# Convergence of probability measures on a generating field of a sigma-field

Let $(\Omega,\mathcal{B})$ be a measurable space and let $\mathcal{F}$ be a generating field of $\mathcal{B}$. Assume $\mathcal{F}$ is standard, i.e. it is countable, and any normalized, non-negative, finitely-additive set function on $\mathcal{F}$ is also countably-additive, thus can be extended to a probability measure on $\mathcal{B}$ uniquely by Caratheodory's extension theorem.

Let $m_n$ be a sequence of probability measures on $\mathcal{B}$. If for all $F\in\mathcal{F}$, $m_n(F)\to\alpha(F)$ for some set function $\alpha$, then $\alpha$ is normalized, non-negative, finitely-additive on $\mathcal{F}$ and so can be extended to a probability measure $m$ on $\mathcal{B}$. Is it true that $m_n(F)\to m(F)$ for all $F\in\mathcal{B}$?

This is related to the following question: if $\mathcal{F}\subset\mathcal{G}\subset\mathcal{B}$ where $\mathcal{G}$ could be countable, and the convergence $m_n(F)\to\alpha(F)$ is on $\mathcal{G}$. Restricting $\alpha$ on $\mathcal{F}$ and extend $\alpha|_{\mathcal{F}}$ to a probability measure $m$ on $\mathcal{B}$. Then is it true that $m(F) = \alpha(F)$ on $\mathcal{G}$? We know it is true if $\mathcal{G}=\mathcal{B}$, since by Vitali–Hahn–Saks theorem, $\alpha$ is already a probability measure.

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No, you can have measures whose supports are finite sets tending towards (atomless) measures. E.g., the average over n equally spaced points tends the uniform distribution (I have cantor space in mind here). But, I am not familar with the term "standard generating field". What kind of examples do you have in mind? – George Lowther Sep 19 '12 at 23:51
I don't see in your example for what event $F$ $m_n(F)\to m(F)$ is false. A "standard field" is a term in R. Gray's book Probability, Random Processes, and Ergodic Properties. A field is standard if it possesses a basis, which is a sequence of finite fields $\mathcal{F}_n$ with the following properties: 1. $\mathcal{F}_n$ asymptotically generates F, that is, $\mathcal{F}_n\subset\mathcal{F}_{n+1}$ and $\mathcal{F} = \bigcup_n\mathcal{F}_n$. 2. If $G_n$ is a sequence of atoms of $\mathcal{F}_n$ such that $G_{n+1}\subset G_n$, then $\bigcup_n G_n\neq\emptyset$. – Wei Mao Sep 20 '12 at 4:55