A Radon measure on $G$ being left-invariant on a dense subgroup $H \subset G$ is a Haar measure on $G$. Let $G$ be a locally compact group, $H$ a dense subgroup, and $μ$ a Radon
measure on $G$ such that $μ(hA) = μ(A)$ holds for every measurable set $A ⊂ G$ and
every $h ∈ H$. Show that $μ$ is a (left-invariant) Haar measure.
Any help with this is appreciated !
 A: 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 $G$.

A: The idea here is to show that $\mu_G(A)=\sup\{\mu_G(K)|K\subseteq A, compact\}$. This helps us because it reduces the problem to showing that $\mu_G(gK)=\mu_G(K)$ holds for all $K\subseteq G$ compact.
First of all note that $\mu_G(K)<\infty$ holds. To see this use that $\mu_G$ is locally finite and therefore there exists a closed unit neighborhood $U$ of finite measure. Then the density of $H$ in $G$ implies that $K\subseteq \bigcup_{h\in H} hU$. Since this is an open covering, one obtains $h_1,\dots,h_n\in H$ such that $K\subseteq\bigcup_{i=1}^n h_iU$ and thus
\begin{align*}
\mu_G(K)\leq\mu_G\left(\bigcup_{i=1}^n h_iU\right)\leq \sum_{i=1}^n \mu_G(h_iU)=n\mu_G(U)<\infty.
\end{align*}
Next we want to show that $g\mapsto \mu_G(K)$ is continuous for every $K\subseteq G$ compact. Therefore for fixed $g_0\in G$ and a compact neighborhood $V$ of $g_0$, it follows that $\textbf{1}_{gK}\leq\textbf{1}_{VK}$ and thus
\begin{align*}
V\ni g\mapsto\mu_G(gK)=\int_{G}\textbf{1}_{gK}(x)d\mu_G(x)
\end{align*}
is continuous by Lebesgue's theorem on dominated convergence as $VK$ is compact and thus $\mathbf{1}_{VK}$ integrable.
Since $g\mapsto\mu_G(gK)$ is continuous in every neighborhood the assertion follows from the density of $H$ in $G$.
