Proof. Let $\varphi :U \to (0,1)\times V \subset \mathbb R^n$ be a local chart of $\mathcal F$ , where $\varphi ^{-1}((0,1) \times \{c\})$ , $c\in V$ , are the plaques of $\mathcal F$ in U. For any $p\in U$ , consider ${\xi : W\subset U \to (0,1)\times K\subset \mathbb R^n}$ another local chart centered in p such that for any

$$ \begin{align*}z\in W \Rightarrow l_z(\mathcal F|U(z) )> 3\cdot l_z(\mathcal F|W(z)).\end{align*} $$

This can be done by the continuity of $\{l_x\}$ . In particular, for any $x\in W, y\in \mathcal F|W(x) = \xi ^{-1}((0,1),c')$ , the simple curve $\gamma (t) = \xi ^{-1}((1-t)\xi (x)+t\xi (y), c' )$ minimizes the $l_x$ -length connecting x and y, that is, $d_x(x,y) = l_x(\gamma )$ .

Therefore, consider $x\in W, y\in \mathcal F|W(x) = \xi ^{-1}((0,1),c')$ and sequences $x_n \in W, y_n \in \mathcal F|W(x_n) = \xi ^{-1}((0,1),c_n)$ with $x_n \rightarrow x$ and $y_n \rightarrow y$ . Let $\gamma $ be defined as in the previous paragraph and $\gamma _n(t):=\xi ^{-1}((1-t)\xi (x_n)+t\xi (y_n), c_n )$ . By the convergence of the sequences, we have $\gamma _n \rightarrow \gamma $ . However, by the previous discussion on the choice of the local chart W,

$$ \begin{align*}d_{x_n}(x_n,y_n) = l_{x_n}(\gamma_n) \quad \text{and} \quad d_x(x,y) = l_x(\gamma).\end{align*} $$

We then conclude by the continuity of $\{l_x\}$ that $\lim _{n\rightarrow \infty } d_{x_n}(x_n,y_n) =\lim _{n\rightarrow \infty } l_{x_n}(\gamma _n)= l_x(\gamma ) = d_x(x,y)$ .

Therefore, $\{d_x\}$ is plaque-continuous as we wanted to show.

Proof. Assume that T is a local transversal associated to a local chart $(U,\varphi )$ . Let $r>0$ be small enough that we have $B_{d_x}(x,r)\subset U$ for every $x\in T$ . In particular, $U\setminus \overline {T}$ has two open connected components $U_1$ and $U_2$ with $\overline {U_1}\cap \overline {U_2} = \overline {T}$ .

Since U is a local chart, we may consider an orientation on the $\mathcal F|U$ -plaques. Assume that S is not open. Then, there exists $y\in S$ and a sequence $y_k \notin S$ , with $y_k\rightarrow y$ .

Consider $x\in T$ such that $y\in B_{d_x}(x,r)$ and denote by $\varphi $ the flow on the $\mathcal F|U$ -plaques induced by the orientation fixed before and such that

$$ \begin{align*}d_p(\varphi_t(p), p)=|t|,\end{align*} $$

whenever $\varphi _t(p)$ is defined. Let $t_0$ be such that $x=\varphi _{t_0}(y)$ . As $y\in B_{d_x}(x,r)$ , there exists $\delta>0$ for which

$$ \begin{align*}\varphi_t(y)\in S, \quad t\in [t_0-\delta, t_0+\delta].\end{align*} $$

Now, by the plaque continuity and the fact that $y_k\rightarrow y$ ,

$$ \begin{align*}\varphi_{t_0-\delta}(y_k)\rightarrow \varphi_{t_0-\delta}(y), \quad \varphi_{t_0+\delta}(y_k)\rightarrow \varphi_{t_0+\delta}(y).\end{align*} $$

Observe that $\varphi _{t_0-\delta }(y)$ and $\varphi _{t_0+\delta }(y)$ belong to different connected components, thus, for k large enough, the same happens for $\varphi _{t_0-\delta }(y_k)$ and $\varphi _{t_0+\delta }(y_k)$ . Since $\gamma _k:=\{\varphi _{t}(y_k): \quad t\in [t_0-\delta , t_0+\delta ]\}$ is an arc with points in the interior and in the exterior of $U_1$ , it must intersect its boundary, namely T. This implies that $y_k\in S$ for large k, yielding a contradiction.

That is, S is open as we wanted to show.

Proof. For each $x\in M$ , take any $U_x\in \mathcal U$ with $x\in U_x$ . There exists $r_x>0$ for which $B_{d_x}(x,r_x) \subset U_x$ . By plaque continuity of $\{d_x\}$ , there exists a neighborhood $x\in V_x \subset U_x$ for which

$$ \begin{align*}y\in V_x \Rightarrow B_{d_y}(y,r_x) \subset U_x.\end{align*} $$

Since M is compact, we may cover M with a finite number of neighborhoods $V_{x_i}$ , $1\leq i \leq l$ . Take $\mathfrak r = \min \{r_{x_i}: 1\leq i \leq l \}$ .

Proof. Let $\mathcal U$ be a finite cover of M by local charts that are continuity-domains of $\{d_x\}$ . Consider $\mathfrak r$ the number given by Proposition 3.10. In particular, the family $\{U_{\mathfrak r/2} : U\in \mathcal U\}$ , defined by

$$ \begin{align*}U_{\mathfrak r/2} = \{x\in U : d_x(x,\partial U)\geq \mathfrak r/2\},\end{align*} $$

is still a cover of M. Let $A\subset M$ be a Borel subset. Observe that

$$ \begin{align*}\Phi_{t}(A\cap U_{\mathfrak r/2}) \subset U, \quad U\in \mathcal U, \, t<\mathfrak r/2.\end{align*} $$

We will prove that for $U\in \mathcal U$ , the subset $\Phi _t(A \cap U_{\mathfrak r/2})$ is measurable.

Let $\varphi _U:U\to B_1^{n-1}(0)\times (0,1)$ be a local chart of $\mathcal F$ and, inside U, consider the orientation in the plaques $\mathcal {F}(x)$ induced by the orientation in the line segments of the form $\{\overline {x}\}\times (0,1)\subset \mathbb R^{n-1}\times \mathbb R$ . This orientation induces, at each plaque, an order relation which we will denote by $\prec $ (the plaque being implicit in the context).

Now, for $s\in [-t,t]$ , with $0\leq t< \mathfrak r/2$ fixed, we define $\phi _s^U:U_{\mathfrak r/2}\to U$ by:

• • for $s>0$ , $\phi ^U_s(x)$ is the only point of the plaque $\mathcal {F}|U(x)$ such that $d_x(x,\phi _s^U(x))=s$ and $x\prec \phi _s^U(x)$ ;

• • for $s\kern1.4pt{<}\kern1.4pt0$ , $\phi _{s}^U(x)$ is the unique point of the plaque $\mathcal {F}|U(x)$ such that ${d_x(x,\phi _s^U(x))\kern1.4pt{=}\kern1.4pt{-}\kern1.4pt s}$ and $\phi _s^U(x)\prec x$ .

Observe that $\phi ^U_s$ is continuous for every $|s|<t$ since U is a continuity-domain of $\{d_x\}$ and, consequently, it is a homeomorphism. Thus, $\phi _s^U(A\cap U_{\mathfrak r/2})$ is a measurable subset of M for every $s\in [-t,t]$ .

Now, for each $1\leq i \leq n$ , take

$$ \begin{align*} \Phi_t^U(A):=\bigcup_{\substack{q\in\mathbb{Q} \\ q<t}}\phi_q^U(A\cap U_{\mathfrak r/2}), \quad 0\leq t <\mathfrak r/2. \end{align*} $$

Notice that $\Phi _t^U(A)$ is a measurable set, since each set in the countable union is measurable as we have proved before. Consequently,

$$ \begin{align*} \Phi_t(A)=\bigcup_{U\in \mathcal U}\Phi_t^U(A) \end{align*} $$

is a measurable set, as we wanted to show.

Proof. Take $x\in U$ , $U\in \mathcal U$ , and $J\subset M$ as in the statement. Let $y_n \rightarrow y$ , $y_n, y \in F_r(x)$ . We want to show that for $r>0$ ,

$$ \begin{align*}\lim_{n\rightarrow \infty} \mu^U_x(J\cap B_{d_{x}}(y_n,r)) = \mu^U_x(J\cap B_{d_x}(y,r)). \end{align*} $$

Given any $k\in \mathbb N$ , since $\mu ^U_x$ is not atomic, we have that

$$ \begin{align*} \mu^U_x(\partial B_{ d_x}(y,r)) & = 0 \quad \text{and} \quad \mu^U_x(\partial B_{d_x}(y_n,r)) =0\quad \text{ for all } n \in \mathbb N, \end{align*} $$

where $\partial B_{d_x}$ denotes the boundary of the set inside the leaf $\mathcal F(x)$ . Now, let ${B_n:= (J\cap B_{d_x}(y_n,r)) \Delta (J\cap B_{d_x}(y,r))}$ , where $Y\Delta Z$ denotes the symmetric difference of the sets Y and Z. From standard measure theory,

$$ \begin{align*} \limsup_{n\rightarrow \infty} \mu^U_x(B_n) \leq \mu^U_x \Big(\limsup_{n \rightarrow \infty} B_n \Big).\end{align*} $$

Thus,

$$ \begin{align*} \limsup_{n\rightarrow \infty} \mu^U_x(B_n) & \leq \mu^U_x \bigg( \bigcap_{m=1}^{\infty} \bigcup_{n\geq m} B_n \bigg).\end{align*} $$

We know that $z\in \bigcap _{m=1}^{\infty } \bigcup _{n\geq m} B_n $ if, and only if, there exists $n_1<n_2<n_3< \cdots $ such that $z \in B_{n_1}\cap B_{n_2} \cap \cdots $ . Therefore, for each i, either:

• • $z \in (J\cap B_{d_x}(y_{n_i},r)) \setminus (J\cap B_{d_x}(y,r)) \Rightarrow z \in J, z\in B_{d_x}(y_{n_i},r)$ and $z\notin B_{d_x}(y,r)$ ; or

• • $z \in (J\cap B_{d_x}(y,r)) \setminus (J\cap B_{d_x}(y_{n_i},r)) \Rightarrow z \in J, z\in B_{d_x}(y,r)$ and $z\notin B_{d_x}(y_{n_i},r)$ .

Observe that z cannot fall in the second case infinitely often since $\bigcap _i (B_{d_x}(y,r)\setminus B_{d_x}(y_{n_i},r)) = \emptyset $ . Therefore, passing to a subsequence if necessary,

$$ \begin{align*} z \in J, z\in B_{d_x}(y_{n_i},r)\quad \text{and} \quad z\notin B_{d_x}(y,r),\end{align*} $$

which implies, by taking $i \rightarrow \infty $ , that $z\in J\cap \partial B_{d_x}(y,r) \subset \partial B_{d_x}(y,r)$ . Consequently,

$$ \begin{align*}\limsup_{n\rightarrow \infty} \mu^U_x(B_n) \leq \mu^U_x(\partial B_{d_x}(y,r)) = 0.\end{align*} $$

Therefore,

$$ \begin{align*}\lim_{n\rightarrow \infty} \mu^U_x (J\cap B_{d_x}(y,r) \setminus J\cap B_{d_x}(y_n,r)) = \lim_{n\rightarrow \infty} \mu^U_x (J\cap B_{d_x}(y_n,r) \setminus J\cap B_{d_x}(y,r))= 0\end{align*} $$

and, consequently,

$$ \begin{align*} \lim_{n\rightarrow \infty} \mu^U_x (J\cap B_{d_x}(y_n,r)) = & \lim_{n\rightarrow \infty}(\mu^U_x(J\cap B_{d_x}(y_n,r)\setminus J\cap B_{d_x}(y,r)) + \mu^U_x (J\cap B_{d_x}(y,r)) \\ & - \mu^U_x(J\cap B_{d_x}(y,r) \setminus J\cap B_{d_x}(y_n,r)) = \mu^U_x (J \cap B_{d_x}(y,r)), \end{align*} $$

as we wanted to show. The second claim is a consequence of the first by taking $J=M$ .

Proof. If x and y belong to the same $\mathcal F$ -plaque in U, then $\mu ^U_x=\mu ^U_y$ . We already know that for all Borel subset $W\subset M$ ,

$$ \begin{align*}x\in V \mapsto \mu^U_x(W \cap U)\end{align*} $$

is Borel measurable.

Consider the local chart be given by the homeomorphism $\varphi :V \to (0,1)\times G$ , where $G\subset \mathbb R^{n-1}$ is an open subset. Setting $g_r:V \to [0,\infty )$ as

$$ \begin{align*}g_r(x) = \mu^U_x(J\cap B_{d_x}(x,r)),\end{align*} $$

we have

$$ \begin{align*}g_r \circ \varphi^{-1}(x_1,x_2) = \mu^U_{\varphi^{-1}(x_1,x_2)}(J\cap B_{d_{\varphi^{-1}(x_1,x_2)}}(\varphi^{-1}(x_1,x_2),r)).\end{align*} $$

Let us prove that this function is continuous in $x_1$ and Borel measurable in $x_2$ .

By Lemma 4.1, restricted to $V\setminus \mathcal A_U$ , we already have the continuity of $g_r\circ \varphi ^{-1}$ the first coordinate, since the second coordinate being fixed means we are evaluating the function on a single plaque where the conditional measure is non-atomic. Now, fix the first coordinate $x_1$ and consider the transversal $T=\{x_1\} \times B_1^{n-1}(0)$ . By Lemma 3.9, the set

$$ \begin{align*}S:=\bigcup_{x\in \varphi^{-1}(T)}B_{d_x}(x,r)\end{align*} $$

is an open subset of M and, consequently, $S\cap J =\bigcup _{x\in \varphi ^{-1}(T)}J\cap B_{d_x}(x,r) $ is Borel. Thus, $y \mapsto \mu ^U_y(S\cap J)$ is a Borel measurable function in V, which implies that $g_r\circ \varphi ^{-1}(x_1, \cdot )$ is $\mathcal B_T$ -measurable. In particular, its restriction to $T\cap \varphi (V\setminus \mathcal A_U) = T\setminus \varphi (\mathcal A_U)$ is a $\mathcal B_{T\setminus \varphi (\mathcal A_U)}$ -measurable map. However, observe that for $x\in T$ ,

$$ \begin{align*}\mu^U_x(S\cap J) = \mu^U_x(J\cap B_{d_x}(x,r)).\end{align*} $$

Therefore, for $x_1$ fixed, the map $x_2 \in G \setminus \pi _2(\varphi (\mathcal A_U)) \mapsto \mu ^U_{\varphi ^{-1}(x_1,x_2)}(J\cap B_{d_{\varphi ^{-1}(x_1,x_2)}} (\varphi ^{-1}(x_1,x_2),r))$ is $\mathcal B_{G \setminus \pi _2(\varphi (\mathcal A_U))}$ -measurable, where $\pi _2:(0,1)\times G \mapsto G$ is the projection onto the second coordinate.

Consequently, $g_r\circ \varphi ^{-1}$ restricted to $((0,1)\times G)\setminus \varphi (\mathcal A_U)$ is a jointly measurable function with respect to the product $\sigma $ -algebra $\mathcal B_{(0,1)} \times \mathcal B_{G \setminus \pi _2(\varphi (\mathcal A_U))}$ (see for example [Reference Charalambos and Kim6, Lemma 4.51]). As $\varphi $ is a homeomorphism, we conclude that $g_r$ is $\mathcal B_{V\setminus \mathcal A_U}$ -measurable, as we wanted to show.

Proof. First, let us give a better formulation for the definition of $\mathcal Z_U$ . Observe that

$$ \begin{align*} \mathcal Z_U =\{x \in U \setminus \mathcal A_U : \mu^U_x(I)=0 \; \text{for some open ball} \; x\in I\subset \mathcal F(x)\}. \end{align*} $$

Consider $\{q_1,q_2,\ldots \}$ an enumeration of $\mathbb Q \cap [0,1]$ and let $\mathcal U$ be the given finite family of local charts covering M. For each $U \in \mathcal U$ and $i\in \mathbb N$ , define $\phi _i^{U}:U_i\setminus \mathcal A_U\to \mathbb {R}$ by

$$ \begin{align*} \phi_i^{U}(x)=\mu^U_{x}(B_{d_x}(x,q_i)), \end{align*} $$

where $U_i=\{x\in U:B_{d_x}(x,q_i) \subset U\} $ . Observe that we may cover $U_i$ with a countable number of local charts $V_i^j\subset U_i$ , $j\in \mathbb N$ and, by Proposition 4.2, $\phi _i^{U} | V_i^j$ is $\mathcal B_{V_i^j\setminus \mathcal A_U}$ -measurable for every j. In particular, $\phi _i^U$ is $\mathcal B_{U_i\setminus \mathcal A_U}$ -measurable for every i. Therefore, we have that $\mathcal {Z}_i^{U}:=(\phi _i^{U})^{-1}(\{0\})\subset M$ is a $\mathcal B_{U_i\setminus \mathcal A_U}$ -measurable subset, in particular, a $\mathcal B_{U\setminus \mathcal A_U}$ -measurable subset. It is not difficult to see that

(2) $$ \begin{align} \mathcal{Z}_U=\bigcup_{i=1}^\infty \mathcal{Z}_i^{U}. \end{align} $$

Therefore, $\mathcal {Z}_U$ is a $\mathcal B_{U\setminus \mathcal A_U}$ -measurable subset as we wanted to show. Moreover, $\mu (\mathcal {Z}_U)=0$ .

Proof. For a certain $0<r<\mathfrak r$ fixed, take any $x \in M_0$ . Let $y\in B_{d_x}(x,\mathfrak r) \cap M_0$ and $U \in \mathcal U$ with $B_{d_x}(y,\mathfrak r) \subset U$ , take $y_n \in B_{d_x}(x,\mathfrak r)\cap M_0$ with $y_n \rightarrow y$ as $n\rightarrow \infty $ and ${B_{d_x}(y_n,\mathfrak r) \subset U}$ .

By definition,

$$ \begin{align*}\mu_y = \mu_y^U(\cdot |B_{d_x}(y,\mathfrak r)), \; \mu_{y_n} = \mu_{y_n}^U(\cdot |B_{d_x}(y_n,\mathfrak r)). \end{align*} $$

Therefore, for $n\in \mathbb N$ such that $B_{d_x}(y,r)\subset B_{d_x}(y_n,\mathfrak r)$ and $B_{d_x}(y_n,r)\subset B_{d_x}(y,\mathfrak r)$ ,

(5) $$ \begin{align} \mu_{y_n}(J\cap B_{d_x}(y_n,r)) = \frac{\mu_{y_n}^U(J\cap B_{d_x}(y_n,r))}{\mu_{y_n}^U(B_{d_x}(y_n,\mathfrak r))} = \frac{\mu_y^U(J\cap B_{d_x}(y_n,r))}{\mu_{y}^U(B_{d_x}(y_n,\mathfrak r))}. \end{align} $$

By Lemma 4.1, we have $\mu _y^U(J\cap B_{d_x}(y_n,r))\rightarrow \mu _y^U(J\cap B_{d_x}(y,r))$ and $\mu _{y}^U( B_{d_x}(y_n,\mathfrak r)) \rightarrow \mu _{y}^U(B_{d_x}(y,\mathfrak r))$ as $n\rightarrow \infty $ . Therefore, $\mu _{y_n}(J\cap B_{d_x}(y_n,r))\rightarrow \mu _{y}(J\cap B_{d_x}(y,r))$ as we wanted to show.

Proof. Let $x\in M_0$ . First, let us prove that $r \in [0,\mathfrak r] \mapsto \mu _x(J\cap B_{d_x}(x,r))$ is a continuous function. Let $0=r<\mathfrak r$ and $r_n \in [0,\mathfrak r] \searrow r$ (if $r=\mathfrak r$ , the argument is analogous). Hence,

$$ \begin{align*}\mu_{x}(J\cap B_{d_x}(x,r_n)) = \mu_x(J\cap B_{d_x}(x,r)) + \mu_x(J\cap B_{d_x}(x,r_n)\setminus J\cap B_{d_x}(x,r)).\end{align*} $$

As $\mu _x$ is non-atomic,

$$ \begin{align*}\lim_{n\rightarrow \infty}\mu_x(B_{d_x}(x,r_n)\setminus B_{d_x}(x,r)) =0.\end{align*} $$

Then,

$$ \begin{align*}\mu_x(J\cap B_{d_x}(x,r_n)\setminus J\cap B_{d_x}(x,r)) =\mu_x(J \cap ( B_{d_x}(x,r_n)\setminus B_{d_x}(x,r))) \rightarrow 0,\end{align*} $$

showing the first part of the statement.

Let us show the second part of the statement. For each $x\in M_0$ , let $x\in V_x$ be a local chart with

$$ \begin{align*}y\in V_x \Rightarrow B_{d_y}(y,\mathfrak r) \subset U_x \quad \text{for some} \; U_x \in \mathcal U.\end{align*} $$

As M is compact, we may cover M with a finite number of such local charts, say $V_1,V_2, \ldots , V_l$ and call $U_1, U_2, \ldots , U_l$ the associated local charts in $\mathcal U$ . For any j, consider

$$ \begin{align*}y\in V_j \mapsto \mu_y(J\cap B_{d_y}(y,r)).\end{align*} $$

Observe that

$$ \begin{align*}\mu_y(J\cap B_{d_y}(y,r)) =\frac{\mu^{U_j}_y(J\cap B_{d_y}(y,r))}{\mu^{U_j}_y(B_{d_y}(y,\mathfrak r))}. \end{align*} $$

Therefore, by Proposition 4.2, $y\in V_j \cap M_0 \mapsto \mu _y(J\cap B_{d_y}(y,r))$ is a $\mathcal B_{U_j\setminus \mathcal A_{U_j}}$ -measurable map. As j is arbitrary, $y\in M_0 \mapsto \mu _y(J\cap B_{d_x}(y,r))$ is a $\mathcal B_{U\cap M_0}$ -measurable map.

Thus, the map given by (6) is jointly measurable as it is continuous in the first coordinate and $\mathcal B_{M_0}$ -measurable in the second. The second conclusion follows from taking $J=M$ .

Proof. Clearly, it is sufficient to check the measurability of $\Delta (x)$ in $M_0$ . By Corollary 4.5, for any fixed $x\in M_0$ , the function $w(x,\cdot ):(0,\mathfrak r) \rightarrow \mathbb (0,\infty )$ is continuous. Thus, for n small enough,

$$ \begin{align*} \sup_{\varepsilon \leq 1/n} \frac{\mu_x(B_{d_x}(x,\varepsilon))}{2\varepsilon} = \sup_{r\in \mathbb Q \cap (0,1/n] } \frac{\mu_x(B_{d_x}(x,r))}{2r} \quad \text{ for all } x\in M_0. \end{align*} $$

For each $n\in \mathbb N$ , the map $L_n:M_0 \to \mathbb R$ , $x\in M_0 \mapsto \sup _{r\in \mathbb Q \cap (0,1/n] } {(\mu _x(B_{d_x}(x,r)))}/{2r}$ is measurable, as it is defined by the supremum over a countable number of measurable maps. Consequently, since $\Delta (x)= \lim _{n\rightarrow \infty } L_n(x)$ for all $x\in M_0$ , we conclude that $\Delta (x)$ is measurable as we wanted.

Proof. Recall that for every $x\in M_0$ (see (3)), the measure $\mu _x$ is defined on $B_{d_x}(x,\mathfrak r)$ by

$$ \begin{align*}\mu_x(J) = \mu_x^U(J | B_{d_x}(x,\mathfrak r)) = \frac{\mu_x^U(J \cap B_{d_x}(x,\mathfrak r))}{\mu_x^U(B_{d_x}(x,\mathfrak r)) }, \quad J\subset B_{d_x}(x,\mathfrak r).\end{align*} $$

Abusing notation, we may define $x\in M_0 \mapsto \mu _x(J)$ for any Borel set $J\subset M$ by taking $\mu _x(J) := \mu _x(J\cap B_{d_x}(x,\mathfrak r))$ . By Corollary 4.5, the map $x \in M_0 \mapsto \mu _x$ is measurable in the sense that for any Borel subset $J \subset M$ , the map $x \in M_0 \mapsto \mu _x(J\cap B_{d_x}(x,\mathfrak r))$ is measurable.

Using Lusin’s theorem, we may take $\{K_i\}$ an increasing sequence of compact subsets of $M_0$ for which the map $x\mapsto \mu _x$ is continuous when restricted to each $K_i$ . Once again, we emphasize that, in this case, it means that for any Borel subset $J\subset M$ , the map $x \in K_i \mapsto \mu _x(J\cap B_{d_x}(x,\mathfrak r))$ is continuous. As $\mu $ is ergodic, for $\mu $ -a.e. point $x\in K_i$ , the orbit of x is dense in a full measurable subset of $K_i$ . Let $P_i=\mathcal N\cap K_i$ .

Let $x\in P_i$ and $y\in V_x \cap \mathcal {F}^c|U_x(x)\cap P_i$ . From Remark 6.2, we know that there exists $F\in \mathscr {L}^+(\mathcal {F}^c(x))$ such that $F(x)=y$ , and since F is a limit center map, there exists a sequence ${n_k}\subset \mathbb {Z}$ with $|n_k|\to \infty $ so that the sequence $f^{n_k}|_{\mathcal {F}^c(x)}$ converges pointwise to F.

By the continuity of $x \in P_i \mapsto \mu _x$ and since $y\in P_i$ ,

$$ \begin{align*} F_*\mu_x=\lim_{k\to \infty}f^{n_k}_*\mu_x=\lim_{k\to \infty}\mu_{f^{n_k}(x)}=\mu_{F(x)}=\mu_y. \end{align*} $$

By taking the limit set of $K_i$ , we conclude that for every point $x\in P$ , where $P=\lim _i P_i$ , and $y\in V_x\cap \mathcal {F}^c|U_x(x)\cap P$ , there exists $F\in \mathscr {L}^+(\mathcal {F}^c(x))$ such that $F(x)=y$ and $F_*\mu _x=\mu _y$ as we wanted.

Proof. Assume that $\mathcal F^c(x)$ is not compact. The argument is identical for the other case.

We know that $\mathcal L^+(\mathcal F^c(x))$ is a closed group as it is isomorphic to either group of translations (see Remark 6.2). Furthermore, for any $J\subset M$ Borel, the map $y \in B_{d_x}(x,\mathfrak r)\cap M_0 \mapsto \mu _y(J\cap B_{d_x}(y,\mathfrak r))$ is continuous by Corollary 4.4.

Let $y_n \in G^{U_x}_x$ with $y_n \to y \in V_x \cap \mathcal F^c|U(x)$ and consider $F_n \in \mathcal L^+(\mathcal F^c(x))$ be given by Lemma 6.3 satisfying $F_n(x) = y_n$ .

Observe that the sequence $\{F_n\}_n$ converges pointwise to a map $F\in \mathcal L^+(\mathcal F^c(x))$ with $F(x)=y$ . Indeed, by Remark 6.2, there are real numbers $t_n$ for which

$$ \begin{align*}\psi_L^{-1}\circ T_{t_n}\circ \psi_L = F_n, \quad n\geq 1,\end{align*} $$

where $L=\mathcal F^c(x)$ . In particular,

$$ \begin{align*}T_{t_n}(\psi_L(x)) = \psi_L(y_n).\end{align*} $$

Since $\psi _L(y_n)$ converges to $\psi _L(y)$ , because $\psi _L$ is a homeomorphism, it follows that $|\psi _L(x) -\psi _L(y_n)| \rightarrow |\psi _L(x) -\psi _L(y)|$ . Since $T_{t_n}$ are translations, we have $t_n = |\psi _L(x) -\psi _L(y_n)| \rightarrow |\psi _L(x) -\psi _L(y)|$ . Consequently, the sequence of translations $\{T_{t_n}\}$ converges to the translation $T_t:\mathbb R\to \mathbb R$ with $t = |\psi _L(x) -\psi _L(y)|$ . In particular, $F_n$ converges to $F:= \psi _L^{-1}\circ T_t\circ \psi _L$ .

Finally, by the continuity mentioned previously,

$$ \begin{align*}F_*\mu_x = \lim_{n\rightarrow \infty}(F_n)_*\mu_x = \lim_{n\rightarrow \infty} \mu_{y_n} = \mu_y.\end{align*} $$

Thus, $y \in G^{U_x}_x$ , showing that $G^{U_x}_x$ is closed in $\mathcal F^c|U_x(s)$ as we wanted.

Proof. Assume that for some $x\in P$ , the support of $\mu ^{U_x}_x$ is a Cantor subset of $\mathcal F^c|U(x)$ . As $V_x$ is compact, the intersection $V_x \cap \text {supp}(\mu ^{U_x}_x)$ is also a Cantor subset and is the support of the restriction $\mu ^{U_x}_x(\cdot | V_x \cap \mathcal F^c|U(x))$ .

As we have seen, for $\mu $ -a.e. $x\in P$ , the set $G^{U_x}_x$ is a full measure subset of $ V_x \cap \mathcal F^c|U(x))$ and is closed by Lemma 6.4; therefore, it must contain the support $V_x \cap \text {supp}(\mu ^{U_x}_x)$ which is a Cantor subset. Consider an ordering of $\mathcal F^c|U(x)$ and consider $a = \text {inf} (V_x \cap \text {supp}(\mu ^{U_x}_x))$ , $b=\text {sup} (V_x \cap \text {supp}(\mu ^{U_x}_x)),$ and $\gamma $ the center arc connecting a and b inside $\mathcal F^c|U(x)$ .

Let $\zeta : \gamma \to [0,1]$ given by $\zeta (a)=0$ , $\zeta (b) =1$ , be increasing and linear with respect to $d_x$ , that is, $\zeta (y) = d_x(a,y)/d_x(a,b)$ . The set $\zeta (\gamma ) \subset [0,1]$ is a Cantor subset, in particular, there exists a strictly increasing homeomorphism $H_0 : \zeta (\gamma ) \to \mathcal C_3$ with $H_0(0)=0$ , $H_0(1)=1$ , where $\mathcal C_3$ here denotes the standard ternary Cantor set in $[0,1]$ . Now, let us extend $H_0$ to a map of $[0,1]$ .

First, we fix some notation. Consider $I^0 = [0,1]$ , $I^1_1=(1/3,2/3)=(a^1_1,b^1_1)$ , and $I^m_k = (a^m_k,b^m_k)$ , with $1\leq k \leq 2^{m-1}$ the open intervals removed at the mth stage of the construction of the ternary Cantor set with $a^m_k < b^m_k < a^m_{k+1}$ for any $m\geq 2$ and $1\leq k \leq 2^{m-1}-1$ . Therefore,

$$ \begin{align*} \mathcal C_3 = I_0 \setminus \bigg(\bigcup_{m=1}^{\infty} \bigcup_{k=1}^{2^{m-1}} I^m_k\bigg). \end{align*} $$

Let also $J^m_k := (H_0^{-1}(a^m_k), H_0^{-1}(b^m_k)) \subset [0,1]$ . Then, $\zeta (\gamma ) = I_0 \setminus ( \bigcup _{m=1}^{\infty } \bigcup _{k=1}^{2^{m-1}} J^m_k)$ . Consider $H:[0,1] \to [0,1]$ defined by:

• • the restriction of H to each $J^m_k$ is the linear map given by

  $$ \begin{align*} H|J^m_k(y) = \bigg(1-\frac{y-H_0^{-1}(a^m_k)}{H_0^{-1}(b^m_k)-H_0^{-1}(a^m_k)}\bigg)\cdot a^m_k+ \frac{y-H_0^{-1}(a^m_k)}{H_0^{-1}(b^m_k)-H_0^{-1}(a^m_k)}\cdot b^m_k; \end{align*} $$

• • the restriction of H to $\zeta (\gamma )$ is $H_0$ , that is, $H|\zeta (\gamma ) = H_0$ .

Observe that H is continuous and strictly increasing; thus, it is a strictly increasing bijection and, consequently, a homeomorphism.

Consider now the homeomorphism $\widetilde {H} := H\circ \zeta : \gamma \to [0,1]$ and denote by $E_3$ the set of all points in $\mathcal C_3$ that are extremes of the removed intervals, that is, $E_3 = \{a^m_k, \; b^m_k : m\in \mathbb Z, \; 1\leq k \leq 2^{m-1} \}$ , or equivalently,