I'm studying Lebesgue Measure. I have a problem on proving that Lebesgue Measure in $\Bbb{R}^k$ is invariant under isometries
Here is my work so far. Let $T$ $\mathbb{R}^k \to \mathbb{R}^k$ is an isometries We know that if A is Lebesgue measurable then T(A) is an isometries
Let $A \subset \mathbb{R}^k$ is measurable Lebesgue then $\forall \epsilon >0, \exists (\Delta^n)_n \subset \mathbb{R}^k , A \subset (\Delta^n)_n$ and $\mu(A)+\epsilon > \displaystyle \sum_{n=1}^{\infty} |\Delta^n| = \displaystyle \sum_{n=1}^{\infty} |T(\Delta^n)| >\mu(T(A)) $
After that, we need to prove that $\mu(T(A)) \ge \mu (A)$.
Let's choose $T'=T^{-1}$, $A'=T(A)$ we got $\mu(T(A)) \ge \mu (A)$.
The problem is $\displaystyle \sum_{n=1}^{\infty} |T(\Delta^n)| >\mu(T(A))$ is true if we got $T(\Delta^n)$ are rectangle. But if T is rotation, $T(\Delta^n)$ aren't rectangles so I am stuck here. Any help is appreciated.