diff --git a/Mathematics/5th/Advanced_dynamical_systems/Advanced_dynamical_systems.tex b/Mathematics/5th/Advanced_dynamical_systems/Advanced_dynamical_systems.tex
index 36333b9..6e4b978 100644
--- a/Mathematics/5th/Advanced_dynamical_systems/Advanced_dynamical_systems.tex
+++ b/Mathematics/5th/Advanced_dynamical_systems/Advanced_dynamical_systems.tex
@@ -556,7 +556,7 @@
   \begin{definition}
     A subset $C\subseteq \RR$ is a \emph{Cantor set} if it is closed, it has no isolated points and it has empty interior.
   \end{definition}
-  \begin{theorem}
+  \begin{theorem}\label{ADS:theorem_irrational_rotation_number}
     Let $F\in\Homeo(\TT^1)$ with $\rho(F)\notin\quot{\QQ}{\ZZ}$. Then, there exists a surjective continuous map $H:\TT^1\to \TT^1$ such that $H\circ F=R_{\rho(F)}\circ H$. Moreover, we have exactly one of the following two properties:
     \begin{enumerate}
       \item $F$ is conjugated to $R_{\rho(F)}$ and in that case $F$ is minimal.
@@ -568,10 +568,82 @@
     $$
       h(f(x))-h(f(0))=\mu([f(0),f(x)))=\mu([0,x))=h(x)
     $$
-    where we have used the invariance of $\mu$. Thus, $h\circ f=R_{h(f(0))}\circ h$ and necessarily we need $h(f(0))=\rho(R_{h(f(0))})=\rho(f)$, by the invariance of the rotation number by semi-conjugacy. This gives $H\circ F=R_{\rho(F)}\circ H$. Now, we can express the dichotomy as follows: either $\supp\mu=\TT^1$ or $\supp\mu=:X\subsetneq \TT^1$. The first case is equivalent to $h$ being strictly increasing and so $h$ is a homeomorphism. Then, $H$ conjugates $F$ and $R_{\rho(F)}$ and so $F$ is minimal because $R_{\rho(F)}$ is minimal. In the second case, we have that $X$ is a nonempty closed invariant set that has no isolated points because $\mu$ has no atoms. To show that $X$ is minimal, let $\TT^1=X\sqcup U$ with $U$ open, and so it can be written as a countable union of open intervals. Let $D\subseteq X$ be the set containing the endpoints of those intervals and let $$
-      Y=\{y\in\TT^1:H^{-1}(\{y\})\text{ is a closed interval}\}
+    where we have used the invariance of $\mu$. Thus, $h\circ f=R_{h(f(0))}\circ h$ and necessarily we need $h(f(0))=\rho(R_{h(f(0))})=\rho(f)$, by the invariance of the rotation number by semi-conjugacy. This gives $H\circ F=R_{\rho(F)}\circ H$. Now, we can express the dichotomy as follows: either $\supp\mu=\TT^1$ or $\supp\mu=:X\subsetneq \TT^1$. The first case is equivalent to $h$ being strictly increasing and so $h$ is a homeomorphism. Then, $H$ conjugates $F$ and $R_{\rho(F)}$ and so $F$ is minimal because $R_{\rho(F)}$ is minimal. In the second case, we have that $X$ is a nonempty closed invariant set that has no isolated points because $\mu$ has no atoms. To show that $X$ is minimal, let $\TT^1=X\sqcup U$ with $U$ open, and so it can be written as a countable union of open intervals. Let $D\subseteq X$ be the set containing the endpoints of those intervals and let
+    \begin{equation}\label{ADS:eq1}
+      Y:=\{y\in\TT^1:H^{-1}(\{y\})\text{ is a closed interval}\}
+    \end{equation}
+    $Y$ is countable. Now take $M\subseteq X$ be nonempty, closed and invariant. We want to prove that $M=X$. Then, $H(M)\subseteq \TT^1$ is nonempty, closed and invariant by $R_{\rho(F)}$. So $H(M)=\TT^1$ because $R_{\rho(F)}$ is minimal. Now, since $H$ restricted to $X\setminus D$ is injective, then $M\supseteq X\setminus D$. Indeed, by contradiction let $x\in X\setminus D$ such that $x\notin M$. We have $H(X\setminus D)=\TT^1\setminus Y$. Thus, $H^{-1}(H(X\setminus D))=H^{-1}(\TT^1\setminus Y)=X\setminus D$. Then, $H(x)\in H(X\setminus D)\subseteq \TT^1=H(M)$. So $\exists y\in M$ such that $H(x)=H(y)$, and so$y\in H^{-1}(H(X\setminus X))=X\setminus D$. But $H|_{X\setminus D}$ is injective, so $x=y\in M$, which is a contradiction because $x\notin M$. Thus, $M\supseteq X\setminus D$, which implies:
+    $$
+      M=\overline{M}\supseteq \overline{X\setminus D}=\overline{X} = X
+    $$
+    So $M=X$ and, thus, $X$ is minimal. Moreover, $X$ has empty interior. Indeed, if that wasn't the case, we would have $\Fr{X}=\overline{X}\setminus\Int(X)=X\setminus \Int(X)\subsetneq X$ and so $\Fr{X}$ would be a nonempty closed invariant set, which is not possible because $X$ is minimal. Finally, to prove $X=\Omega(F)$, by minimality it suffices to show that $\Omega(F)\subseteq X$. Let $x\in U$, where $\TT^1=X\sqcup U$, with $U=\bigsqcup_{i=1}^\infty I_i$ invariant and $I_i$ intervals. We need to see that $x$ is wandering. Let $I$ be one of such intervals. We may have either $F^n(I)=I$ for some $n\geq 1$ or $F^n(I)\cap I=\varnothing$ for all $n\geq 1$. But in the first case, we would have that the extremities of $I$ are periodic points, which is not possible because $\rho(F)\notin\quot{\QQ}{\ZZ}$. So we must have the second case, which implies that $I$ is a wandering domain, and thus so is $U$.
+  \end{proof}
+  \subsubsection{Unique ergodicity}
+  \begin{definition}
+    A homeomorphism $F:\TT^1\to\TT^1$ is \emph{uniquely ergodic} if it has a unique invariant probability measure.
+  \end{definition}
+  \begin{lemma}
+    If $\alpha\notin \QQ$, then $R_\alpha$ is uniquely ergodic and $\mathcal{M}_{R_\alpha}=\{\text{Leb}\}$.
+  \end{lemma}
+  \begin{proof}
+    Let $\mu\in\mathcal{M}_{R_\alpha}$. We want to see that $\forall \varphi\in \mathcal{C}(\TT^1)$:
+    $$
+      \int_{\TT^1}\varphi(x)\dd{\mu}=\int_{\TT^1}\varphi(x)\dd{x}
+    $$
+    An easy check shows that if $P_n=\sum_{k=-n}^na_k\exp{2\pi\ii k x}$ is a trigonometric polynomial, then $\int_{\TT^1}P_n(x)\dd{x}=a_0$. Moreover, if $k\ne 0$:
+    $$
+      \int_{\TT^1}\exp{2\pi\ii kx}\dd{\mu}=\exp{2\pi\ii k\alpha}\int_{\TT^1}\exp{2\pi\ii kx}\dd{\mu}\implies \int_{\TT^1}\exp{2\pi\ii kx}\dd{\mu}=0
+    $$
+    where the equality is due to the invariance of $\mu$. So, we also have $\int_{\TT^1}P_n(x)\dd{\mu}=a_0$. Now consider the Féjer trigonometric polynomial that converge uniformly to $\varphi$ and use the \mnameref{RFA:domianted}.
+  \end{proof}
+  \begin{proposition}
+    Let $F\in\Homeo(\TT^1)$ with $\rho(F)\notin\quot{\QQ}{\ZZ}$. Then, $F$ is uniquely ergodic.
+  \end{proposition}
+  \begin{proof}
+    Let $H$ be such that $H\circ F=R_\rho\circ H$ (by \mcref{ADS:theorem_irrational_rotation_number}) and so $F^{-1}(H^{-1}(A))=H^{-1}({R_\rho}^{-1}(A))$. Define:
+    $$
+      H_*\mu(A):=\mu(H^{-1}(A))\qquad \forall A\subseteq \TT^1\text{ Borel}
+    $$
+    We have:
+    \begin{multline*}
+      H_*\mu(A)=\mu(H^{-1}(A))=\mu(F^{-1}(H^{-1}(A)))=\\=\mu(H^{-1}({R_\rho}^{-1}(A)))=H_*\mu({R_\rho}^{-1}(A))
+    \end{multline*}
+    where the second equality is due to the invariance of $\mu$. Hence, $H_*\mu$ is invariant by $R_\rho$, and so $H_*\mu=\text{Leb}$. That, is $\mu(H^{-1}(A))=\text{Leb}(A)$. Recall again the set $Y$ of \mcref{ADS:eq1} and $\TT^1=X\sqcup U$, with $H^{-1}(Y)=\overline{U}$. Since $Y$ is countable, $0=\text{Leb}(Y)=\mu(H^{-1}(Y))$. Now since $H|_X$ is a homeomorphism (??is it true??), we have that $\mu(B)=\text{Leb}(H(B))$, and so $\mu$ is uniquely determined.
+  \end{proof}
+  \begin{proposition}
+    Let $F:\TT^1\to\TT^1$ be a homeomorphism. Then, $F$ is uniquely ergodic if and only if $\forall \varphi\in\mathcal{C}(\TT^1)$, $\exists c_\varphi\in\RR$ such that $\frac{1}{n}\sum_{i=0}^n\varphi\circ F^i$ converge uniformly to $c$.
+  \end{proposition}
+  \begin{proof}
+    Assume first that $\mathcal{M}_F=\{\mu\}$ and argue by contradiction. That, is $\exists \varepsilon>0$, $(n_k)\in\NN$ with $n_k\nearrow +\infty$ and $(x_k)\in\TT^1$ such that $\forall k\geq 0$:
+    \begin{equation}\label{ADS:nuk_mu}
+      \abs{\frac{1}{n_k}\sum_{i=0}^{n_k-1}\varphi\circ F^i(x_k)-\int_{\TT^1}\varphi\dd{\mu}}=\abs{\int_{\TT^1}\varphi\dd{\nu_k}-\int_{\TT^1}\varphi\dd{\mu}}>\varepsilon
+    \end{equation}
+    $$
+    $$
+    where $\nu_k=\frac{1}{n_k}\sum_{i=0}^{n_k-1}F_*^i\delta_{x_k}$. Note that $\nu_k\in\mathcal{\TT^1}$ and since $\mathcal{M}(\TT^1)$ is compact, after extracting a subsequence, $(\nu_k)$ converges to $\nu\in\mathcal{M}(\TT^1)$. Now, $\nu$ is invariant. Indeed:
+    $$
+      F_*\nu_k-\nu_k=\frac{1}{n_k}(F_*^{n_k}\delta_{x_k}-\delta_{x_k})\overset{k\to\infty}{\longrightarrow} 0
     $$
-    $Y$ is countable. Now take $M\subseteq X$ be nonempty, closed and invariant. We want to prove that $M=X$. Then, $H(M)\subseteq \TT^1$ is nonempty, closed and invariant by $R_{\rho(F)}$. But $R_{\rho(F)}$ is minimal, so $H(M)=\TT^1$.
+    because $\forall \varphi\in\mathcal{C}(\TT^1)$, $(F_*^{n_k}\delta_{x_k}-\delta_{x_k})(\varphi)$ is bounded by $2\norm{\varphi}$. So $\nu =\mu$, but this is a contradiction with \mcref{ADS:nuk_mu}. Now we prove the converse. Let $\mu\in\mathcal{M}_F(\TT^1)$. Then:
+    \begin{multline*}
+      c_\varphi =\int_{\TT^1}c_\varphi\dd{\mu} =\int_{\TT^1}\lim_{n\to\infty}\frac{1}{n}\sum_{i=0}^{n-1}\varphi\circ F^i\dd{\mu} =\\
+      =\lim_{n\to\infty}\frac{1}{n}\sum_{i=0}^{n-1}\int_{\TT^1}\varphi\circ F^i\dd{\mu}=\lim_{n\to\infty}\frac{1}{n}\sum_{i=0}^{n-1}\int_{\TT^1}\varphi\dd{\mu} =\int_{\TT^1}\varphi\dd{\mu}
+    \end{multline*}
+    where the third equality is due to the uniform convergence and the penultimate equality is due to the invariance of $\mu$. This implies that $\mu$ is uniquely determined.
   \end{proof}
+  \begin{definition}
+    For $k\in\NN\cup\{0\}$ we define the set $\mathcal{D}^k(\TT^1)$ as:
+    $$
+      \mathcal{D}^k(\TT^1):=\{f:\RR\to\RR \text{ $\mathcal{C}^k$-diffeomorphism such that }f(x+1)=f(x)+1\}
+    $$
+    Note that $f\in \mathcal{D}^k(\TT^1)$ if and only if $f=\id +\varphi$, with $\varphi\in\mathcal{C}^k(\TT^1)$.
+    We also define the set $\Diffplus^k(\TT^1)$ as:
+    $$
+      \Diffplus^k(\TT^1):=\{F:\TT^1\to\TT^1 \text{ $\mathcal{C}^k$-diffeomorphism with orientation preserving}\}
+    $$
+  \end{definition}
+  \begin{proposition}
+    Let $F\in\Diffplus^1(\TT^1)$ with $\rho(F)\notin\quot{\QQ}{\ZZ}$, $\mu$ be the unique invariant probability measure of $F$ and $f\in\mathcal{D}^1(\TT^1)$ be a lift of $F$. Then, $\displaystyle \lim_{n\to \infty}\frac{1}{n}\log Df^n(x)=\int_{\TT^1}\log(Df)\dd{\mu}=0$
+  \end{proposition}
 \end{multicols}
 \end{document}
\ No newline at end of file
diff --git a/preamble_formulas.sty b/preamble_formulas.sty
index 72e07f3..9385b5e 100644
--- a/preamble_formulas.sty
+++ b/preamble_formulas.sty
@@ -303,7 +303,8 @@
 \newcommand{\topo}{\tau} % symbol for the topology. Feasible options are: \tau, \mathcal{T}...
 \newcommand{\conn}{\mathrel{\#}} % connected sum. \mathrel gives the space of a relation (like +,-,...) while \mathbin gives the space of a binary operator (like =).
 \renewcommand{\S}{S} % S of the S ^ n (n-th dimensional sphere)
-\newcommand{\Homeo}{\mathrm{Homeo}^+}
+\newcommand{\Homeoplus}{\mathrm{Homeo}^+} % set of orientation-preserving homeomorphisms
+\newcommand{\Diffplus}{\mathrm{Diff}_+} % set of orientation-preserving diffeomorphisms
 
 %%% GALOIS THEORY
 \newcommand{\FF}{\ensuremath{\mathbb{F}}} % finite fields