Any translation invariant Borel measure such that $\mu([0,1))=1$ is identical to the Lebesgue measure

Question

I want to show that if $\mu$ is a Borel measure with $\mu([0,1))=1$ that is a translation invariant, then $\mu$ is the same as the Lebesgue measure.

How can I do that?

What properties does this measure have to have in order to be Lebesgue?

Answer 1

It's hard to answer without further information. A few things come to mind:

• The Lebesgue measure is the unique translation invariant regular Radon measure $\mu$ (translation invariant means s.t. $\mu(A)=\mu(x+A)$ for $x\in\mathbb{R}^n$ and $A\subset\mathbb{R}^n$ measurable) on $\mathbb{R}^n$ such that $\mu([0,1]^n)=1$ (it is the Haar measure on $\mathbb{R}^n$).

• Two measures agree, if they agree on a generating set of the $\sigma$-algebra, i.e. boxes, Euclidean balls, etc. Take your pick.

• More generally, two measures agree if their integrals wrt. to "enough" functions agree (in the previous point, characteristic functions).