A Radon measure on $G$ being left-invariant on a dense subgroup $H \subset G$ is a Haar measure on $G$.

Solution 1:

This is Exercise 1.6 in Principles of Harmonic Analysis by Deitmar & Echterhoff.

Let $G$ be a locally compact group, let $H \subset G$ be a dense subgroup, and let $\mu_G$ be a Radon measure on $G$ such that $\mu_G (h A) = \mu_G(A)$ whenever $A \subset G$ is measurable and $h \in H$. Since $H$ is dense in $G$, there exists for all $g \in G$ a net $(h_j)_{j \in J} \subset H$ with $h_j \to g$. The continuity of the map $G \to [0, \infty], \;\; x \mapsto \mu_G (x A)$ implies next that $$ \mu_G (A) - \mu_G (gA) = \mu_G (A) - \lim_j \mu_G (h_j A) = 0, $$ and hence $\mu_G (A) = \mu_G (gA)$ for all $g \in G$. Thus $\mu_G$ is a Haar measure on $G$.