# The invariant measure (reading / replicating parts of the paper by Levitt)

## From K to L

• $$\mathcal{K} = \{\phi_i :A_i \to B_i \}_{i=1, … , k}$$ be a non-nesting closed system on a finite tree with at least one infinite orbit.
• Non nesting forces a special structure for the set of finite regular $$\mathcal{K}$$ – orbits
1. Union of all finite regular orbits = a finite union of open intervals
2. Its complement $$K_1$$ is a finite union of closed subtrees, not all of them points.
• Disregard isolated points of $$K_1$$
• D = disjoint union of all closed edges of $$K_1$$
• D is a multi – interval (a finite disjoint union of compact intervals)
• $$\delta D$$ = set of endpoints of compact intervals in D.
• int D = $$D \backslash \delta D$$
• The system of maps $$\mathcal{K}$$ naturally induces a system $$\mathcal{L}$$ on D
• Replace each $$\phi_i$$ by the collection of its restrictions to edges of $$K_i$$, keeping only those maps whose domains contains more than a single point.
• We may split domains at preimages of vertices of $$B_i$$
• The system $$\mathcal{L}$$ is closed and non-nesting
• Every regular orbit of $$\mathcal{L}$$ is infinite
• Every $$\mathcal{L}$$ invariant probability measure without atoms corresponds to a $$\mathcal{K}$$ invariant probability measure (supported on $$K_1$$ )
• Hence it is sufficient to find an $$\mathcal{L}$$ invariant measure

## From L to M

• Let $$\overset{o}{\mathcal{L}}$$ be the open system obtained by replacing each element $$\phi : A \to B$$ of $$\mathcal{L}$$ to the open interval A \ {endpoints}.
• Hardest case: every minimal closed(?) $$\overset{o}{\mathcal{L}}$$ invariant set is a finite singular orbit (we need dense orbits).
• Why is this hardest? What is singular orbit?
• We want every orbit to be dense (and the system to be open) to apply Sacksteders theorem.
• In the worst case, none-of the orbits are dense. Since they are infinite, there must be accumulation points (every infinite sequence has a convergent subsequence). These accumulation points could be inside or at the end point
• If we look at the accumulation points they form a minimal invariant set. If they are finite and singular, it is hardest to get and infinite dense orbit out of them.
• There are finitely many minimal sets, among which are the endpoints of $$d \in \delta D$$
• By splitting D (thus changing $$\mathcal{L}$$ and $$\overset{o}{\mathcal{L}}$$, we may assume that the points of $$\delta D$$ are the only minimal sets of $$\overset{o}{\mathcal{L}}$$
• Note that the closure of every $$\overset{o}{\mathcal{L}}$$-orbit contains an end point of D.
• For every $$d \in \delta D$$, we choose an $$\mathcal{L}$$ word $$w_d$$ defined on a non – degenerate closed interval $$I_d$$ containing d and sending d to a point $$\tilde {d} \in int D$$
• Such a word exists because otherwise non-nesting of $$\mathcal{L}$$ would imply finiteness of orbits near d.
• $$E = \{ p \in \delta D\}$$ such that some $$\overset{o}{\mathcal {L}}$$ – orbit accumulates on p.
• This implies we can get points of some orbit arbitrarily close to p.
• Choose $$x_p , y_p$$ from these points (see next).
• For each $$p \in E$$ choose $$x_p, y_p \in I_p$$ such that some $$\overset{o}{\mathcal {L}}$$ – word $$\tau_p$$ sends $$x_p$$ to $$y_p$$ in an orientation preserving way.
• Since every $$\overset{o}{\mathcal {L}}$$ – orbit accumulates on E, we may require, furthermore, that for every $$d \in \delta D$$ the orbit of $$\tilde{d}$$ meets the union of all intervals $$(x_p , y_p ), p \in E$$.
• Notice that previously we found every end point d went to some interior point $$\tilde {d}$$
• Look at the orbit of $$\tilde {d}$$. Clearly it accumulates at some end point $$p \in E$$
• There are finitely many d (hence $$\tilde {d}$$ ). Thus for these we may arrange $$(x_p , y_p ), p \in E$$ such that orbit of d intersects the union of $$(x_p , y_p ), p \in E$$
• After changing $$w_d$$, we may assume that every $$\tilde {d}$$ belongs to some $$(x_p , y_p ), p \in E$$.
• Note that every $$\overset{o}{\mathcal {L}}$$ orbit in int D meets some $$(x_p , y_p )$$ infinitely often (why?)
• For $$p \in E$$, $$C_p$$ = circle obtained by identifying the endpoints of $$[x_p , y_p]$$.
• C = disjoint union of these circles.

Goal: Construct a non-nesting system of maps $$\mathcal{M}$$ on C with every orbit infinite, in such a way that an $$\mathcal{M}$$ -invariant measure on C provides an $$\mathcal{L}$$ invariant measure on D

• Construct a finite family $$\mathcal {P}$$ of homeomorphisms $$\lambda : U \to V$$ between open intervals $$U \subset int D$$ and intervals $$V \subset C$$. It will have three types of maps:
1. For each $$p \in E$$, include the natural map $$\pi_p : (x_p , y_p) \to C_p$$
2. Then for each p, we consider the points $$(x_p , y_p )$$ and their common image $$z_p \in C$$. Restrict $$\tau_p$$ to a homeomorphism between a small neighborhood $$X_p$$ of $$x_p$$ and a neighborhood $$Y_p$$ of $$y_p$$. Since $$\tau_p$$ preserves orientation, both these neighborhoods are naturally homeomorphic to a neighborhood $$Z_p$$ of $$z_p$$ in $$C_p$$. We include the maps $$X_p \to Z_p$$ and $$Y_p \to Z_p$$
3. The ranges of the maps constructed so far cover C. If domains cover int D then we may stop. Otherwise, we need a third type of maps. Let $$x \in int D$$. Some $$\overset{o}{\mathcal {L}}$$ word $$\alpha_x$$ sends a neighborhood $$U_x$$ of x to an open interval $$V_x$$ contained in some $$(x_p , y_p)$$. The word $$\alpha_x$$ may be chosen to be constant near the end point of D. But compactness (?), we deduce that int D may be covered by finitely many $$U_x$$. We include the corresponding finite set of maps from $$U_x \to \pi_p (V_x) \subset C_p$$
• This defines the family $$\mathcal {P}$$. We now use it to carry $$\overset{o}{\mathcal {L}}$$ over to a system $$\mathcal {M}$$ on C.
• Let $$\gamma_1 : U_1 \to V_1$$ and $$\gamma_2 : U_2 \to V_2$$ be two elements of $$\mathcal{P}$$. If $$U_1 \cap U_2 \neq \phi$$, then $$\gamma_2 \gamma_1^{-1}$$ is a homeomorphism between two subintervals of C. We include it in $$\mathcal{M}$$.
• Similarly, for $$\theta \in \overset{o}{\mathcal {L}}$$ , we include $$\gamma_2 \theta \gamma_1^{-1}$$ is its domain in non-empty.
• $$\mathcal {M}$$ = set of all maps thus obtained, for all possible choices of $$\gamma_1, \gamma_2 \in \mathcal {P}$$ and $$\theta \in \overset{o}{\mathcal {L}}$$ .
• It is an open system of maps in C.
• The construction of $$\mathcal {P}$$ and $$\mathcal{M}$$ was done in such a way that the following properties hold:
• Given $$\gamma_1, \gamma_2 \in \mathcal {P}$$ and an $$\overset{o}{\mathcal {L}}$$ word w, given x in the fomain of $$\gamma_2 w \gamma_1^-1$$, there is an $$\mathcal {M}$$ word equal to $$\gamma_2 w \gamma_1^{-1}$$ on the neighborhood of x.
• Given an $$\mathcal {M}$$ word w, and y in the domain of $$\gamma_2^{-1} w \gamma_1$$ some $$\overset{o}{\mathcal {L}}$$ word coincides with $$\gamma_2^{-1} w \gamma_1$$ near y.
• Every $$\mathcal {M}$$ orbit is infinite, because every $$\overset{o}{\mathcal {L}}$$ orbit in int D meets some $$(x_p , y_p)$$ infinitely often.
• Also note that $$\mathcal{M}$$ is non nesting because every orientation preserving $$\mathcal{M}$$ word that has a fixed point is a restriction of the identity (why?).
• Furthermore, any $$\mathcal {M}$$ invariant probability measure with no atom lifts to an $$\overset{o}{\mathcal {L}}$$ invariant measure on D.
• The existence of the words $$w_d$$ (which depends on the non nesting of $$\mathcal {L}$$ implies that this measure has finite total mass, hence extends to $$\mathcal{L}$$ invariant measure on D.

We have now reduced our problem to finding an $$\mathcal{M}$$ invariant measure with no atom (assuming that every minimal set of $$\overset{o}{\mathcal {L}}$$ is infinite )

## From M to N

• Let $$F \subset C$$ be a minimal set of $$\mathcal {M}$$.
• It is infinite and has no isolated point.
• For each component $$C_p$$ of C that meets F, we choose a collapsing map $$\rho_p : C_p \to C_p’$$ where $$C_p’$$ is another circle and $$\rho_p$$ sends each component of $$C_p \backslash (F \cap C_p )$$ to a point.
• Let C’ be the union of circles $$C_p’$$
• Given an element $$\phi : A \to B$$ of $$\mathcal {M}$$ whose domain meets F, we consider images A’, B’ of A, B in C’ and the natural homeomorphism $$\phi’$$ between interiors of A’ and B’ (note that A’ , B’ are non degenerate intervals, but they need not be open).
• $$\mathcal{N}$$ = collection of these $$\phi’$$ . It is an open system of maps on C’.
• $$\mathcal {N}$$ is non -nesting and every regular orbit is dense, because every $$\mathcal{M}$$ orbit contained in F is dense in F (why?) but there may be finite singular orbits.

## From N to O

• Choose an interval [x, y] disjoint from all finite singular orbits such that some orientation preserving $$\mathcal {N}$$ word sends x to y.
• Perform the same operation as in (L to M) so as to obtain a system $$\mathcal{O}$$ on the circle obtained by identifying the endpoints of [x, y]
• This last system is open, non-nesting with every orbit dense.
• By Sacksteder’s theorem it admits an invariant measure.
• This measure first lifts to an $$\mathcal {N}$$ invariant measure, then to the required $$\mathcal {M}$$ invariant measure.
• This completes the proof when every minimal set of the original system $$\overset{o}{\mathcal {L}}$$ is finite.
• If $$\overset{o}{\mathcal {L}}$$ has an infinite minimal set F, we first collapse to a point every component of D \F (as in M to N) .
• We obtain a system $$\mathcal {N}$$ on a multi-interval with every regular orbit dense and we deal with it as before.