If $A$ is measurable, is $TA, T\in\text{End}(\mathbb{R}^n)$ measurable? Let us define, as Kolmogorov-Fomin's Элементы теории функций и функционального анализа does, the definition of outer measure of a bounded set $A\subset  \mathbb{R}^n$ as $$\mu^{\ast}(A):=\inf_{A\subset \bigcup_k P_k}\sum_k m(P_k)$$where the infimum is extended to all the possible covers of $A$ by finite or countable families of $n$-paralleliped $P_k=\prod_{i=1}^n I_i$, where $I_i\subset\mathbb{R}$ are finite intervals, whose measure $m(P_k)$ is the product of the length of the intervals $I_i$.
A set is said to be elementary when it is the union of a finite number of such $n$-parallelipeds and a set $A$ is said to be measurable if, for any $\varepsilon>0$, there is an elementary set $B$ such that $$\mu^{\ast}(A\triangle B)<\varepsilon.$$The function $\mu^{\ast}$ defined on measurable sets only is called Lebesgue measure and the notation $\mu$ is used for it.
Let us come to the question. I have been able to understand, by following F. Jones's Lebesgue Integration on Euclidean Space, that for any open subset $G\subset\mathbb{R}^n$ and any linear transformation $T\in\text{End}(\mathbb{R}^n)$ $$\mu^{\ast}(TG)=|\det(T)|\mu^{\ast}(G)$$. By taking the outer regularity described here into account, I think that the one-to-one correspondence between open sets containing $A$ and open sets containing $TA$ if $T$ is invertible (while $TA$ belongs to a subspace of dimension $<n$ if $T$ is singular), can allow us to generalise $$\mu^{\ast}(TA)=|\det(T)|\mu^{\ast}(A)$$ for any subset $A\subset\mathbb{R}^n$.
What I cannot derive is: if $A$ is measurable (definition above), is $TA$ measurable, and, if it is, how can it be proved? I thank any answerer very much!
 A: Let us try to do this without Fubini, as you request:
If $T:\mathbb R^{d}\to \mathbb R^{d}$, we will also denote the matrix corresponding to $T$ by $T$.
Withut loss of generality, we may assume that $T$ has the form $T(x_1,x_2,\cdots, x_d)=(x_1+x_2,x_2,\cdots x_d)$, from which it is easy to see that $\det T=1$.
We will use the following facts:
Lebesgue measure is translation-invariant:
$\tag1 m(E+a)=m(E)\  \text {for all}\  E\in \mathscr M, a\in \mathbb R.$
Every open set is a disoint union of cubes:
$\tag2 U=\bigcup^{n}_{i=1}Q_i \ \text {where $U$ is open and $Q_i$ are cubes in $\mathbb R^{d}$ with disjoint interiors}.$
The Lebesgue measure of the boundary of a cube is equal to $0$:
$\tag3 \text { m$(\partial Q_i)=0$}$.
We do the proof in steps:
$a)$: If $Q=I_1\times I_2\times ,\cdots ,\times I_d$ is a cube in $\mathbb R^{d}$, then using the definition of Lebesgue measure on elementary sets, together with $(1)$ above, we get
$m(T(Q))=m(I_1+x^{1}_{2})m(I_2+x^{2}_{2})\cdots m(I_d+x^{d}_{2})=m(I_1)m(I_2)\cdots m(I_d)=m(Q)$, so the result is true for cubes.
$(b)$: Now we use $(2)$ and $(3)$ to obtain the result on any open set $U$:
$m(T(U))=m(T(\bigcup^{n}_{i=1}Q_i))=\sum_{i=1}^{n}m(T(Q_i))=\sum_{i=1}^{n}m(Q_i)=m(\bigcup^{n}_{i=1}Q_i)=m(U)$.
Replacing $T$ by $T^{-1}$, we now have 
$\tag4 m(T^{-1})(U)=m(U)$ for any open set $U\subset \mathbb R^{d}$.
$(c)$: Now let $B$ be a Borel set. Then for each $\epsilon >0$, there is an open set $U$ such that $B\subseteq U$ and $m(B)+\epsilon>m(U)$, from which it follows that
 $m(T(B))\leq m(T(U))=m(U)<m(B)+\epsilon$ so $m(T(B))\leq m(B)$.
On the other hand, there is an open set $V$ such that $T(B)\subseteq V$ and $m(T(B))+\epsilon>m(V)$, from which it follows, using $(4)$, that $m(B)\leq m(T^{-1}(V))=m(V)<m(T(B))+\epsilon$, so $m(B)\leq m(T(B))$
Thus our claim is true for all Borel sets.
$(d)$: Finally. let $E\in \mathscr M^{d}$. Then $E=N\cup F$ where $F$ is Borel and $N$ is null, disjoint from $F$. This fact follows because Lebesgue measure is regular. Indeed, $F$ is readily constructed as a countable union of compact sets. Now, it is an easy exercise to show that $m(T(N))=0$. Then, $m(T(E))=m(T(N\cup F)=m(T(N))+m(T(F))=m(T(F))=m(F),$ and we are done.
