most recent 30 from http://mathoverflow.net 2013-05-19T06:52:20Z http://mathoverflow.net/feeds/question/111949 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/111949/weak-convergence-and-cesaro-convergence-of-mu-n-e-imply-convergence-of-mu FelipeG 2012-11-09T23:16:57Z 2012-11-12T19:55:25Z

<p>Let $X$ be a compact metric Borel space. Suppose $\mu_{n}(A)\rightarrow\mu(A)$ for all $\mu-$continuity sets $A$ (sets with zero boundary measure), where $\mu_{n}$ is a sequence of probability measures. (some people call it weak other weak* convergence)</p> <p>If $E$ is a measurable set such that $\mu(E)>0$ and the Cesaro average of $\mu_{n}(E)$ converges; can we conclude that $\mu_{n}(E)$ converges? </p> <p>Can we conclude this with extra hypothesis?</p> <p>I am particularly interested in the case when $T:X\rightarrow X$ is a continuous transformation, $\mu_{n}=T^{n}\mu_{1},$ and $E=\cap T^{-i}A_{i}$ where $A_{i}$ is a sequence of $\mu-$continuity sets. </p>

http://mathoverflow.net/questions/111949/weak-convergence-and-cesaro-convergence-of-mu-n-e-imply-convergence-of-mu/111960#111960 Alexandre Eremenko 2012-11-10T03:17:18Z 2012-11-10T03:17:18Z

<p>In general, the answer is "no", because $\mu_n(E)$ can be an arbitrary sequence of zeros and ones. Take on the real line $E$={$0$}, and $\mu_n=\delta_{x_n}$, where $x_n$ tends to $0$. We have $\mu_n\to\delta_0$ weakly. Then $\mu_n(E)=1$ or $0$ depending on whether $x_n=0$ or not. So you can choose such sequence $x_n$ that Cesaro means do not converge.</p> <p>In the dynamical setting, I don't know the answer, but I suppose that in such generality it will be also "no".</p>