# Weak convergence, and Cesaro convergence (of mu_n (E) ) imply convergence (of mu_n (E))?

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)

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?

Can we conclude this with extra hypothesis?

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.

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In your assumption do you mean that the Cesaro averages of $\mu_n(E)$ converge to $\mu(E)$? –  R W Nov 10 '12 at 1:00
And what is here a "μ−continuity set A" ? –  Pietro Majer Nov 10 '12 at 8:38
@Pietro I think a $\mu$-continuity set is a set whose boundary has $\mu$ measure $0$ (and I think it should have been stated in the OP). –  Davide Giraudo Nov 10 '12 at 11:34

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.
Thanks, I was not looking for the case when the cesaro means do not converge. I wanted that as a part of the hypothesis. But I think your counter example works, if you take $x_{n}$=0 only sporadically, then you will have cesaro mean convergence but not convergence. –  FelipeG Nov 12 '12 at 20:05