Translation invariant measures on $\mathbb R$.

Solution 1:

Here is a way to argue out. I will let you fill in the details.

1. If we let $\mu([0,1))=C$, then $\mu([0,1/n)) = C/n$, where $n \in \mathbb{Z}^+$. This follows from additivity and translation invariance.
2. Now prove that if $(b-a) \in \mathbb{Q}^+$, then $\mu([a,b)) = C(b-a)$ using translation invariance and what you obtained from the previous result.
3. Now use the monotonicity of the measure to get lower continuity of the measure for all intervals $[a,b)$.

Hence, $\mu([a,b)) = \mu([0,1]) \times(b-a)$.

Solution 2:

Let $\lambda$ be a translation-invariant measure on the Borel sets that puts positive and finite measure on the right-open unit interval $[0,1)$ then $\lambda$ is a positive multiple of Lebesgue measure. Here is an outline of the proof: Every Borel measure is determined by its behavior on finite intervals. By translation invariance, you know that a right-open interval of length $1/2^n$ has measure $1/2^n \lambda[0,1)$, since $2^n$ such pieces form a disjoint cover over $[0,1)$ and every such piece can be translated into every other other such piece. Now you can approximate every interval by such pieces to pin down the measure of each interval.

Solution 3:

$\mathbb{R}$ is a locally compact group with respect to addition and the translation invariant measures are the Haar measures on this group. A general theorem by Von Neumann states that such a measure is unique up to a multiplicative constant.

Solution 4:

Lebesgue measure and its multiples are not the only measures that are invariant under translations. Take, for example, the counting measure over the integers or over the rationals. Sure, they are infinite, but your question doesn't mention finiteness.