Congruent measurable sets I have a question regarding Congruent relations:
In Euclidean geometry, two subsets of $\mathbb{R}^{d}$ are said to be congruent if one set can be mapped onto the other by translations and rotations.
Claim. Two congruent measurable sets must have the same Lebesgue measure.
I need to prove two things:


*

*If B is a rotation of a rectangle A then $λ^{*}(B) = λ(A)$

*If C is congruent to D then $λ^{*}(C) = λ^{*}(D)$


Edit:
for the first question I can use the fact that if A is a rotation of B then w.l.o.g:
$
-a\leq x_{A}\leq a
$
$
-b\leq y_{A}\leq b
$
and the Lebesgue measure is therefore (2a)*(2b)=4ab
then we can get that:
$
x_{B}=x_{A}cos(\alpha)+y_{A}sin(\alpha)
$
$
y_{B}=y_{A}cos(\alpha)-x_{A}sin(\alpha)
$
then, if we isolate $x_{A}, y_{A}$ from both equations we get that the Lebesgue measure (area of the rectangle) is 4ab.
But, I am having a problem with the second part of the question. is there some theorem that says that each sub-set in $\mathbb{R}^2$ can be written as a union of disjoint rectangles?
can someone help me with this question?
Thanks.
 A: You can use the change of variables theorem and the fact that given J Jacobian matrix of the composition of a rotation and a translation , |det(J)|=1
A: If $D$ is a translation of $C$, it is obvious that $\lambda^*(D) = \lambda^*(C)$. Hence, we only need to consider the case of $D$ being a rotation from $C$, i.e. $D = \{Rx: x \in C\}$ where $R$ is the rotation matrix.
By the definition of $\lambda^*(\cdot)$, and for any $\epsilon > 0$, we can find a disjoint sequence of rectangle cover $C_i$ of $C$, such that
$$ \sum_i \lambda(C_i) \le \lambda^*(C) + \epsilon. $$
Let $D_i = \{ Rx: x \in C_i \}$. Then $D_i$ is a disjoint sequence of rotated rectangle cover of $D$. By monotonicity and subadditivity of $\lambda^*(\cdot)$, 
$$\lambda^*(D) \le \sum_i \lambda^*(D_i) = \sum_i\lambda(C_i) \le \lambda^*(C) + \epsilon,$$
where the equality, $\lambda^*(D_i) = \lambda(C_i)$ , follows from the conclusion of (1).
Hence $\lambda^*(D) \le \lambda^*(C)$. By using inverted matrix $R^T$, we can similarly get $\lambda^*(C) \le \lambda^*(D).$ Hence $\lambda^*(C) = \lambda^*(D)$.
