I wish to show that any Borel measure which is finite over compact sets and is invariant under Euclidean isometrics, is in fact Lebesgue's measure multiplied by a positive factor. In other words: $\forall E \in \mathcal{B_{\mathbb{R}}} : \mu(E) = c\cdot \lambda(E)$, when $\lambda$ is Lebesgue's measure.
I proved that for $c = \mu ([0,1])$ and for any open set in $\mathbb{R}$. Now defining $\mathcal{E} = \{E : \mu(E) = \mu([0,1])\cdot \lambda(E) \}$ I wish to demonstrate that $\mathcal{E}$ is a $\sigma$ algebra and then $\mathcal{B}_{\mathbb{R}}\subseteq\mathcal{E}$.
I don't succeed to show closure under complement, and I would like some help with that.
For non-emptiness and closure under countable unions:
- $\mathcal{E}$ is not empty , because it contains all open sets.
- $\{E_n\}_1^\infty$ a disjoint collection in $\mathcal{E}$ : $\mu(\biguplus E_n) = \sum_1^\infty\mu(E_n)=\sum_1^\infty \mu([0,1])\cdot \lambda(E_n) = \mu([0,1])\cdot \lambda(\biguplus E_n)$. Thus $\mathcal{E}$ is closed to countable unions (I am using here a lemma which promises it's enough to demonstrate it for disjoint collection).