Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
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
add comment

1 Answer

up vote 1 down vote accepted

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".

share|improve this answer
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
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.