Let $G$ be a Lie group equipped with a Haar measure $\mu$ and $H$ be a closed subgroup of $G$, equipped with a Haar measure $\nu$. We do not assume that there exist a $G$-invariant measure on $H\backslash G$ and therefore there may not be a quotient integral formula.
However, I wonder if it is still possible to prove the following decomposition:
There exists a Borel subset $Y$ of $G$ such that the product map $$H\times Y \to G: (h,y)\mapsto hy$$ is a Borel isomorphism (inverse exists and Borel measurable), such that there exists a Borel measure $\mu_Y$ on $Y$, satisfying $$d\mu(hy)=d\nu(h)d\mu_Y(y).$$
I believe this should be true but don't know how to find such a $Y$.