About the measurable subsets and the Lipschitz condition I have, again, a doubt with the measurable subsets.
If I have that $T\colon\mathbb{R}^n\longrightarrow \mathbb{R}^n$ is Lipschitz, does $T$ send Lebesgue measurable sets in Lebesgue measurable sets. If it is, if $T$ is linear, how can I compute or have a formula to $|T(E)|_n$?
 A: Let $\Omega = \{E | T(E) \text{ is measurable} \}$.
We note that $\mathbb{R}^n$ is $\sigma$-compact. We will use the $\|\cdot\|_\infty$ norm on $\mathbb{R}^n$.
If $K$ is compact, then $T(K)$ is compact since $T$ is continuous. Hence $\Omega$ contains all the compact sets. Since $T(\cup_n A_n) = \cup_n T(A_n)$, $\Omega$ also contains all closed and open sets (since any open set in $\mathbb{R}^n$ is the union of a countable collection of compact sets). Hence $\Sigma$ contains all the $F_\sigma$ sets (countable unions of closed sets). 
Suppose $y \in B(x,r)$. Since $T$ is Lipschitz, we have $T(y) \in B(T(x),Lr)$, hence $T(B(x,r)) \subset B(T(x),Lr)$, and so $m(T(B(x,r))) \leq L^n m(B(T(x),r))= L^n m(B(x,r))$ (the latter equality from translation invariance). Note that the same result is true for the closed ball.
Let $Q = [0,1)^n$ and note that $Q \in \Omega$ (it is a $F_\sigma$). Similarly scalings and translations of $Q$ are also in $\Omega$. Let ${\cal Q}_1 = \{ \{z\}+Q\}_{z \in \mathbb{Z}^n}$, and ${\cal Q}_k = \frac{1}{2^k} {\cal Q}_1$.  Define $U_1 = \cup \{R \in {\cal Q}_1 | R \subset U\}$, and define $U_{k+1} = U_k\cup \{R \in {\cal Q}_{k+1} | R \subset U \setminus U_k \}$. Note that $U = \cup_k U_k$, and hence $U$ can be written as $U = \cup_k Q_k$ a countable, disjoint union of members of ${\cal Q}_1,{\cal Q}_2,...$. Now note that $B((\frac{1}{2},...,\frac{1}{2}), \frac{1}{2}) \subset Q \subset \overline{B((\frac{1}{2},...,\frac{1}{2}), \frac{1}{2})}$, and hence we have $m(T(Q)) \leq L^n m(Q)$. A similar argument shows that the same relationship holds for each $Q_k$. Hence we have $m(U) = \sum_k m(Q_k)$, and $T(U) = \cup_k T(Q_k)$, from which we have $m(T(U)) = m(\cup_k T(Q_k)) \leq \sum m(T(Q_k))\leq L^n \sum_k m(Q_k) = L^n m(U)$.
Now suppose $Z$ has measure zero. Since $m$ is outer regular, we can find a collection of open sets $U_k$ such that $Z \subset U_{k+1} \subset U_k$, and $m(U_k) < \frac{1}{k}$. We have $m(T(U_k)) \leq L^n m(U_k) < L^n\frac{1}{k}$, and $T(Z) \subset T(U_k)$ for all $k$. Hence $T(Z) \subset \cap_k T(U_k)$, and we have $m(\cap_k T(U_k)) = 0$. Since $m$ is complete, it follows that $Z$ is measurable, and hence $m(Z) = 0$.
Finally, suppose $E$ is a measurable set. Since $m$ is inner regular, we have $E = F \cup N$, where $F$ is a $F_\sigma$ and $N$ has measure zero. Since $T(E) = T(F) \cup T(N)$, we have that $E \in \Omega$. Hence $\Omega$ contains all the measurable sets.
