Linear map of borel set is a measurable set? Let $T\colon\mathbb{R}^n \to \mathbb{R}^m$ ($n>m$) be a linear map. Is it true that $T(A)$ is a Lebesgue measurable set for the Borel set $A$?
$T(A)$ for the compact set $A$ is compact set, so is measurable. If $A$ is an open set, then $A$ is countable sum of compact sets and $T(A)$ is measurable. For closed set is the same situation. I don't know what else for other type of Borel sets?
Of course, it may by used the open mapping theorem if we assume that $T$ is surjective.
 A: If by measurable set you mean a  Borel set then the answer is NO.  There exist Borel sets in $\mathbb R^{2}$ whose images under the projection $(x,y) \to x$ are not Borel. 
See A Borel set whose projection onto the first coordinate is not a Borel set
Answer for Lebesgue mesaurability: YES, $T(A)$ is Lebesgue measurable. This can be proved using the theory of analytic sets. $A$ is analytic and its image under any Borel measurable map is analytic. So $T(A)$ is analytic. This implies that $T(A)$ is universally measurable, in particular Lebesgue measurable. [Reference: Measure Theory by Cohn]. 
PS
There is easier proof using the fact that $T$ is Lipschitz.
I fact we can write $A$ is the union of a null set and a sigma compact set. Since $T$ maps null sets to null sets it follows that $T(A)$ is also of the same type. 
A: If $T$ is not of full rank, then its image is a null set, thus $T(A)$ is contained in a null set and, in particular, it is measurable. If $T$ is full rank, then it is surjective, and you already know how to answer in this case.
