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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.

Carlos Esparza
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1 Answers1

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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 $f(x) = f(0)$.

Yanko
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  • As a minor point you are also assuming that the observable is square integrable. – Alp Uzman Dec 05 '22 at 02:24
  • @AlpUzman It is bounded and measurable, I think this should be enough to imply square integrability. – Yanko Dec 05 '22 at 03:16
  • So the whole business where you consider $K ⊆ G$ compact is so that you can apply Fubini's theorem, right? Don't we need a similar argument for $X$? (since $X \times K$ is not guaranteed to be compact) – Carlos Esparza Dec 06 '22 at 19:22
  • I think the proof that $G \curvearrowright L^2(X)$ is stronly continuous only requires approximating $L^2$ functions by compactly supported continuous functions. I think that this should work in any $σ$-compact space. – Carlos Esparza Dec 06 '22 at 19:40
  • The bigger problem seems to be that this requires $f \in L^2(X)$, as @AlpUzman pointed out. I don't think this is minor because it means that the proof only works for finite measure spaces $X$. – Carlos Esparza Dec 06 '22 at 19:41
  • Ok I get it now, but then I do not really use the fact that $f$ is square integrable (if $g$ is zero a.e. then $\int |g|^2 d\mu = 0$ always). Why is Fubini does not work? The space $X$ is $\sigma$-finite. – Yanko Dec 06 '22 at 21:23
  • @CarlosEsparza actually it seems like you are right and $K$ is not needed here at all, you can just integrate over $G$. – Yanko Dec 06 '22 at 21:25
  • @Yanko, oh right since the integrand is non-negative we don't need any additional hypothese for Fubini's (Tonelli's) theorem. – Carlos Esparza Dec 06 '22 at 23:26
  • @Yanko At some point in your argument you claim that we have a co-null subgroup. The subset in question is co-null, but why is it a subgroup? If it were, then we are done since a co-null subgroup $H$ of $G$ is equal to $G$: Pick $g\in G$, then $gH$ is co-null as well, so $gH\cap H$ is non-empty, so there are $h_1,h_2\in H$ such that $gh_1=h_2$, so $g\in H$. – user446046 Jan 03 '23 at 16:38
  • Ye it is just a set. If $G$ is polish I could improve it to be a group. Let $X_0$ be a co-null set and let $A\subseteq G$ be a co-null set so that $f(x)=f(gx)$ for all $g\in A$ and $x\in X_0$. Then, choose a countable set $a_n\in A$ which is dense in $G$ (here you need that $G$ is separable). Now, $\bigcap_n a_n X_0$ is also co-null and the same goes for all words of elements in $A$. Therefore we can find some $X_1$ co-null and ${a_n,a_n^{-1}}$ invariant. It then follows that $f(x) = f(gx)$ for all $g\in A$ implies $f(x)=f(gx)$ for all $g$ in the group generated by $A$. – Yanko Jan 03 '23 at 23:24