Radon-Nikodym derivative in a compact Hausdorff space

Question

Let $X$ be a compact Hausdorff space where $X$ have infinitely many points and the topology is non-discrete, $m$ be a regular probability measure defined on the Borel $\sigma$-algebra of $X$, and $g$ an homeomorphism of $X$. Suppose that the image measure $g_{\ast}m$ (defined by $g_{\ast}m(O) = m\big[g^{-1}(O)\big]$ for each Borel subset $O$) is equivalent to $m$. Then, for each $F\in C(X)$, I will have:

$$ \begin{aligned} &\hspace{0.6cm} \int_X F(x)\frac{d\,g_{\ast}m}{d\,m}(x)\,dm(x) = \int_X F(x)\,d(g_{\ast}m)(x)\\ & = \int_X F(x)\, dm(g^{-1}(x)) = \int_X F\circ g(x)\,dm(x) \end{aligned} $$

Define $\phi_m$ to be the positive linear functional defined by $m$ (and defined on $C(X)$) and $\phi_{g_{\ast}m}$ similarly, $I = \{F\in C(X): \phi_m(\vert\,f\,\vert)=0\}$ and $I_g = \{F\in C(X): \phi_{g_{\ast}m}(\vert\,f\,\vert) = 0\}$. Then the image of the following mapping:

$$ M: I_g\rightarrow I,\hspace{0.3cm} F(x) \mapsto F(x)\frac{d\,g_{\ast}m}{d\,m}(x) $$ is equal to the image of the action (restrcited on $I_g$) defined by $g\bullet F(x) = F\circ g(x)$. Obviously both mappings are invertible and the action is a isometric isomorphism between $I_g$ and $I$. My questions are:

1. Since the action $F\mapsto g\bullet F$ is multiplicative, $M$ will never coincide with the action unless $g$ is the identity. In this case what can we tell about the local range of $(d\,g_{\ast}m/d\,m)$?
2. Is it true that, when $X$ has non-empty homeomorphism group, for each regular probability measure $m$, there exists a non-identity element $g\in\operatorname{Homeo}(X)$ (the homeomorphism group of $X$) such that $m$ is equivalent to $g_{\ast}m$?
3. Given a non-trivial countable subgroup $G\subseteq\operatorname{Homeo}(X)$, will there exist a regular non-atomic probability measure $m$ such that $g_{\ast}m\sim m$ for each $g\in G$?

Update: Let $m$ be a probability regular non-atomic measure and assume $g\in\operatorname{Homeo}(X)$ satisfies $m\sim gm$. According to this post, we can find a compact set $C_{\epsilon}$ such that both $m(C_{\epsilon})$ and $gm(C_{\epsilon})$ are less than $\epsilon$. I tried to use this and the Urysohn's Lemma to decide the local behavior of the derivative but failed. Any hints will be appreciated.

Update2: From the Corollary 6.2.2 in Measures on topological space written by V.I.Bogachev (link), for a fixed $g\in\operatorname{Homeo}(X)$, one can find a Radon probability measure $m$ (and hence regular when $X$ is compact) such that $g_{\ast}m = m$. In such case, compared to my original question, I suppose it would not be too greedy to ask, given $m$ a probability measure, is it possible to find a subgroup $G\subseteq\operatorname{Homeo}(X)$ such that $g_{\ast}m\sim m$ for each $g\in G$.

Answer 1

I think the answer for 2 is no. I will state some counter-examples below, after some partial results.

A Necessary condition involving the support

From now on, all measures are Borel probability measures. For every measure $\mu$, define its support $$\mathrm{Supp}(\mu) := \overline{\{ x : \forall V \text{ open with } x \in V, \mu(V)>0 \}} .$$

We have $\mathrm{Supp}(g_*\mu) = g(\mathrm{Supp}(\mu))$. Indeed, let $x$ be in $\mathrm{Supp}(g_*\mu)$, then for every $V \ni x$ open, $0<g_*\mu(V) = \mu(g^{-1}(V))$. From there we can see that $g^{-1}(x) \in \mathrm{Supp}(\mu)$.

An immediate consequence of the definition is that two equivalent measures have the same support. This gives us the following Lemma:

Lemma: For every homeomorphism $g$, if $\mu$ and $g_*\mu$ are equivalent, then $g(\mathrm{Supp}(\mu)) = \mathrm{Supp}(\mu)$.

Partial answer for measures supported on a strict subspace

Proposition 1 (positive answer to 2.): If for every open set $V \notin \{\emptyset, X\}$, there is a homeomorphism $g$ of $X$ such that $g(V) = V$ and $g|_V \neq id$, then for every $\mu$ with $ \mathrm{Supp}(\mu) \neq X$ we have a non-identity $g$ such that $\mu$ and $g_*\mu$ are equivalent.

Indeed, fix $V = X \setminus \mathrm{Supp}(\mu)$, and take $g$ given by the assumption. The assumption of Proposition 1 is true on many obvious examples. Note the condition that $\mathrm{Supp}(\mu)\neq X$. See the conjecture below that relates to the case where we do not have this condition. The next proposition provides a partial converse to Proposition 1.

Proposition 2 (negative answer to 2.): Assume that $X$ is separable. If there exists an open set $V \notin \{\emptyset, X\}$ such that the only homeomorphism $g$ with $g(V)=V$ is the identity, then there is a measure $\mu$ such that the only homeomorphism $h$ with such that $\mu$ and $h_*\mu$ are equivalent is the identity.

We use the separability of $X$ (more precisely of $V^c$) to explicitely construct $\mu$. Maybe we can still construct $\mu$ under weaker conditions, or conversely find some non-separable $X$ for which Proposition 2 fails. A look at examples of non-separable compact spaces (which are necessarily not metrizable) might help.

proof: The challenge consists in finding a suitable $\mu$. Since $X$ is separable, so is the compact space $V^c$: let $(v_n)_{n\geq 1}$ be sequence of $V^c$ that is dense in $V^c$, and $\mu = \sum_{n\geq 1} 2^{-n} \delta_{v_n}$. We can check that $\mu$ is a regular Borel measure with support $V^c$. Any homeomorphism $g$ such that $\mu$ and $g_*\mu$ are equivalent must have $V^c = g(V^c) \iff V = g(V)$, thus are the identity. $\square$

Examples of $X$ satisfying the assumption of Proposition 2 include spaces with an isolated point. Here is a somewhat complicated example without isolated point. We fix a numbering $(q_n)_{n\geq 1}$ of $\mathbb Q \cap (0,1)$, and construct a space $T$ as the limit of the following inductive construction --- which should remind you of Koch’s snowflake:

• start with $T = [0,1]$, and "distinguish" the sequence $(q_n)$.

• For every $n\geq 1$, glue $2n$ copies of $[0,2^{-n}]$ (with distinguished vertices $(2^{-n} q_k)_{k\geq 1}$) to the point $q_n$, by identifying the zeroes of the copies with $q_n$. This way, the $q_n$ all have non-homeomorphic neighborhoods ($q_n$’s neighborhood looks like a $2n+2$-pointed star), so any homeomorphism of $T$ must send $q_n$ to $q_n$ for every $n$. This ensures that the homeomorphisms of $T$ are the identity on the original copy of $[0,1]$.

• Iterate this process on the glued copies, making sure that the added copies are small enough so that everything converges and that no distinguished vertex ever has the same number of branches than some other distinguished vertex.

You should obtain $T$ that is a compact tree with infinitely many branches, with a countable number of branching points (the distinguished vertices) that are dense in $T$, such that the only homeomorphism of $T$ is the identity.

Now build $X$ by taking two copies $T_1$ and $T_2$ of $T$ and identifying the zero of $T_1$ with the one of $T_2$ and the one of $T_1$ with the zero of $T_2$. There are two homeomorphisms of $X$: the identity, and the exchange sending $T_1$ onto $T_2$. But once we fix any measure with support included in $T_1$, the only homeomorphism $g$ that ensures that $\mu$ and $g_*\mu$ are equivalent is the identity.

Conjectured negative example with full support

Finally, here is a conjecture that, if true, provides a "natural" X and a rather natural $\mu$ with support $X$ for which 2. fails.

Conjecture: Let $X=[0,1]$ equipped with the usual topology. The only homeomorphism that leaves $K = (\mathbb Q \cap [0,1]) \cup \{e^q, q\in \mathbb Q \cap (-\infty, 0]\}$ is the identity.

Note that $X$ has a non-trivial homeomorphism group. $K$ is countable; let $(k_n)_{n\geq 1}$ be an enumeration of $K$, and define $\mu = \sum_{n\geq 1} 2^{-n} \delta_{k_n}$ --- we can check that it is Borel and regular. For every homeomorphism $g$ of $X$, $\mu$ and $g_*\mu$ are equivalent if and only if $g(K) = K$. Then assuming this conjecture, the only homeomorphism with $\mu$ and $g_*\mu$ equivalent is the identity, providing a negative answer to 2. even for a "nice" $X$.