 A: To get the measurable subsets of $\mathbb{R} \times \mathbb{R}$, you do the following steps:


*

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

*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 $\mathbb{R}$.

*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 $\mathcal{F}$.
A: 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.
