Preimage of Lebesgue null set under a singular linear map Let $T:\mathbb{R}^n\to \mathbb{R}^n$ ($n\in\mathbb{Z}_+$) be a linear map, but not necessarily invertible. Let $N\subset\mathbb{R}^n$ be a Lebesgue measurable null set. Now, is it true that the preimage of $N$ under $T$—that is, $T^{-1}(N)$—is Lebesgue measurable?
If it is, how can I go about proving it? If not, along what steps can a counterexample be constructed?
Thank you very much.
 A: To try to prove it, simplify the problem. You can assume that the nullity of $T$ is at most 1, since any linear map is a composition of such maps. In fact, you might as well assume that $T$ is a projection along one axis. You can assume that $N$ is bounded, etc.
To try to construct a counterexample, take the first nontrivial case, $n=2$. Assume that you have a non-measurable set $V\subset\mathbb R$ to play with. Is there a way to disguise $V$ as a null set in  $\mathbb R^2$, so that its true colors are revealed by some $T^{-1}$?
A: Chris, thanks so much once again for your help. Now I think that the statement is FALSE. To see this, let $n=2$, let $V\subset\mathbb{R}$ be your favorite Lebesgue non-measurable set, and let $N=V\times\{0\}$. Now, $N$ is Lebesgue measurable in $\mathbb{R}^2$, since $N\subset\mathbb{R}\times\{0\}$, and the latter is a Borel set of vanishing measure. But let now $T:\mathbb{R}^{2}\to\mathbb{R}^{2}$ be represented by the matrix
\begin{equation}
\left[\begin{array}{c}1&0\\0&0\end{array}\right]
\end{equation}
In this case, $T^{-1}(N)=V\times\mathbb{R}$, which is not measurable in $\mathbb{R}^2$.
