Here $G$ is a locally compact second countable topological group with left haar measure $\mu$, and $\varGamma$ is a discrete subgroup with a borel subset $\varOmega \subseteq G$ s.t. $G=\biguplus_{\gamma\in\varGamma}\varOmega\gamma$.
Claim: If $ \mu (\varOmega)<\infty $ then there exists a $G$ invariant finite non-trivial radon measure $\nu$ on $G/\varGamma$
Remarks on what I managed:
My main issue is that it is NOT assumed that the ambient group is unimodular and infact we don't assume anything on the modular function.
Ideally I would hope that $\nu(A)=\mu(\pi^{-1}(A)\cap\Omega)$ for borel $A\subseteq G/\varGamma$ is left invariant. If this measure $\nu$ is left invariant without assuming the condition on the modular function, then we're done. Is this the case? I did not manage to show this.
[Though I did manage to show that it is true in case we know in addition that $\triangle_G|_\varGamma=1$. But I cannot assume this.]
In general, homogeneous spaces do not necessarily admit an invariant measure, and in our case they will admit a semi invariant measure with character $\triangle_G$ by some famous criterion. But since the fundamental domain is finite measure, then I would like to believe that it will have an invariant one, and many documents claim this without proof.
Extra failed attempt:
In the original claim, I have also managed to show that $\nu(A)=\lambda(\pi^{-1}(A)\cap\Omega)$ where $\lambda(B)=\mu(B^{-1}$) is a left semi-invariant measure, and that in this case $\nu (G/\varGamma)=\mu(\varOmega^{-1})$.
If I knew $\mu(\varOmega^{-1})<\infty$ then the semi-invariance is actually an invariance, because $\nu (G/\varGamma)=\nu (gG/\varGamma)=u(g)\nu (G/\varGamma)$ so $u=1$ and thus $\nu$ is left invariant. But that assumption may not be true, as inverse map in general does not necessarily preserve finite measure sets.
