How to show that x-y is Lebesgue-Lebesgue measurable Which is the cleanest way to show that the difference, $d:R^n\times R^n\rightarrow R^n$, $d(x,y)= x-y$, is Lebesgue-Lebesgue measurable? (i.e. foreach A lebesgue measurable in $R^n$, $d^{-1}(A)$ is Lebesgue measurable in $R^n\times R^n$).
Thanks in advance.
 A: Unitary matrices preserve measure. A diagonal matrix of full rank is a Lesbesgue-Lesbesgue measurable transformation. Linear maps over the reals have a singular value decomposition.
A: Nicolo is asking about functions where the inverse image
of a Lebesgue measurable set is Lebesgue measurable. This
is stronger than the usual definition of measurability
where it is required only the inverse image of each Borel
set must be Lebesgue measurable. Continuous functions need not
be measurable by this stronger criterion. If $B$ has zero
Lebesgue measure and $A=f^{-1}(B)$ has nonzero measure then each
subset of $B$ is Lebesgue measurable but its inverse image may
be non-measurable. A simple example is given by $f:x\mapsto (x,0)$
from $\mathbb{R}$ to $\mathbb{R}^2$. Taking $A$ to be a
non-measurable subset of $\mathbb{R}$ and $B=f(A)$ we see this
$f$ is not Lebesgue-Lebesgue measurable. More interesting examples
occur on the real line when there are continuous homeomorphisms
from $\mathbb{R}$ to itself taking Cantor sets of positive measure
to Cantor sets of zero measure.
To return to Nicolo's example. Each surjective linear map
from $\mathbb{R}^m\to\mathbb{R}^n$ is Lebesgue-Lebesgue measurable
as it can be decomposed as a composition of linear bijections
and the projection map $\mathbb{R}^m\to\mathbb{R}^n$ mapping onto
the first $n$ coordinates (both these types of maps can be seen to
be Lebesgue-Lebesgue measurable). By definition, the class
Lebesgue-Lebesgue measurable maps is closed under composition
(unlike the class of Lebesgue-measurable maps!).
