[proofplan] We construct the Bowen-Ruelle measure as the equilibrium state of the unstable-Jacobian potential $-\log J^u f$ on the transitive hyperbolic basic set $\Lambda$. The Bowen-Ruelle-Sinai theorem for $C^{1+\alpha}$ Axiom A attractors identifies this equilibrium state with the unique SRB measure and also gives the required full unstable-volume set of typical points on every local unstable plaque. Absolute continuity of stable holonomy then saturates this leafwise full-measure set to a full ambient-volume subset of an attracting neighbourhood, and finite-time invariance of Birkhoff averages pulls the result back to the whole topological basin. [/proofplan]

[step:Define the unstable-Jacobian potential on the attractor] Because $\Lambda$ is a hyperbolic set for the $C^{1+\alpha}$ diffeomorphism $f: M \to M$, the tangent bundle over $\Lambda$ has a continuous $df$-invariant splitting \begin{align*} T_\Lambda M = E^s \oplus E^u. \end{align*} The unstable distribution $E^u$ is Hölder continuous on $\Lambda$ for a $C^{1+\alpha}$ hyperbolic diffeomorphism. For each $x \in \Lambda$, define the unstable Jacobian $J^u f(x)$ by \begin{align*} J^u f(x) := |\det(df_x|_{E^u_x}: E^u_x \to E^u_{f(x)})|, \end{align*} where the determinant is computed using the Riemannian volume densities induced by $g$ on the unstable subspaces. This defines a map \begin{align*} J^u f: \Lambda \to (0,\infty). \end{align*} The Hölder regularity of $df$ and of $E^u$ implies that $J^u f$ is Hölder continuous. Define \begin{align*} \varphi^u: \Lambda \to \mathbb{R}, \quad x \mapsto -\log J^u f(x). \end{align*} Since $J^u f$ is positive and Hölder continuous on compact $\Lambda$, the potential $\varphi^u$ is Hölder continuous. [/step]

[step:Apply thermodynamic formalism on the transitive basic set] We use the standard thermodynamic-formalism theorem for transitive Axiom A basic sets: if $X$ is a transitive Axiom A basic set for a $C^1$ diffeomorphism and $\phi: X \to \mathbb{R}$ is Hölder continuous, then there is a unique invariant Borel probability measure attaining the topological pressure [variational principle](/theorems/7763) for $\phi$. The hypotheses apply with $X=\Lambda$ and $\phi=\varphi^u$: by assumption $\Lambda$ is a transitive Axiom A basic set, and the preceding step proves that $\varphi^u$ is Hölder continuous. Hence there exists a unique $f|_\Lambda$-invariant Borel probability measure $\mu_\varphi$ on $\Lambda$ such that \begin{align*} h_{\mu_\varphi}(f|_\Lambda) + \int_\Lambda \varphi^u(x)\,d\mu_\varphi(x) = P(f|_\Lambda,\varphi^u), \end{align*} where $h_{\mu_\varphi}(f|_\Lambda)$ denotes measure-theoretic entropy and $P(f|_\Lambda,\varphi^u)$ denotes topological pressure. We call $\mu_\varphi$ the equilibrium state for $\varphi^u$. [/step]

[step:Identify the equilibrium state as the unique SRB measure]We use the following standard Bowen-Ruelle-Sinai theorem for $C^{1+\alpha}$ Axiom A attractors. Let $A$ be a transitive Axiom A attractor for a $C^{1+\alpha}$ diffeomorphism, and let $\phi^u=-\log J^u f$ be its unstable-Jacobian potential. Then the unique equilibrium state for $\phi^u$ has conditional measures on measurable partitions subordinate to local unstable plaques equivalent to the induced unstable Riemannian volumes. Conversely, every invariant probability measure supported on $A$ with absolutely continuous conditional measures on such unstable plaques is this equilibrium state. Moreover, this measure is ergodic and, in every sufficiently small local product rectangle for $A$, its Birkhoff basin meets each local unstable plaque in a set of full induced unstable volume. The hypotheses of this theorem are satisfied: $f$ is $C^{1+\alpha}$, $\Lambda$ is a transitive Axiom A attractor, and $\varphi^u=-\log J^u f$ is the Hölder unstable-Jacobian potential constructed above. Therefore the conditional measures of $\mu_\varphi$ on subordinate unstable plaques are absolutely continuous with respect to the induced unstable Riemannian volumes, so $\mu_\varphi$ is an SRB measure supported on $\Lambda$. Define \begin{align*} \mu_{\mathrm{SRB}} := \mu_\varphi. \end{align*} If $\nu$ is any other SRB probability measure supported on $\Lambda$, the converse part of the same theorem makes $\nu$ an equilibrium state for $\varphi^u$. The uniqueness of the equilibrium state gives \begin{align*} \nu = \mu_\varphi = \mu_{\mathrm{SRB}}. \end{align*} Thus the SRB measure supported on $\Lambda$ is unique.[/step]

[guided]The point of introducing $\varphi^u=-\log J^u f$ is that this potential exactly compensates for expansion of Riemannian volume along unstable directions. We now invoke the standard Bowen-Ruelle-Sinai theorem in the precise form needed here. The theorem says that for a $C^{1+\alpha}$ diffeomorphism and a transitive Axiom A attractor, the equilibrium state of the unstable-Jacobian potential has unstable conditional measures equivalent to induced Riemannian volume on unstable plaques; it also says that any invariant probability measure with this unstable absolute-continuity property must be the same equilibrium state. We verify the hypotheses. The map $f: M \to M$ is $C^{1+\alpha}$ by the theorem statement. The set $\Lambda$ is a transitive Axiom A attractor by hypothesis. In the first step we defined \begin{align*} \varphi^u: \Lambda \to \mathbb{R}, \quad x \mapsto -\log J^u f(x), \end{align*} and proved that it is Hölder continuous. Therefore the Bowen-Ruelle-Sinai theorem applies to $f|_\Lambda$ and $\varphi^u$. The theorem gives that the conditional measures of the equilibrium state $\mu_\varphi$ on measurable partitions subordinate to local unstable plaques are equivalent to the Riemannian volume measures induced by $g$ on those plaques. Equivalence of measures is stronger than absolute continuity, and the SRB definition in the statement only asks for absolute continuity. Hence $\mu_\varphi$ is an SRB probability measure supported on $\Lambda$. We set \begin{align*} \mu_{\mathrm{SRB}} := \mu_\varphi. \end{align*} Now suppose $\nu$ is another SRB probability measure supported on $\Lambda$. By the definition of SRB measure used in the statement, the conditional measures of $\nu$ on subordinate local unstable plaques inside $\Lambda$ are absolutely continuous with respect to the induced unstable Riemannian volumes. The converse direction of the Bowen-Ruelle-Sinai theorem applies to this measure and concludes that $\nu$ is an equilibrium state for the same potential $\varphi^u$. But thermodynamic formalism on the transitive basic set $\Lambda$ gives only one equilibrium state for this Hölder potential. Therefore \begin{align*} \nu = \mu_\varphi = \mu_{\mathrm{SRB}}. \end{align*} This proves both existence and uniqueness of the SRB measure supported on $\Lambda$.[/guided]

[step:Obtain full volume in a local attracting neighbourhood] Choose finitely many sufficiently small local product rectangles $R_1,\dots,R_q \subset M$ covering $\Lambda$. For each index $i \in \{1,\dots,q\}$, the rectangle $R_i$ is parametrized by local unstable plaques and local stable plaques whose dimensions add to $\dim M$. For an unstable plaque $W \subset R_i$, let $m^u_W$ denote the Riemannian volume induced by $g$ on $W$. The local product coordinates are chosen so that the restriction $m|_{R_i}$ disintegrates into conditional measures equivalent to $m^u_W$ along unstable plaques and transverse measures equivalent to induced stable plaque volume. By the leafwise physicality part of the Bowen-Ruelle-Sinai theorem quoted above, for every such unstable plaque $W$ we have \begin{align*} m^u_W(W \setminus B(\mu_{\mathrm{SRB}})) = 0. \end{align*} Indeed, the theorem supplies convergence of Birkhoff averages to $\int_M \psi\,d\mu_{\mathrm{SRB}}$ for every $\psi \in C(M)$ on a full $m^u_W$-measure subset of each local unstable plaque. We also use the standard absolute [continuity theorem](/theorems/1145) for stable holonomies of $C^{1+\alpha}$ hyperbolic systems: in a local product rectangle, stable holonomy maps between local unstable plaques send induced unstable-volume null sets to induced unstable-volume null sets. Its hypotheses hold because $f$ is $C^{1+\alpha}$ and the rectangles lie in the hyperbolic set associated to the attractor $\Lambda$. Let $S_i$ be the stable saturation inside $R_i$ of $B(\mu_{\mathrm{SRB}}) \cap \Lambda$, namely \begin{align*} S_i := \{x \in R_i : x \in W^s_{\mathrm{loc}}(z) \text{ for some } z \in B(\mu_{\mathrm{SRB}}) \cap \Lambda \cap R_i\}. \end{align*} Stable contraction implies that $S_i \subset B(\mu_{\mathrm{SRB}})$. To see this, take $x \in W^s_{\mathrm{loc}}(z)$ with $z \in B(\mu_{\mathrm{SRB}})$. Then \begin{align*} \operatorname{dist}_g(f^k(x),f^k(z)) \to 0 \end{align*} as $k \to \infty$. If $\psi \in C(M)$, compactness of $M$ makes $\psi$ uniformly continuous, so \begin{align*} \frac{1}{N}\sum_{k=0}^{N-1}|\psi(f^k(x))-\psi(f^k(z))| \to 0. \end{align*} Thus $x$ and $z$ have the same Birkhoff limits for every $\psi \in C(M)$, and $x \in B(\mu_{\mathrm{SRB}})$. The full unstable-volume statement on every unstable plaque, the absolute continuity of stable holonomy, and the product disintegration of $m|_{R_i}$ imply \begin{align*} m(R_i \setminus S_i) = 0 \end{align*} for every $i \in \{1,\dots,q\}$. Since $\Lambda$ is an attractor, after shrinking the rectangles if necessary there exists an open neighbourhood $V$ of $\Lambda$ contained in $\bigcup_{i=1}^q R_i$ such that every point of $V$ lies on a local stable plaque through a point of $\Lambda$. Hence \begin{align*} m(V \setminus B(\mu_{\mathrm{SRB}})) = 0. \end{align*} Choose an attracting neighbourhood $U \subset V$ satisfying $f(U) \subset U$ and \begin{align*} \bigcap_{n=0}^{\infty} f^n(U)=\Lambda. \end{align*} Then \begin{align*} m(U \setminus B(\mu_{\mathrm{SRB}})) = 0. \end{align*} [/step]

[step:Pull the local full-measure basin back to the full topological basin] The attracting neighbourhood $U$ satisfies \begin{align*} B_{\mathrm{top}}(\Lambda)=\bigcup_{n=0}^{\infty} f^{-n}(U). \end{align*} If $x \in B_{\mathrm{top}}(\Lambda)$, then $f^n(x) \in U$ for some $n \geq 0$ because the forward orbit approaches $\Lambda \subset U$ and $U$ is a neighbourhood of $\Lambda$. Conversely, if $f^n(x) \in U$ for some $n \geq 0$, then $f^k(x)$ approaches $\Lambda$ because $f(U) \subset U$ and $\bigcap_{j=0}^{\infty} f^j(U)=\Lambda$. We record the finite-time invariance of the physical basin. If $n \geq 0$ and $f^n(x) \in B(\mu_{\mathrm{SRB}})$, then $x \in B(\mu_{\mathrm{SRB}})$. Indeed, for every $\psi \in C(M)$ and every $N>n$, \begin{align*} \frac{1}{N}\sum_{k=0}^{N-1}\psi(f^k(x)) - \frac{1}{N}\sum_{k=0}^{N-n-1}\psi(f^k(f^n(x))) = \frac{1}{N}\sum_{k=0}^{n-1}\psi(f^k(x)) - \frac{1}{N}\sum_{k=N-n}^{N-1}\psi(f^k(x)). \end{align*} The absolute value of the right-hand side is at most \begin{align*} \frac{2n\|\psi\|_{C^0(M)}}{N}, \end{align*} which tends to $0$ as $N \to \infty$. Since $f^n(x) \in B(\mu_{\mathrm{SRB}})$, the second Cesaro average converges to $\int_M \psi(y)\,d\mu_{\mathrm{SRB}}(y)$, and therefore the first Cesaro average has the same limit. This proves $x \in B(\mu_{\mathrm{SRB}})$. It follows that, for every $n \geq 0$, \begin{align*} f^{-n}(U) \setminus B(\mu_{\mathrm{SRB}}) \subset f^{-n}(U \setminus B(\mu_{\mathrm{SRB}})). \end{align*} Since $f^n: M \to M$ is a $C^{1+\alpha}$ diffeomorphism, it preserves the smooth volume null class. Hence \begin{align*} m(f^{-n}(U \setminus B(\mu_{\mathrm{SRB}})))=0 \end{align*} for every $n \geq 0$. Taking the countable union over $n$ gives \begin{align*} m(B_{\mathrm{top}}(\Lambda) \setminus B(\mu_{\mathrm{SRB}})) = 0. \end{align*} Together with the construction and uniqueness of $\mu_{\mathrm{SRB}}$, this proves the theorem. [/step]