Let $T:\mathbb{R^2} \to \mathbb{R^2}$ be Lipschitz function.
Then,
(a) If $E$ is a set in $\mathbb{R}^2$ with Lebesgue measure zero, then $T(E)$ has measure zero in $\mathbb{R}^2$.
(B) If $A$ is a measurable set in $\mathbb{R}^2$, then $T(A)$ is also a measurable in $\mathbb{R}^2$.
How I can approach to this problem?
This is true in general for $T:\mathbb R^{d}\to R^{d}$. Let us do this for $d=1$; the extension to $R^{d}$ is more or less obvious. Let $T$ have Lipschitz constant $C>0.$
a). For $\epsilon >0 $ we can find a disjoint union of intervals $\left \{ I_n \right \}_{n}$ such that $E\subseteq \cup^{m}_{n=1} I_n$ and $m(\cup^{m}_{n=1} I_n)=\sum ^{m}_{n=1} m(I_n)<\epsilon$. But then
$m(T(E))\leq m(T(\cup^{m}_{n=1} I_n))=m(\cup^{m}_{n=1} T(I_n))=\sum ^{m}_{n=1} m(T(I_n))\leq C\sum ^{m}_{n=1} m(I_n)=C\epsilon$, so $T(E)$ has measure zero.
b). If $E$ is Lebesgue measureable then its regularity implies that it is the union of a null set $N$ and an $F_{\sigma } $ set $F$. Indeed, $F$ can easily be constructed as a countable union of compact sets.
So by part a) it will suffice to prove that $m(F)$ is measureable. By the comment, write $F=\bigcup^{\infty }_{n=1}K_n$ where each $K_n$ is compact. Now since $T$ is Lipschitz, it is continuous, so $T(K_n)$ is compact, hence closed, which implies that $T(F)=T(\bigcup^{\infty }_{n=1}K_n)=\bigcup^{\infty }_{n=1}T(K_n)$, is an $F_{\sigma} $, being a countable union of closed sets.
