Lebesgue measure as Haar measure

Example

Let $n \ge 1$ and let $G = (\R^n, +)$, a locally compact abelian group with the usual topology. Let $\lambda$ be Lebesgue measure restricted to the Borel sets of $\R^n$. Then $\lambda$ is a Haar measure of $G$.

Check of the three conditions. It is a Radon measure: compact sets are bounded, hence have finite measure, and Lebesgue measure is outer regular and inner regular by compact sets. It is nonzero: the unit cube $[0, 1]^n$ has measure $1$. It is invariant: for $x \in \R^n$ and a box $B = \prod_i [a_i, b_i]$, the translate $x + B = \prod_i [a_i + x_i, b_i + x_i]$ has the same volume $\prod_i (b_i - a_i)$. Boxes are closed under finite intersections, generate the Borel sets, and countably many of them cover $\R^n$. So the Borel measures $A \mapsto \lambda(x + A)$ and $\lambda$, equal and finite on boxes, agree on all Borel sets: $\lambda(x + A) = \lambda(A)$ for every Borel set $A$.

By the existence and uniqueness of Haar measure, every nonzero translation-invariant Radon measure on $\R^n$ is $c\lambda$ for some $c > 0$. So Lebesgue measure is characterized, up to a constant, by invariance under translations.