Consider the linear transformation $T(x,y) = (x-y,y)$ on $\mathbb{R^{2n}}$. Prove that that the pre-image of $T$ maps measurable sets to measurable sets.
Its clear that T is linear, invertible, and continuous on $\mathbb{R^{2n}}$, which implies that $T$ and $T^{-1}$ are measurable on $\mathbb{R^{2n}}$.
I need to prove that $T^{-1}(E)$ is measurable for an arbitrary measurable $E$.
My attempt: E is measurable which means it can be written as the union of a measure zero set, $N$, and a borel set, $G$.
So, since $T^{-1}(N\cup G) = T^{-1}(N)\cup T^{-1}(G)$, and $T^{-1}$ measurable implies $T^{-1}(G)$ is measurable, then we only need to prove that $T^{-1}$ maps null sets to measurable sets.
$T^{-1}(N) = \{ (x,y)\in \mathbb{R^n} \vert T(x,y)\in N \}$
$= \{(x,y)\in \mathbb{R^{2n}} \vert (x-y,y)\in N\} $ $= \{(x,y)\in \mathbb{R^{2n}} \vert [(x,y)-(y,0)]\in N\} $ $= \{(x,y)\in \mathbb{R^{2n}} \vert (x,y)\in [N+\{(y,0)\}]\} $
Which is simply equivalent to the set $[N +\{(y,0)\}]$
Since Lebesgue measure is translation invariant, we must have $m(N) = m([N-\{(y,0\}])$, which implies that $T^{-1}$ maps null sets to null sets.
This doesn't seem right, and issues with it?