diff --git a/Mathematics/5th/Advanced_dynamical_systems/Advanced_dynamical_systems.tex b/Mathematics/5th/Advanced_dynamical_systems/Advanced_dynamical_systems.tex
index 62554c5..c61f884 100644
--- a/Mathematics/5th/Advanced_dynamical_systems/Advanced_dynamical_systems.tex
+++ b/Mathematics/5th/Advanced_dynamical_systems/Advanced_dynamical_systems.tex
@@ -663,11 +663,15 @@
     $$
       \int_{\TT^1}\varphi(x)\dd{\mu}=\int_{\TT^1}\varphi(x)\dd{x}
     $$
-    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 $k\ne 0$:
+    First we consider the case of trigonometric polynomials, i.e. $P_n(x) =\sum_{k=-n}^na_k\exp{2\pi\ii k x}$, where $a_k = \bar a_{-k}$, 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}.
+    where the equality is due to the invariance of $\mu$. 
+    Thus $\int_{\TT^1}\varphi(x)\dd{\mu}=\int_{\TT^1}\varphi(x)\dd{x}$, and the result holds for trigonometric polynomials. 
+    In the general case of a continuous function $\phi \in \mathcal{C}(\TT^1)$, we recall that the set of trigonometric polynomials is dense in $\mathcal{C}(\TT^1)$.
+    Thus the result for the case of trigonometric polynomials passes to the limit and we have that $\int_{\TT^1}\varphi(x)\dd{\mu}=\int_{\TT^1}\varphi(x)\dd{x}$.
   \end{proof}
   \begin{proposition}\label{ADS:uniquely_ergodic}
     Let $F\in\Homeoplus(\TT^1)$ with $\rho(F)\notin\quot{\QQ}{\ZZ}$. Then, $F$ is uniquely ergodic.