## Tag Archives: unstable saturate

### Stable and unstable saturation

Let $(X,d)$ be a compact metric space, $f:X\to X$ be a homeomorphism preserving some probability measure $\mu$. The stable set $W^s(x)$ is defined by the formula
$W^s(x)=\{y\in X: d(f^nx,f^ny)\to 0\text{ as }n\to+\infty\}.$ For convenience denote the collection of null sets as $\mathcal{N}=\{N\subset X:\exists \tilde{N}\supset N, \mu(\tilde{N})=0\}$.

[Lemma 6.3.2. Brin M., Stuck G. Introduction to dynamical systems]
Let $\phi:X\to [0,1]$ be an $f$-invariant measurable function. Then there exists a subset $N\in\mathcal{N}$ such that for each $x\in X\backslash N$, $\phi$ is constant on $W^s(x)\backslash N$.

Proof. Pick a sequence of continuous function $\phi_k, k\ge1$ on $X$ such that
$\phi_k\to \phi$ both $L^1(\mu)$ and $\mu-a.e.$.

By Birkhoff Ergodic Theorem, the limit $\phi_k^+(x)=\lim_{n\to+\infty}\frac{1}{n}\sum_{i=0}^{n-1}\phi_k(f^ix)$ exists for $\mu-a.e.\; x.$ So there exists $N_k\in\mathcal{N}$ such that the limit exists for each $x\in X\backslash N_k$. Pick such a point $x$. Then for each $y\in W^s(x)$, the limit also exists at $y$ and has the same value as at $x: \phi_k^+(y)=\phi_k^+(x)$. (both $N_k,X\backslash N_k$ are stable-saturated)

By the invariance of $\mu$ and $\phi$, we have
$\int|\phi(y)-\frac{1}{n}\sum\limits_{i=0}^{n-1}\phi_k(f^iy)|d\mu(y)\le\frac{1}{n}\sum\limits_{i=0}^{n-1}\int|\phi(y)-\phi_k(f^iy)|d\mu(y)=\|\phi-\phi_k\|.$

By Dominated Convergent Theorem, $\|\phi-\phi_k^+\|\le\|\phi-\phi_k\|\to 0.$ Passing to a subsequence if necessary, we assume that $\phi_k^+\to\phi,\:\mu-a.e.\:x$. So there exists $N_+\in\mathcal{N}$ such that for each $x\in X\backslash N_+,$ the limit $\lim_{k\to\infty}\phi_k^+(x)$ exists and equals to $\phi(x).$

Now let $N=\bigcup_{k\ge1}N_k\cup N_+$. Then $N\in\mathcal{N}$ and for each point $x\in X\backslash N$, each point $y\in W^s(x)\backslash N$, $\phi_k^+(y)=\phi_k^+(x)$ for each $k\ge1$ and
1. the limits exist (hence are equal): $\lim_{k\to\infty}\phi_k^+(y)=\lim_{k\to\infty}\phi_k^+(x)$.
2. they equal to $\phi$ respectively: $\lim_{k\to\infty}\phi_k^+(x)=\phi(x),$ $\lim_{k\to\infty}\phi_k^+(y)=\phi(y)$.

So we have $\phi(y)=\phi(x)$ for each $y\in W^s(x)\backslash N$ and $x\in X\backslash N$.

A more interesting application:
Corollary 1. Let $(X,f)$, $\mu\in\mathcal{M}(f)$ and $E\subset X$ be an invariant subset. Then there exists a null set $N\in\mathcal{N}_\mu$ such that for each $x\in E\backslash N$, $W^s(x)\backslash N \subset E$. (equivalently, $W^s(x)\backslash E \subset N$)

Proof. The characteristic function $\chi_E$ is invariant and measurable. Let $N$ be given by Lemma. Then if $x\in E\backslash N$, we have $\chi_E(y)=\chi_E(x)=1$ for each $y\in W^s(x)\backslash N$. So $W^s(x)\backslash N\subset E$.

Similarly result holds for unstable sets. The results can be imporved if the stable/unstable foliations are absolutely continuous. This is the case for nonuniformly hyperbolic invariant sets.

Corollary 2. Let $f\in \mathrm{Diff}^r_\mu(M)$ for some $r>1$ as above. Let $\Lambda\subset M$ be an invariant nonuniform hyperbolic subset. Then there exists a null set $N\in\mathcal{N}_\mu$ such that for each $x\in \Lambda\backslash N$, $\mu_{W^s(x)}(W^s(x)\backslash \Lambda)=0$ and $\mu_{W^u(x)}(W^u(x)\backslash \Lambda)=0$.

Equivalently, we have for $\mu$-a.e. $x\in\Lambda$, $W^s(x)\subset \Lambda\; \mathrm{mod}\mu_{W^s(x)}$ and $W^u(x)\subset \Lambda\; \mathrm{mod}\mu_{W^u(x)}$.