I am trying to prove non-ergodicity of a certain map, and it boils down to the following.
Suppose we have a compact connected Lie group $G$ with Haar measure $\mu$, and suppose that $G$ is non-abelian. Consider a non-empty abelian Lie subgroup of $G$, let's call it $H$. Can we conclude that $0<\mu(H)<\mu(G)$?
Apologies if this is trivial, this is quite a journey from my field which is why I am consulting the MSE experts :)
Edit: Is $H$ even Borel (or open) in the first place? The $H$ I have in mind is the group generated by some $g\in G$ which is not the identity. It is not clear to me!
Any proper Lie subgroup $H$ of a connected Lie group $G$ has measure $0$. First of all, $H$ is indeed Borel in $G$, since every compact subset of $H$ is closed in $G$ and so since $H$ is $\sigma$-compact it is $F_\sigma$ in $G$.
Now since $G$ is connected, we must have $\dim H<\dim G$. Covering $H$ by compact coordinate charts, we then see that $H$ is a countable union of compact subsets of $G$ with empty interior, and so is meager in $G$. Since every coset of $H$ is also meager in $G$, it follows that $H$ must have uncountable index in $G$ (otherwise $G$ would be a countable union of cosets of $H$, contradicting the Baire category theorem). Since all the cosets of $H$ must have the same measure, this implies they must all have measure $0$ (otherwise the measure on $G$ would not be $\sigma$-finite).
(There are other arguments you can give as well; for instance, you can say that each compact coordinate chart in $H$ is a smoothly embedded submanifold of $G$ of positive codimension and so must have measure $0$ because Haar measure on $G$ has the same null sets as Lebesgue measure in any coordinate chart.)