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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.

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All you need is translation invariance, because Lebesgue measure is Haar Measure.

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Could you explain that better? What you mean by Haar Measure? –  checkmath Mar 7 '12 at 2:43
    
Every Lie group admits a unique (up to scaling) left-invariant measure. This measure is called a Haar measure. (c.f. en.wikipedia.org/wiki/Haar_measure) –  Neal Mar 7 '12 at 2:46
    
If you have a locally compact topological group, there is a translation-invariant measure on it that is unique up to scale. –  ncmathsadist Mar 7 '12 at 2:47
    
en.wikipedia.org/wiki/Haar_measure see this. –  ncmathsadist Mar 7 '12 at 2:47
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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.

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