Consider a locally compact Hausdorff $σ$-compact topological space $X$ and a locally compact Hausdorff $σ$-compact topological group $G$ acting continuously and transitively on $X$ such that there exists a $G$-invariant Radon measure $\mu$ on $X$.
Now suppose $f: X \to [0, 1]$ is a Borel-measurable function on $X$ that is a.e. $G$-invariant. That means for every $g \in G$ we have$f(g x) = f(x)$ for $\mu$-almost every $x \in X$.
How do you show (if true) that such an $f$ will be constant outside of a set of measure zero?
I'm trying to use this to show that any $G$-invariant measure on $X$ would be unique up to scale using the approach outlined in this answer.
This answer assumes that $X$ is metric.
By assumptions we have:
$$ f(x) = f(gx) \text { } \mu-\text{a.e.} x$$ $$ \Rightarrow \int_X |f(x)-f(gx)|^2 dx=0 $$ $$\Rightarrow \int_G \int_X |f(x)-f(gx)|^2dxdg=0$$ By Fubini (see e.g. When can a sum and integral be interchanged?) $$\int_X \int_G |f(x)-f(gx)|^2 dg dx=0$$
We therefore deduce that for $\mu$-a.e. $x\in X$ we have for Haar-a.e. $g\in G$ that $f(x) = f(gx)$.
Now a co-null subset(!) must be dense (if the complement contains an open subset it would have positive measure). It is then follows that
Lemma: for $\mu$-a.e. $x\in X$ we have that $f(x)=f(gx)$ for all $g\in G$.
The proof uses the fact that the action defined by $G$ on $L^2(X)$ is strongly continuous. It is surely true for a metrix $X$, but I do not know if it is true in the generality you need. See this answer and the paper linked there. Ergodic action of dense subgroup.
Of course now the proof is complete because for $\mu$-a.e. $x\in X$ we have $f(x) = f(0)$.