How to show product of two nonmeasurable sets is nonmeasurable?

Solution 1:

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 $\mathcal{F}$.