This is a question very similar to I recently asked on mathexchange, but different enough to get its own entry in MO.

The setting is still the same. I consider the metric space $\mathbb{R}$ and the Lebesgue-measure $\lambda$ on $\mathbb{R}$ as well as the transitive group action $+:(\mathbb{R},+)\times\mathbb{R}\to\mathbb{R}$, for which the measure and the metric are both invariant. When integrating over a set $M$ with respect to $\lambda$ I can fix a point $x\in\mathbb{R}$ (the metric space rather than the group) and, since the action is transitive, integrate over a set $M'$ with respect to some measure $\mu$ on the additive group $(\mathbb{R},+)$, such that $M'+x=M$ and $μ(M')=λ(M)$. This is possible as well as completely independent of the choice of $x$ by choosing $μ$ to be the Lebesgue-measure on $(\mathbb{R},+)$.

Now I want to generalise this statement to arbitrary metric spaces and group actions. For this I need a radon-measure $\mu$ on $X$ and a locally compact group $G$ with group action $G\times X\to X$ such that the measure and metric are both invariant. Furthermore I require the action to be transitive and continuous.

In order to have the same properties like the example above on the real number line I need a measure $\nu$ on $G$ such that $\nu(U)=\mu(Ux)$ for every measurable $U\subseteq G$ and this should be independent of $x\in X$. When trying to simply construct such a measure I ran into problems because I couldn't guarantee that $Ux$ would be measurable for all $U$, only for compact $U$ (since the action is continuous and the image of $G\to X,g\mapsto gx$ would therefore be compact), but the compact subsets of $G$ do not form a $\sigma$-algebra, as far as I know. Also, I couldn't prove that $Ux$ and $Uy$ would have the same measure, which would make defining such a measure on $G$ impossible.

I now wonder whether the haar-measure on $G$ could actually fulfill this role or at least what kind of conditions I'd need so that the haar measure could work. I would much rather not require the action to be free since this would be too strong. Any ideas are much appreciated!