Let $T : \mathbb R^d \to \mathbb R^d$ is a linear invertible map where $d$ is a positive integer. If $E \subseteq \mathbb R^d$ is Lebesgue measurable then $T(E)$ is Lebesgue measurable.
I am not sure how can I prove the proposition above. I can see that $T$ and $T^{-1}$ are comtinuous maps thus $T$ is a homeomorphism. Also I can see that proposition will be satisified for Borel sets but I cannot prove it for Lebesgue measure.
I appreciate any reference, hint etc.
Thanks in advance.
On
It's a fact that for any measurable $E$ there is a $B \subseteq E$ Borel such that $E - B$ has measure 0. As you said, $T$ is a homeomorphism so $T[B]$ is Borel. You now only have to show that $T[E - B]$ has measure 0 to conclude that $T[E] = T[B] \cup T[E - B]$ is a union of a Borel set and a set of measure 0, hence measurable. To show this, you can use the fact that linear maps are Lipschitz (use the norm of $T$ as the Lipschitz constant).
By the way, you can also compute $m(T[E]) = |det(T)| m(E)$. Check this on rectangles like $[0, 1]^d$ first to conclude that.
Let $\Sigma = \{ E| T(E) \text{ is Lebesgue measurable} \}$. It is straightforward to show that $\Sigma$ is a $\sigma$ algebra.
We want to show that $\Sigma$ contains the Lebesgue measurable sets.
Suppose $U$ is open, then since $(T^{-1})^{-1} U = TU$ we see that $TU$ is open and hence $\Sigma$ contains the open sets and hence the Borel sets.
All that remains is to show that $\Sigma$ contains all Lebesgue null sets.
Suppose $N$ is a Lebesgue null set. Since $T$ is Lipschitz it maps Lebesgue null sets to Lebesgue null sets. Hence $T(N)$ is a Lebesgue null set and hence $N \in \Sigma$.