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. 



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


In the dynamical setting, I don't know the answer, but I suppose that in such generality it will
be also "no".


