I a having some difficulty with this question. It is somewhat alien to me and in my linear analysis course Fubini's theorem was not covered so I would prefer an answer that doesn't use that.
My question is this:
Let $\lambda_2=\lambda \otimes \lambda$ be the product Lebesgue measure in $\mathbb{R}^2$ and $D$ be the closed disk of radius 1 centred at $(0,0)$. Is $D$ $\lambda_2$-measurable? Justify your answer.
Let $\mathcal M$ be the $\sigma$-algebra of $\lambda$-measurable sets on $\mathbb R$. The $\sigma$-algebra of measurable sets in $\mathbb R \times \mathbb R$ is the smallest $\sigma$-algebra on $\mathbb R \times \mathbb R$ that contains all rectangles (that is sets of the form $A\times B$, where $A,B \in \mathcal M$). Note that all open intervals in $\mathbb R$ are members of $\mathcal M$ . Now let $\{x_n\}$ be a enumaration of points $\{ (x,y) \in \mathbb Q \times \mathbb Q | (x,y) \not \in D\}$. Now for each $n$ let $R_n$ be the cartesian product of two open intervals such that $R_n\cap D = \emptyset$ (can be done since $\mathbb R^2 \setminus D$ is open). Note now that $\cup_{n=1}^\infty R_n = \mathbb R^2 \setminus D$. So $\mathbb R^2 \setminus D $ is product-measurable. Thus so is $D = (\cup_{n=1}^\infty R_n)^C$.
Edit: The sets $R_n$ are defined to have center $x_n$, which is needed. Also I might need to control the size of the rectangles are not too small in order for them to corver $\mathbb R^2 \setminus D$. Say $r_n = \frac{|x_n|-1}{2}$ and let $R_n = (x_n^1-r_n;x_n^1+r_n)\times(x_n^2-r_n;x_n^2+r_n)$, where $x_n = (x_n^1,x_n^2)$.