modified thm

Mathematics/2nd/Mathematical_analysis/Mathematical_analysis.tex

@@ -175,7 +175,7 @@
   \begin{theorem}
     Let $(f_n)$ be a sequence of continuous functions defined on $D\subseteq\RR $. If $(f_n)$ converges uniformly to $f$ on $D$, then $f$ is continuous on $D$, that is, for any $x_0\in D$, it satisfies: $$\lim_{n\to\infty}\left(\lim_{x\to x_0} f_n(x)\right)=\lim_{x\to x_0} f(x)$$
   \end{theorem}
-  \begin{theorem}
+  \begin{theorem}\label{MA:uniformintegral}
     Let $(f_n)$ be a sequence of functions defined on  $I=[a,b]\subseteq\RR $. If $(f_n)$ are Riemann-integrable on $I$ and $(f_n)$ converges uniformly to $f$ on $I$, then $f$ is Riemann-integrable on $I$ and $$\int_a^b\lim_{n\to\infty} f_n(x) \dd{x}=\lim_{n\to\infty} \int_a^bf_n(x) \dd{x}$$
   \end{theorem}
   \begin{theorem}

Mathematics/5th/Advanced_dynamical_systems/Advanced_dynamical_systems.tex

@@ -663,15 +663,11 @@
     $$
       \int_{\TT^1}\varphi(x)\dd{\mu}=\int_{\TT^1}\varphi(x)\dd{x}
     $$
-    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 $k\ne 0$:
+    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 $\mu$. 
-    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 $\int_{\TT^1}\varphi(x)\dd{\mu}=\int_{\TT^1}\varphi(x)\dd{x}$.
+    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 means, which converge uniformly to $\varphi$ (recall \mnameref{MA:fejerthm}) and use the \mcref{MA:uniformintegral}.
   \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.