If a measure is absolutely continuous with respect to the Lebesgue measure, invariant under orthogonal transformations and translations then this measure is a multiple of the Lebesgue measure.
- Anybody can ask a question
- Anybody can answer
- The best answers are voted up and rise to the top
All you need is translation invariance, because Lebesgue measure is Haar Measure.
Let $m$ be the measure given by your measure to the unit cube Then translation invariance allows us to compute the measure of all dyadic cubes —those whose faces are parallel to the coordinate planes and vertices with rational coordinates of the form $n2^m$ with $n$, $m\in\mathbb Z$.
In particular, the area of all these cubes is $m$ times their Lebesgue measure.
Since the completion of the $\sigma$-algebra generated by the dyadic cubes is the Lebesgue $\sigma$-algebra, and your measure agrees (up to a scalar) with the Lebesgue measure on this generating set, we obtain the equality you want by the uniqueness of the procedure of extension of measures —see Halmos's or any other good measure theory book.