## Gibbs measures have local product structure

Let ${M}$ be a compact smooth manifold and ${f\colon M\rightarrow M}$ a transitive Anosov ${C^{1+\alpha}}$ diffeomorphism. If ${\mu}$ is an ${f}$-invariant Borel probability measure on ${M}$ that is absolutely continuous with respect to volume, then the Hopf argument can be used to show that ${\mu}$ is ergodic. In fact, recently it has been shown that even stronger ergodic properties such as multiple mixing can be deduced; for example, see Coudène, Hasselblatt, and Troubetzkoy [Stoch. Dyn. 16 (2016), no. 2].

A more general class of measures with strong ergodic properties is given by the theory of thermodynamic formalism developed in the 1970s: given any Hölder continuous potential ${\varphi\colon M\rightarrow{\mathbb R}}$, the quantity ${h_\mu(f) + \int\varphi\,d\mu}$ is maximized by a unique invariant Borel probability measure ${\mu = \mu_\varphi}$, which is called the equilibrium state of ${\varphi}$; the maximum value of the given quantity is the topological pressure ${P = P(\varphi)}$. The unique equilibrium state has the Gibbs property: for every ${\varepsilon>0}$ there is a constant ${K>0}$ such that

$\displaystyle K^{-1} \leq \frac{\mu(B_n(x,\varepsilon))}{e^{S_n\varphi(x) - nP}} \leq K, \ \ \ \ \ (1)$

where ${B_n(x,\varepsilon) = \{y\in M : d(f^kx,f^ky) < \varepsilon\ \forall 0\leq k < n\}}$ is the Bowen ball around ${x}$ of order ${n}$ and radius ${\varepsilon}$, and we write ${S_n\varphi(x) = \sum_{k=0}^{n-1} \varphi(f^kx)}$ for the ${n}$th ergodic sum along the orbit of ${x}$.

Historically, strong ergodic properties (mixing, K, Bernoulli) for equilibrium states have been established using methods such as Markov partitions rather than via the Hopf argument. However, in the more general non-uniformly hyperbolic setting, it can be difficult to extend these symbolic arguments, and so it is interesting to ask whether the Hopf argument can be applied instead, even if it only recovers some of the strong ergodic properties. The key property of absolutely continuous measures that is needed for the Hopf argument is the fact that they have a local product structure, which we define below. It was shown by Haydn [Random Comput. Dynam. 2 (1994), no. 1, 79–96] and by Leplaideur [Trans. Amer. Math. Soc. 352 (2000), no. 4, 1889–1912] that in the uniformly hyperbolic setting, the equilibrium states ${\mu_\varphi}$ have local product structure when ${\varphi}$ is Hölder continuous; thus one could apply the Hopf argument to them.

This post contains a direct proof that any measure ${\mu}$ with the Gibbs property (1) has local product structure; see Theorem 3 below. (This will be a bit of a longer post, since we need to recall several different concepts and then do some non-trivial technical work.) Since Bowen’s proof of uniqueness of equilibrium states using specification [Math. Systems Theory 8 (1974/1975), no. 3, 193–202] establishes the Gibbs property, this means that the equilibrium states produced this way could be addressed with the Hopf argument (I haven’t carried out the details yet, so I claim no formal results here). I should point out, though, that even without the use of Markov partitions, Ledrappier showed that these measures have the K property, which in particular implies multiple mixing. Since multiple mixing is the strongest thing we might hope to get from the Hopf argument, my primary motivation for the present approach is that Dan Thompson and I recently generalized Bowen’s result to systems satisfying a certain non-uniform specification property [Adv. Math. 303 (2016), 745–799], and the unique equilibrium states we obtain satisfy a non-uniform version of the Gibbs property (1), so it is reasonable to hope that they also have local product structure and can be studied using the Hopf argument; but this is beyond the scope of this post and will be addressed in a later paper.

1. Local product structure

Before defining local product structure for ${\mu}$, we recall some definitions. Since ${f}$ is Anosov, every ${x\in M}$ has local stable and unstable manifolds ${W^{s,u}_x}$, which have the following properties.

1. There are ${C \geq 1}$ and ${\lambda < 1 }$ such that for all ${x\in M}$, ${y\in W^s_x}$, and ${n\geq 0}$, we have ${d(f^n x, f^n y) \leq C\lambda^n d(x,y)}$; a similar contraction bound holds going backwards in time when ${y\in W^u_x}$.
2. There is ${\delta>0}$ such that if ${d(f^n x, f^ny) \leq \delta}$ for all ${n\geq 0}$, then ${y\in W^s_x}$; similarly for ${W^u_x}$ with ${n\leq 0}$.
3. There is ${\varepsilon>0}$ such that if ${d(x,y) \leq \varepsilon}$, then ${W_x^s \cap W_y^u}$ is a single point, which we denote ${[x,y]}$. Moreover, there is a constant ${Q}$ such that ${d([x,y],x) \leq Q d(x,y)}$, and similarly for ${d([x,y],y)}$.

A set ${R\subset M}$ is a rectangle if it has diameter ${\leq\varepsilon}$ and is closed under the bracket operation: in other words, for every ${x,y\in R}$, the intersection point ${[x,y] = W_x^s \cap W_y^u}$ exists and is contained in ${R}$.

Lemma 1 For every ${x\in M}$, there is a rectangle ${R}$ containing ${x}$.

Proof: Write ${V_x^s = \{y\in W_x^s : d(x,y) \leq \frac{\varepsilon}{2(Q+1)}\}}$ and similarly for ${V_x^u}$. Consider the set ${R = \{[y,z] : y\in V_x^u, z\in V_x^s\}}$ and observe that for every ${[y,z]\in R}$ we have

$\displaystyle d([y,z],x) \leq d([y,z],y) + d(y,x) \leq (Q+1)d(y,x) \leq \tfrac\varepsilon2.$

Thus ${R}$ has diameter ${\leq \varepsilon}$, and for every ${[y,z], [y',z']\in R}$ we have

$\displaystyle [[y,z],[y',z']] = W_{[y,z]}^s \cap W_{[y',z']}^u = W^s_y \cap W^u_{z'} = [y,z'] \in R,$

so ${R}$ is indeed a rectangle. $\Box$

Given ${x\in M}$ and a rectangle ${R\ni x}$, let ${V^{s,u}_x = W^{s,u}_x \cap R}$. Then ${R}$ can be recovered from ${V^{s,u}_x}$ via the method in the proof above, as the image of the map ${[\cdot,\cdot] \colon V_x^u \times V_x^s \rightarrow M}$. Given measures ${\nu^{s,u}}$ on ${V^{s,u}_x}$, let ${\nu^u\otimes\nu^s}$ be the pushforward of ${\nu^s\times \nu^u}$ under this map; that is, for every pair of Borel sets ${A\subset V_x^u}$ and ${B\subset V_x^s}$, we put ${(\nu^u\otimes \nu^s)(\{[y,z] : y\in A, z\in B\}) = \nu^u(A)\nu^s(B)}$.

Definition 2 A measure ${\mu}$ has local product structure with respect to ${W^{s,u}}$ if for every ${x\in M}$ and rectangle ${R\ni x}$ there are measures ${\nu^{s,u}}$ on ${V^{s,u}_x}$ such that ${\mu|_R \ll \nu^u \otimes \nu^s}$.

Theorem 3 Let ${f\colon M\rightarrow M}$ be a ${C^1}$ Anosov diffeomorphism, ${\varphi\colon M\rightarrow {\mathbb R}}$ a Hölder continuous function, and ${\mu}$ an ${f}$-invariant Borel probability measure on ${M}$ satisfying the Gibbs property (1) for some ${P\in {\mathbb R}}$. Then ${\mu}$ has local product structure in the sense of Definition 2. Moreover, there is ${\bar K \geq 1}$ such that for all ${R}$, the measures ${\nu^{s,u}}$ can be chosen so that the Radon–Nikodym derivative ${\psi = \frac{d\mu}{d(\nu^u\otimes \nu^s)}}$ satisfies ${\bar K^{-1} \leq \psi \leq \bar K}$ at ${\mu}$-a.e. point.

A quick side remark: here the diffeomorphism is only required to be ${C^1}$. The reason for the ${C^{1+\alpha}}$ hypothesis at the beginning of this post was so that the geometric potential ${\varphi(x) = -\log |\det Df|_{E^u_x}|}$ is Hölder continuous and its unique equilibrium state (the absolutely continuous invariant measure if it exists, or more generally the SRB measure) has the Gibbs property; this may not be true if ${f}$ is only ${C^1}$.

2. Conditional measures

In order to prove Theorem 3, we must start by recalling the notion of conditional measures; see Coudène’s book (especially Chapters 14 and 15) or Viana’s notes for more details than what is provided here.

Let ${(X,\mathop{\mathcal A},\mu)}$ be a Lebesgue space. A partition of ${X}$ is a map ${\xi \colon X\rightarrow \mathop{\mathcal A}}$ such that for ${\mu}$-a.e. ${x,y\in X}$, the sets ${\xi(x)}$ and ${\xi(y)}$ either coincide or are disjoint. Write ${\Xi = \{\xi(x) : x\in X\}}$ for the set of partition elements, and say that the partition ${\xi}$ is finite if ${\Xi}$ is finite.

Given a finite partition ${\xi}$, it is easy to define conditional measures ${\mu_{\xi(x)}}$ on the set ${\xi(x)}$ for ${\mu}$-a.e. ${x}$ by writing

$\displaystyle \mu_{\xi(x)}(A) = \frac{\mu(A \cap \xi(x))}{\mu(\xi(x))} \ \ \ \ \ (2)$

when ${\mu(\xi(x))>0}$, and ignoring those partition elements with zero measure. One can recover the measure ${\mu}$ from its conditional measures by the formula

$\displaystyle \mu(A) = \sum_{C\in \Xi} \mu_C(A) \mu(C). \ \ \ \ \ (3)$

If we write ${\hat\mu}$ for the measure on ${\Xi}$ defined by putting ${\hat\mu(\{C\}) = \mu(C)}$ for all ${C\in \Xi}$, then (3) can be written as

$\displaystyle \mu(A) = \int_\Xi \mu_C(A) \,d\hat\mu(C). \ \ \ \ \ (4)$

Even when the partition ${\xi}$ is infinite, one may still hope to obtain a formula along the lines of (4).

Example 1 Let ${X}$ be the unit square, ${\mu}$ be two-dimensional Lebesgue measure, and ${\xi(x)}$ be the horizontal line through ${x}$. Then Fubini’s theorem gives (4) by taking ${\mu_C}$ to be Lebesgue measure on horizontal lines, and defining ${\hat\mu}$ on ${\Xi}$, the set of horizontal lines in ${[0,1]^2}$, in one of the two following (equivalent) ways:

1. given ${E \subset \Xi}$, let ${\hat\mu(E) = \mu(\bigcup_{C\in E} C)}$;
2. identify ${\Xi}$ with the interval ${\{0\}\times [0,1]}$ on the ${y}$-axis, and define ${\hat\mu}$ as the image of one-dimensional Lebesgue measure on this interval.

Note that ${\hat\mu}$ must satisfy the first of these no matter what the partition ${\xi}$ is, while the second is a convenient description of ${\hat\mu}$ in this particular example.

A similar-looking example (and the one which is most relevant for our purposes) comes by letting ${R\subset M}$ be a rectangle and letting ${\xi(y) = V_y^u = W_y^u\cap R}$ be the partition into local unstable leaves. To produce conditional measures ${\mu_y^u = \mu_{V_y^u}}$, we need to use the fact that the partition ${\xi}$ is measurable. This means that there is a sequence of finite partitions ${\xi_n}$ that refines to ${\xi}$ in the sense that for ${\mu}$-a.e. ${y}$, we have

$\displaystyle \xi_{n+1}(y) \subset \xi_n(y) \text{ for all } n, \text{ and } \xi(y) = \textstyle\bigcap_{n=1}^\infty \xi_n(y).$

Lemma 4 Given any rectangle ${R}$, the partition of ${R}$ into local unstable leaves is measurable.

Proof: Fix ${x\in R}$ and let ${\{\eta_n(y)\}}$ be a refining sequence of finite partitions of ${V_x^s}$ with the property that ${\bigcap_n \eta_n(y) = \{y\}}$ for all ${y\in V_x^s}$; then let ${\xi_n(y) = \bigcup_{z\in \eta_n(y)} V_z^u}$, and we are done. $\Box$

Whenever ${\xi}$ is a measurable partition of a compact metric space ${X}$, we can define the conditional measures ${\mu_{\xi(y)}}$ as the limits of the conditional measures ${\mu_{\xi_n(y)}}$. Indeed, one can show (we omit the proofs) that for ${\mu}$-a.e. ${y}$, the limit ${I_y(\varphi) := \lim_{n\rightarrow\infty} \int \varphi \,d\mu_{\xi_n(y)}}$ exists for every continuous ${\varphi\colon X\rightarrow{\mathbb R}}$ and defines a continuous linear functional ${I_y\in C(X)^*}$; the corresponding measures ${\mu_{\xi(y)}}$ satisfy

$\displaystyle \int\varphi\,d\mu = \int_\Xi \int_C \varphi\,d\mu_C \,d\hat\mu(C) \ \ \ \ \ (5)$

for every ${\varphi\in C(X)}$, where once again we put ${\hat\mu(E) = \mu(\bigcup_{C\in E} C)}$.

The key result that we will need to describe properties of ${\mu_y^u}$ when ${\xi}$ is the partition of ${R}$ into local unstable leaves is that for ${\mu}$-a.e. ${y\in R}$ and every ${\varphi\in C(X)}$, we have

$\displaystyle \int \varphi\,d\mu_y^u = \lim_{n\rightarrow\infty} \frac 1{\mu(\xi_n(y))}\int_{\xi_n(y)} \varphi\,d\mu, \ \ \ \ \ (6)$

where ${\xi_n}$ is any sequence of finite partitions that refines to ${\xi}$.

In order to establish the local product structure for ${\mu}$ that is claimed by Theorem 3, we will show that the measures ${\mu_y^u}$ vary in an absolutely continuous manner as ${y}$ varies within ${R}$. That is, we consider for every ${x,y\in R}$ the holonomy map ${\pi_{x,y} \colon V_x^u \rightarrow V_{y}^u}$ defined by moving along local stable manifolds, so

$\displaystyle \pi_{x,y}(z) = [z,y] = V_z^s \cap V_{y}^u \text{ for all } z\in V_x^u.$

Our goal is to use the Gibbs property (1) for ${\mu}$ to prove that for every rectangle ${R}$ and ${\mu}$-a.e. ${x,y\in R}$, the conditional measures ${\mu_x^u}$ and ${\mu_{y}^u}$ satisfy

$\displaystyle (\pi_{x,y})_* \mu_x^u \ll \mu_{y}^u \text{ with } \bar{K}^{-1} \leq \frac{d(\pi_{x,y})_* \mu_x^u}{d\mu_{y}^u} \leq \bar{K}. \ \ \ \ \ (7)$

Once this is established, we can proceed as follows. Consider a rectangle ${R}$ with the decomposition ${\xi}$ into local unstable manifolds, let ${x\in R}$ be such that (7) holds for ${\mu}$-a.e. ${y\in R}$, and then identify ${\Xi}$ with ${V_x^s}$, as in the second characterization of ${\hat\mu}$ in Example 1. Let ${\nu^s}$ be the measure on ${V_x^s}$ corresponding to ${\hat\mu}$ under this identification, and let ${\nu^u = \mu_x^u}$. Given ${y\in V_x^s}$, let ${\psi_y = \frac{d(\pi_{x,y})_* \mu_x^u}{d\mu_{y}^u} \colon V_y^u \rightarrow [\bar{K}^{-1},\bar{K}]}$, so that in particular we have

$\displaystyle \int_{V_y^u} \varphi\,d\mu_y^u = \int_{V_y^u} \frac{\varphi}{\psi_y}\, d(\pi_{x,y})_*\mu_x^u = \int_{V_x^u} \frac{\varphi([z,y])}{\psi_y([z,y])} \,d\mu_x^u$

for every continuous ${\varphi\colon R\rightarrow {\mathbb R}}$. Then by (5), we have

\displaystyle \begin{aligned} \int\varphi\,d\mu &= \int_{\Xi} \int_C \varphi \,d\mu_C \,d\hat\mu(C) = \int_{V_x^s} \int_{V_y^u} \varphi \,d\mu_y^u \,d\nu^s(y) \\ &= \int_{V_x^s} \int_{V_x^u} \frac{\varphi([z,y])}{\psi_y([z,y])} \,d\nu^u(z) \,d\nu^s(y). \end{aligned}

By Definition 2, this shows that ${\mu}$ has local product structure with respect to ${W^{s,u}}$. Thus in order to prove Theorem 3, it suffices to shows that Gibbs measures satisfy the absolute continuity property (7).

It is worth noting quickly that our use of the term “absolute continuity” here has a rather different meaning from another common concept, which is that of a measure with “absolutely continuous conditional measures on unstable manifolds”. This latter notion is essential for the definition of SRB measures (indeed, in the Anosov setting it is the definition), and involves comparing ${\mu_x^u}$ to volume measure on ${V_x^u}$, instead of to the pushforwards of other conditional measures under holonomy.

In order to prove the absolute continuity property (7), we need to obtain estimates on ${\mu_y^u}$. We start by getting estimates on ${\mu}$ from the Gibbs property (1), and then using these to get estimates on ${\mu_y^u}$ using (6).

We will need a family of partitions of ${R}$ that refines to the partition into points. Fix a reference point ${q\in R}$, and suppose we have chosen partitions ${\eta_n^s}$ of ${V_q^s}$ and ${\eta_m^u}$ of ${V_q^u}$ for ${m,n\geq 0}$. Then we can define a partition ${\xi_{m,n}}$ of ${R}$ by taking the direct product of these two partitions, using the foliations of ${R}$ by local stable and unstable leaves: that is, we put

$\displaystyle \xi_{m,n}(y) = \{z\in R : \eta_m^u([z,q]) = \eta_m^u([y,q]) \text{ and } \eta_n^s([q,z]) = \eta_n^s([q,y]) \} \ \ \ \ \ (8)$

In order to obtain information on ${\mu(\xi_{m,n}(y))}$ using the Gibbs property (1), we need to put an extra condition on the partitions we use; we need to them to be adapted, meaning that each partition element both contains a Bowen ball and is contained within a larger Bowen ball. Most of the ideas here are fairly standard in thermodynamic formalism, but it is important for us to work separately on the stable and unstable manifolds, then combine the two, so we describe things explicitly. Fix ${\varepsilon>0}$. Given ${y\in V_q^u}$ and ${m\geq 0}$, let

\displaystyle \begin{aligned} B_m^u(y,\varepsilon) &= \{y'\in V_q^u : d(f^ky' , f^k y) \leq \varepsilon\ \forall 0\leq k \leq m \} \\ &= \{y'\in R : d(f^k y', f^ky) \leq \varepsilon\ \forall -\infty < k \leq m \}. \end{aligned}

Similarly, given ${z\in V_q^s}$ and ${n\geq 0}$, let

\displaystyle \begin{aligned} B_n^s(z,\varepsilon) &= \{z'\in V_q^s : d(f^{-k} z', f^{-k}z) \leq \varepsilon\ \forall 0\leq k \leq n\} \\ &= \{z'\in R : d(f^k z', f^kz) \leq \varepsilon\ \forall -n\leq k < \infty\}. \end{aligned}

Given ${\varepsilon > 0}$, we say that the partitions ${\eta_m^u}$ are ${\varepsilon}$-adapted if for every partition element ${\eta_m^u(y')}$ there is ${y\in V_q^u}$ such that

$\displaystyle B_m^u(y,\varepsilon/2) \subset \eta_m^u(y') \subset B_m^u(y,\varepsilon).$

We make a similar definition for ${\eta_n^s}$ using ${B_n^s}$. Note that we can produce an ${\varepsilon}$-adapted sequence of partitions ${\eta_n^s}$ as follows:

1. say that ${E\subset V_q^s}$ is ${(n,\varepsilon)}$-separated if for every ${x\neq y \in E}$ there is ${0\leq k\leq n}$ such that ${d(f^{-k}x,f^{-k}y) > \varepsilon}$;
2. let ${E \subset V_q^s}$ be a maximal ${(n,\varepsilon)}$-separated set and observe that ${\bigcup_{x\in E} B_n^s(x,\varepsilon) = V_q^s}$, while the sets ${\{B_n^s(x,\varepsilon/2) : x\in E \}}$ are disjoint;
3. enumerate ${E}$ as ${x_1,\dots, x_\ell}$, and build an ${\varepsilon}$-adapted partition ${\eta_n^s}$ by considering the sets

$\displaystyle C_j = B_n^s(x_j,\varepsilon/2) \cup \Big( V_q^s \setminus \bigcup_{i>j} B_n^s(x_i,\varepsilon)\Big). \ \ \ \ \ (9)$

Lemma 5 Let ${\mu}$ be a measure satisfying the Gibbs property (1). Then there are ${\varepsilon > 0}$ and ${K_0>0}$ such that if ${\eta_m^u}$ and ${\eta_n^s}$ are ${\varepsilon}$-adapted partitions of ${V_q^u}$ and ${V_q^s}$, then the product partition ${\xi_{m,n}}$ defined in (8) satisfies

$\displaystyle K_0^{-1} \leq \frac{\mu(\xi_{m,n}(x))}{e^{-(n+m)P + \sum_{k=-n}^{m-1} \varphi(f^k x)}} \leq K_0 \ \ \ \ \ (10)$

for every ${x\in R}$.

Proof: First we show that the upper bound in (10) holds whenever ${\varepsilon>0}$ is sufficiently small. Fix ${\varepsilon'>0}$ such that the Gibbs bound (1) holds for Bowen balls of radius ${2\varepsilon'}$, and such that ${\varepsilon'}$ is significantly smaller than the size of any local stable or unstable leaf. It suffices to show that for every ${x\in R}$ we have ${\xi_{m,n}(x) \subset B_{-n,m}(y,2\varepsilon') := f^{-n}(B_{n+m}(f^n y, 2\varepsilon'))}$ for some ${y}$, since then (1) gives the upper bound, possibly with a different constant; note that replacing ${x}$ with ${y}$ in the denominator changes the quantity by at most a constant factor, using the fact that ${\varphi}$ is Hölder continuous together with some basic properties of Anosov maps. (See, for example, Section 2 of this previous post for a proof of this Bowen property.)

Given ${x,y\in M}$ such that ${x}$ and ${y}$ lie on the same local stable leaf with ${d(x,y) \geq \varepsilon'}$, let

$\displaystyle r^u(x,y) = \inf\{d(x',y') : x'\in W_x^u, y'\in W_y^u \cap W_{x'}^s\}$

be the closest that any holonomy along local unstable leaves can bring ${x}$ and ${y}$. Note that ${r^u}$ is positive and continuous in ${x}$ and ${y}$; by compactness there is ${\bar{r}^u > 0}$ such that ${r^u(x,y) \geq \bar{r}^u}$ for all ${x,y}$ as above. In particular, this means that if ${x,y}$ are the images under an (unstable) holonomy of some ${x',y'}$ with ${d(x',y') < \bar{r}^u}$, then we must have ${d(x,y) <\varepsilon}$.

Choose ${\bar{r}^s}$ similarly for stable holonomies, and fix ${\varepsilon <\min(\bar{r}^s,\bar{r}^u)}$. Fix ${y\in V_q^u}$, ${z\in V_q^s}$, ${y'\in B_m^u(y,\varepsilon)}$, and ${z'\in B_n^s(z,\varepsilon)}$. Then for every ${-n\leq k\leq m}$ we have

$\displaystyle d(f^k[y',z'],f^k[y,z]) \leq d(f^k[y',z'],f^k[y',z]) + d(f^k[y',z],f^k[y,z]). \ \ \ \ \ (11)$

These distances can be estimated by observing that ${f^k[y',z']}$ and ${f^k[y',z]}$ are the images of ${f^k z'}$ and ${f^kz}$ under a holonomy map along local unstables, and similarly ${f^k[y',z]}$ and ${f^k[y,z]}$ are images of ${y}$ and ${y'}$ under a holonomy map along local stables. Then our choice of ${\varepsilon}$ shows that both quantities in the right-hand side of (11) are ${\leq \varepsilon'}$, which gives the inclusion we needed. This proves the upper bound in (10); the proof of the lower bound is similar. $\Box$

4. A refining sequence of adapted partitions

Armed with the formula from Lemma 5, we may look at the characterization of conditional measures in (6) and try to prove that the absolute continuity bound (7) holds whenever ${x,y\in R}$ both satisfy (6). There is one problem before we do this, though; the formula in (6) requires that the sequence of partitions ${\eta_n^s}$ be refining, and there is no a priori reason to expect that the adapted partitions produced by the simple argument before Lemma 5 refine each other. To get this additional property, we must do some more work. To the best of my knowledge, the arguments here are new.

4.1. Strategy

Start by letting ${E_n \subset V_q^s}$ be a maximal ${(n, 10\varepsilon)}$-separated set. We want to build a refining sequence of adapted partitions using the sets ${E_n}$, where instead of using Bowen balls with radius ${\varepsilon/2}$ and ${\varepsilon}$ we will use Bowen balls with radius ${\varepsilon}$ and ${20\varepsilon}$; this does not change anything essential about the previous section. We cannot immediately proceed as in the argument before Lemma 5, because if we are not careful when choosing the elements of the partition ${\eta_n^s}$, their boundaries might cut the Bowen balls ${B_{n'}^s(x,\varepsilon)}$ for ${x\in E_{n'}}$ and ${n' > n}$, spoiling our attempt to build an adapted partition ${\eta_{n'}^s}$ that refines ${\eta_n^s}$.

A first step will be to only build ${\eta_n^s}$ for some values of ${n}$. More precisely, we fix ${\ell\in {\mathbb N}}$ such that ${C\lambda^\ell < \frac 13}$, so that for every ${x\in M}$ and ${y\in W_x^s}$, we have ${d(f^\ell x, f^\ell y) < \frac 13 d(x,y)}$; then writing

$\displaystyle d_n^s(x,y) := \max_{0\leq k \leq n} d(f^{-k}x, f^{-k}y), \ \ \ \ \ (12)$

we have the following for every ${n,t\in {\mathbb N}}$:

$\displaystyle d_n^s(x,y) \leq 3^{-t} d_{n+t\ell}^s(x,y) \text{ whenever } d_{n+t\ell}^s(x,y) < \varepsilon. \ \ \ \ \ (13)$

We will only build adapted partitions for those ${n}$ that are multiples of ${\ell}$. We need to understand when the Bowen balls associated to points in the sets ${E_n}$ can overlap. We write ${\mathcal{E} = \{(x,n) \in V_q^s \times \ell{\mathbb N} : x\in E_n \}}$.

Definition 6 An ${\varepsilon}$-path between ${(x,n)\in \mathcal{E}}$ and ${(x',n')\in \mathcal{E}}$ is a sequence ${\{(z_i, k_i) \in \mathcal{E} : 0\leq i\leq m \}}$ with ${(z_0,k_0) = (x,n)}$ and ${(z_m,k_m) = (x',n')}$ such that ${B_{k_i}^s(z_i,\varepsilon) \cap B_{k_{i+1}}^s(z_{i+1},\varepsilon) \neq \emptyset}$ for all ${0\leq i < m}$. A subset ${\mathcal{C} \subset \mathcal{E}}$ is ${\varepsilon}$-connected if there is an ${\varepsilon}$-path between any two elements of ${\mathcal{C}}$.

Given ${n\in {\mathbb N}}$, let ${\mathcal{E}_{\geq n} = \{(x,k)\in \mathcal{E} : k\geq n\}}$. In the next section we will prove the following.

Proposition 7 If ${\mathcal{C} \subset \mathcal{E}_{\geq n}}$ is ${\varepsilon}$-connected and ${(x,n)\in \mathcal{C}}$, then

$\displaystyle U(\mathcal{C}) := \bigcup_{(y,k) \in \mathcal{C}} B_k^s(y,\varepsilon) \subset B_n^s(x,5\varepsilon). \ \ \ \ \ (14)$

For now we show how to build a refining sequence of adapted partitions assuming (14) by modifying the construction in (9). The key is that we build our partitions ${\eta_n^s}$ so that for every ${\varepsilon}$-connected ${\mathcal{C} \subset \mathcal{E}_{\geq n}}$, the set ${U(\mathcal{C})}$ is completely contained in a single element of ${\eta_n^s}$.

Suppose we have built a partition ${\eta_{n-\ell}^s}$ with this property; we need to construct a partition ${\eta_n^s}$ that refines ${\eta_{n-\ell}^s}$, still has this property, and also has the property that every partition element ${A}$ has some ${(x,n)\in \mathcal{E}}$ such that ${B_n^s(x,\varepsilon) \subset A \subset B_n^s(x,20\varepsilon)}$. To this end, let ${Z}$ be an element of the partition ${\eta_{n-\ell}^s}$, and enumerate the ${\varepsilon}$-connected components of ${\mathcal{E}_{\geq n}}$ as ${\mathcal{C}_1,\dots, \mathcal{C}_m, \mathcal{D}_1,\mathcal{D}_2,\dots}$, where each ${\mathcal{C}_j}$ contains some ${(x_j,n)}$, while each ${\mathcal{D}_i}$ is in fact a subset of ${\mathcal{E}_{\geq n+\ell}}$.

It follows from (14) that ${U(\mathcal{C}_j) \subset B_n^s(x_j,5\varepsilon)}$ for all ${1\leq j\leq m}$. Given ${i\in {\mathbb N}}$, we can observe that ${U(\mathcal{D}_i) \subset B_{n_i}^s(y_i,5\varepsilon) \subset B_n^s(y_i,5\varepsilon)}$ for some ${(y_i,n_i)\in \mathcal{E}}$ with ${n_i > n}$. Moreover, the sets ${\{B_n^s(x_j,10\varepsilon) : 1\leq j\leq m\}}$ cover ${Z}$, so every ${U(\mathcal{D}_i)}$ must intersect some set ${B_n^s(x_j,10\varepsilon)}$, and hence be contained in some ${B_n^s(x_j,20\varepsilon)}$. Given ${i\in {\mathbb N}}$, let ${J(i) = \min \{j : U(\mathcal{D}_i) \subset B_n(x_i,20\varepsilon)\}}$. Then for each ${1\leq j\leq m}$, let ${I_j = \{i\in {\mathbb N} : J(i) = j\}}$. Define sets ${A_1,\dots, A_m}$ by

$\displaystyle A_j = U(\mathcal{C}_j) \cup \Big( \bigcup_{i\in I_j} U(\mathcal{D}_i) \Big) \cup \Big( Z\setminus \bigcup_{j' > j} B_n(x_{j'},20\varepsilon) \Big).$

It is not hard to verify that the sets ${A_j}$ form a partition of ${Z}$ such that ${B_n^s(x_j,\varepsilon) \subset A_j \subset B_n^s(x_j,20\varepsilon)}$, and such that every ${\varepsilon}$-connected component ${\mathcal{C}}$ of ${\mathcal{E}_{\geq n}}$ that has ${U(\mathcal{C}) \cap Z \neq\emptyset}$ is completely contained in some ${A_j}$. Repeating this procedure for the other elements of ${\eta_{n-\ell}^s}$ produces the desired ${\eta_n^s}$.

4.2. Proof of the proposition

Now we must prove Proposition 7. Let ${\mathcal{E} \subset V_q^s\times \ell{\mathbb N}}$ be as in the previous section, so that in particular, every ${(x,n)\in \mathcal{E}}$ has that ${n}$ is a multiple of ${\ell}$, and given any ${(x,n)\neq (y,n)\in \mathcal{E}}$ we have ${d_n^s(x,y) \geq 10\varepsilon}$.

The following concept is essential for our proof.

Definition 8 Given an ${\varepsilon}$-path ${\{(z_i,k_i)\}_{i=0}^m \subset \mathcal{E}}$, a ford is a pair ${0\leq i k_i}$ for all ${i. Think of “fording a river'' to get from ${B_{k_i}^s(z_i,\varepsilon)}$ to ${B_{k_j}^s(z_j,\varepsilon)}$ by going through deeper levels of the ${\varepsilon}$-path; the alternative is that ${k_a \leq k_i}$ for some ${i, in which case ${B_{k_a}^s(z_a,\varepsilon)}$ is a sort of “bridge'' between ${z_i}$ and ${z_j}$.

Lemma 9 Suppose that ${\{(z_i,k_i)\}_{i=0}^m \subset \mathcal{E}}$ is an ${\varepsilon}$-path without any fords, and that ${k_i \geq k_0}$ for all ${1\leq i \leq m}$. Then ${d_{k_0}^s(z_0,z_m) \leq 4\varepsilon}$.

Proof: By the definition of ${\varepsilon}$-path, for every ${0\leq a < m}$ there is a point ${y_a \in B_{k_a}^s(z_a,\varepsilon) \cap B_{k_{a+1}}^s(z_{a+1},\varepsilon)}$. Thus

\displaystyle \begin{aligned} d_{k_0}^s(z_0,z_m) &\leq d_{k_0}^s(z_0,y_0) + \Big( \sum_{a=1}^{m-1} d_{k_0}^s(y_{a-1}, z_a) + d_{k_0}^s(z_a,y_a) \Big) + d_{k_0}^s(y_{m-1},z_m) \\ &\leq 2\varepsilon + \sum_{a=1}^{m-1} (\tfrac 13)^{k_a - k} \big( d_{k_a}^s(y_{a-1}, z_a) + d_{k_a}^s(z_a,y_a) \big) \\ &\leq 2\varepsilon \Big( 1 + \sum_{a=i+1}^{j-1} (\tfrac13)^{k_a - k} \Big), \end{aligned} \ \ \ \ \ (15)

where the second inequality uses (13). Now we need to estimate how often different values of ${k_a}$ can appear. Let ${A = \{1, \dots, m-1\}}$; we claim that for every ${t\in {\mathbb N}}$, we have

$\displaystyle \#\{a\in A : k_a = k_0 + t\ell \} \leq 2^{t-1}. \ \ \ \ \ (16)$

First note that ${k_a > k_0}$ for all ${a\in A}$, because otherwise ${(0,a)}$ would be a ford. Since the ${\varepsilon}$-path has no fords, every ${i with ${k_i = k_j}$ must be separated by some ${a}$ with ${k_a < k_i}$, and we conclude that for every ${t\in {\mathbb N}}$, we have

\displaystyle \begin{aligned} \#\{a \in A : &\, k_a = k_0 + t\ell\} \leq 1 + \#\{a\in A : k_0 < k_a < k_0 + t\ell\} \\ &= 1 + \sum_{q=1}^{t-1} \#\{a\in A : k_a = k_0 + q\ell\}. \end{aligned}

For ${t=1}$, this establishes (16) immediately. If (16) holds up to ${t-1}$, then this gives

$\displaystyle \#\{a\in A : k_a = k_0 + t\ell\} \leq 1 + \sum_{q=1}^{t-1} 2^{q-1} = 2^{t-1},$

which proves (16) for all ${t\in {\mathbb N}}$ by induction. Combining (15) and (16), we have

\displaystyle \begin{aligned} d_k^s(z_i,z_j) &\leq 2\varepsilon \Big( 1 + \sum_{t=1}^\infty \Big(\frac13 \Big)^t \#\{a\in A : k_a = k_i + t\ell\}\Big) \\ &\leq 2\varepsilon + \varepsilon \sum_{t=1}^\infty \Big(\frac23 \Big)^t =4\varepsilon, \end{aligned}

which proves Lemma 9. $\Box$

Now we show that the ${10\varepsilon}$-separation condition rules out the existence of fords entirely.

Lemma 10 If ${\{(z_i,k_i)\}_{i=0}^m \subset \mathcal{E}}$ is an ${\varepsilon}$-path, then it has no fords.

Proof: Fix an ${\varepsilon}$-path ${\{(z_i,k_i)\}_{i=0}^m}$. Denote the set of fords by

$\displaystyle I = \{(i,j)\in \{0,1,\dots,m\}^2 : i < j \text{ and } k_i = k_j \leq k_a \text{ for all } i

our goal is to prove that ${I}$ is empty. Put a partial ordering on ${I}$ by writing

$\displaystyle (i',j') \preceq (i,j)\ \Leftrightarrow\ [i',j'] \subset [i,j] \ \Leftrightarrow\ i \leq i' < j' \leq j.$

If ${I}$ is non-empty, then since it is finite it must contain some element ${(i,j)}$ that is minimal with respect to this partial ordering. In particular, the ${\varepsilon}$-path ${(z_i,k_i),\dots, (z_j,k_j)}$ contains no fords, and so by Lemma 9 we have ${d_{k_i}^s(z_i,z_j) \leq 4\varepsilon}$, contradicting the assumption that ${d_{k_i}^s(z_i,z_j) \geq 10\varepsilon}$ (since ${E_k}$ is ${(k,10\varepsilon)}$-separated), and we conclude that ${I}$ must be empty, proving the lemma. $\Box$

Now we prove Proposition 7. Let ${\mathcal{C} \subset \mathcal{E}_{\geq n}}$ be ${\varepsilon}$-connected and fix ${(x,n) \in \mathcal{C}}$. Given any ${(y,k) \in \mathcal{C}}$, there is an ${\varepsilon}$-path ${\{(z_i,k_i)\}_{i=0}^m}$ such that ${(z_0,k_0) = (x,n)}$ and ${(z_m, k_m) = (y,k)}$. By Lemma 10, this path has no fords, and so Lemma 9 gives ${d_n^s(x,y) \leq 4\varepsilon}$. We conclude that ${B_n^s(y,\varepsilon) \subset B_n^s(x,5\varepsilon)}$, which proves Proposition 7.

5. Completion of the proof

Thanks to the previous section, we can let ${\eta_m^u}$ and ${\eta_n^s}$ be ${\varepsilon}$-adapted partitions of ${V_q^u}$ and ${V_q^s}$ such that ${\eta_{n'}^s}$ refines ${\eta_n^s}$ whenever ${n'>n}$ and ${n,n'}$ are both multiples of ${\ell}$. By Lemma 5, we have good lower and upper estimates on ${\mu(\xi_{m,n}(x))}$ for all ${x\in R}$, where ${\xi_{m,n}}$ is the product partition defined in (8). By (6), there is ${R'\subset R}$ with ${\mu(R\setminus R')=0}$ such that for every ${x\in R'}$, we have

$\displaystyle \int \psi\,d\mu_x^u = \lim_{n\rightarrow\infty} \frac 1{\mu(\xi_{0,n\ell}(x))}\int_{\xi_{0,n\ell}(x)} \psi\,d\mu \ \ \ \ \ (17)$

for every continuous ${\psi\colon M\rightarrow {\mathbb R}}$. The next step towards proving Theorem 3 is the following result.

Proposition 11 There is a constant ${K_1}$ such that for any ${x,y\in R'}$ and every continuous ${\psi\colon V_y^u \rightarrow (0,\infty)}$, we have ${\int \psi\,d(\pi_{x,y})_*\mu_x^u \leq K_1 \int\psi\,d\mu_y^u}$, where ${\pi_{x,y} \colon V_x^u \rightarrow V_y^u}$ is the holonomy map along local unstables.

Proof: Start by fixing for each ${m\in {\mathbb N}}$ a set ${D_m^q \subset V_q^u}$ such that every element of ${\eta_m^s}$ contains exactly one point in ${D_m^q}$. Then let ${D_m^x = \pi_{q,x}D_m^q}$ for every ${x\in R}$.

Given a positive continuous function ${\psi}$, there is ${\delta>0}$ such that if ${d(z,z')<\delta}$, then ${\psi(z) \leq 2\psi(z')}$. Thus for all sufficiently large ${m,n}$, we have ${\psi(z) \leq 2\psi(z')}$ whenever ${\xi_{m,n\ell}(z) = \xi_{m,n\ell}(z')}$. For any such ${m,n}$ and any ${x,y\in R'}$, we conclude that

\displaystyle \begin{aligned} \int_{\xi_{0,n\ell}(x)} \psi\circ \pi_{x,y}\,d\mu &= \sum_{z\in D_m^x} \int_{\xi_{m,n\ell}(z)} \psi\circ \pi_{x,y}\,d\mu \leq \sum_{z\in D_{m}^x} 2\psi(\pi_{x,y}z) \mu(\xi_{m,n\ell}(z)) \\ &\leq \sum_{z\in D_{m}^x} 2\psi(\pi_{x,y}z) K_0 e^{-(n+m)P + S_{n,m}\varphi(z)}, \end{aligned}

where we write ${S_{n,m}\varphi(z) = \sum_{k=-n\ell}^{m-1} \varphi(f^kz)}$ and where the last inequality uses Lemma 5. Similarly,

$\displaystyle \mu(\xi_{0,n\ell}(x)) = \sum_{z\in D_m^x} \mu(\xi_{m,n\ell}(z)) \geq K_0^{-1} e^{-(n+m) P + S_{n,m}\varphi(f^kz)},$

and we conclude from (6) that

$\displaystyle \int \psi\circ \pi_{x,y} \,d\mu_x^u \leq \varlimsup_{n\rightarrow\infty} \frac{\sum_{z\in D_m^x} 2\psi(\pi_{x,y}z) K_0 e^{S_{n,m}\varphi(z)}}{\sum_{z\in D_m^x} K_0^{-1} e^{S_{n,m}\varphi(z)}}. \ \ \ \ \ (18)$

Note that since ${\varphi}$ is Hölder continuous, there are ${\beta>0}$ and ${|\varphi|_\beta > 0}$ such that ${|\varphi(z)-\varphi(z')| \leq |\varphi|_\beta d(z,z')^\beta}$ for all ${z,z'\in M}$. Thus given any ${z\in V_x^u}$ and any ${n\geq 0}$, we have

\displaystyle \begin{aligned} \Big|\sum_{k=-n\ell}^{-1} \varphi(f^kx) &- \sum_{k=-n\ell}^{-1} \varphi(f^kz)\Big| \leq \sum_{k=1}^{n\ell} |\varphi(f^kx) - \varphi(f^kz)| \\ &\leq \sum_{k=1}^{n\ell} |\varphi|_\beta (C\lambda^k \varepsilon)^\beta < |\varphi|_\beta (C\varepsilon)^\beta \sum_{k=1}^\infty (\lambda^\beta)^k =: L < \infty. \end{aligned} \ \ \ \ \ (19)

Thus (18) gives

\displaystyle \begin{aligned} \int \psi \,d(\pi_{x,y})_*\mu_x^u &\leq 2K_0^2 \varlimsup_{n\rightarrow\infty} \frac{\sum_{z\in D_m^x} \psi(\pi_{x,y}z) e^{L \sum_{k=-n\ell}^{-1} \varphi(f^k x)} e^{S_m\varphi(z)}} {\sum_{z\in D_m^x} e^{L^{-1} \sum_{k=-n\ell}^{-1} \varphi(f^k x)} e^{S_m\varphi(z)}} \\ &\leq 2K_0^2 e^{2L} \frac{\sum_{z\in D_m^x} \psi(\pi_{x,y}z) e^{S_m\varphi(z)}}{\sum_{z\in D_m^x} e^{S_m\varphi(z)}}. \end{aligned} \ \ \ \ \ (20)

A similar set of computations for ${\mu_y^u}$ shows that

$\displaystyle \int\psi\,d\mu_y^u \geq \frac 1{2K_0^2 e^{2L}} \frac{\sum_{z'\in D_m^y} \psi(z') e^{S_m\varphi(z')}}{\sum_{z'\in D_m^y} e^{S_m\varphi(z')}}. \ \ \ \ \ (21)$

Since ${D_m^y = \pi_{x,y} D_m^x}$, we can rewrite the sums over ${D_m^y}$ as sums over ${D_m^x}$; for example,

$\displaystyle \sum_{z' \in D_m^y} e^{S_{m}\varphi(z')} = \sum_{z\in D_m^x} e^{S_{m}\varphi(\pi_{x,y} z)} \leq e^L \sum_{z\in D_m^x} e^{S_m\varphi(z)},$

where the inequality uses the fact that the estimate (19) also holds for forward ergodic averages of two points on the same local stable manifold. Using a similar estimate for the numerator in (21) gives

$\displaystyle \int \psi\,d\mu_y^u \geq \frac 1{2K_0^2 e^{4L}} \frac{\sum_{z\in D_m^x} \psi(\pi_{x,y}z) e^{S_m\varphi(z)}}{\sum_{z\in D_m^x} e^{S_m\varphi(z)}}.$

Together with (20), this gives

$\displaystyle \int \psi \,d(\pi_{x,y})_*\mu_x^u \leq 4K_0^4 e^{6L} \int\psi\,d\mu_y^u,$

which completes the proof of Proposition 11. $\Box$

To complete the proof of Theorem 3, we first observe that for every open set ${U \subset V_y^u}$, there is a sequence of continuous functions ${\psi_n\colon V_y^u \rightarrow (0,1]}$ that converge pointwise to the indicator function ${\mathbf{1}_U}$; applying Proposition 11 to these functions and using the dominated convergence theorem gives

\displaystyle \begin{aligned} d(\pi_{x,y})_*\mu_x^u(U) &= \int \mathbf{1}_U \,d(\pi_{x,y})_*\mu_x^u = \lim_{n\rightarrow\infty} \int \psi_n \,d(\pi_{x,y})_*\mu_x^u \\ &\leq \lim_{n\rightarrow\infty} K_1 \int\psi_n \,d\mu_y^u = K_1 \int \mathbf{1}_U \,d\mu_y^u = K_1\mu_y^u(U). \end{aligned}

Then for every measurable ${E \subset V_y^u}$, we have

\displaystyle \begin{aligned} d(\pi_{x,y})_*\mu_x^u(E) &= \inf \{ d(\pi_{x,y})_*\mu_x^u(U) : U \supset E \text{ is open} \} \\ &\leq K_1 \inf \{ \mu_y^u(U) : U \supset E \text{ is open} \} = K_1 \mu_y^u(E). \end{aligned}

This proves that ${d(\pi_{x,y})_*\mu_x^u \ll \mu_y^u}$ and that the Radon–Nikodym derivative is ${\leq K_1}$ ${\mu_y^u}$-a.e. The lower bound ${K_1^{-1}}$ follows since the argument is symmetric in ${x}$ and ${y}$. This proves (7) and thus completes the proof of Theorem 3.