I'm studying Measure Theory. After reading my teacher's lecture notes, it is not clear for me the statement:
"In $\Bbb{R}^k$, It's impossible to define a $\sigma$ finite measure the satisfies all the following conditions:
- $m(E)$ $\in$ [$0$,$\infty$] for every E (E is a subsets of $\Bbb{R}^K$)
- $m$ is invariant under isometries
- $m(R)=|R|$ (for all rectangles R in $\Bbb{R}^K$)
I don't know how to prove this statement :( Any help is appreciated
There's an easy way to prove this without appealing to Banach-Tarski. It relies on the following two facts:
The Lebesgue measure is the unique measure on the Borel $\sigma$-algebra $\mathcal{B}(\mathbb R^k)$ of $\mathbb R^k$ satisfying the latter two properties.
The Lebesgue measure does not extend to a complete $\sigma$-algebra on the power set $\mathcal P(\mathbb R^k)$ that is translation invariant.
The first property is a standard measure theory result, and can be proved using the Dynkin $\pi$-$\lambda$ lemma applied to successively larger balls. For the latter note that the Vitali set cannot be measurable which gives the case $k=1$ (the non-measurability proof provided on the wiki page will give you a contradiction). For general $k,$ one can apply the same argument to the product $V \times [0,1]^{k-1}$ where $V$ is the Vitali set.