I have posted the following question also [here](https://math.stackexchange.com/questions/362457/existence-of-dominating-measure-for-weak-compact-set-of-measures) a longer time ago, but due to no answers I thought it might fit better to MO.


Let $(\Omega,\mathcal F)$ be a measurable space and $\mathcal P$ a weak\*-compact subset of the set of all probability measures on $\mathcal F$. Does there exist a probability measure $\mathbb Q$ such that every measure $\mathbb P\in\mathcal P$ is absolutely continuous to $\mathbb Q$, i.e. such that $\mathbb P$ dominates the mesures in $\mathcal P$?


The weak\*-topology is the weakest topology such that all the linear functionals $L\_Z:\mathbb P\mapsto \int\_\Omega Zd\mathbb P$ for $Z:\Omega\rightarrow\mathbb R$ $\mathcal F$-measurable and bounded, are continuous.


Note that when the set $\mathcal P$ is countable one can easily find a dominating measure for $\mathcal P$.



Existence of dominating measure for weak*-compact set of measures
There always exists a dominating measure.


First, given two finite measures $\mu,\nu$ on $(\Omega,\mathcal{F})$, the Lebesgue decomposition theorem says that there is an $A\in\mathcal{F}$ such that $1\_{\Omega\setminus A}\mu$ is absolutely continuous with respect to $\nu$ and $1\_A\mu,\nu$ are singular. This can also be constructed using the Radon-Nikodym derivative as $A=\lbrace d\nu/d(\mu+\nu)=0\rbrace$, and $A$ is uniquely defined up to a $\mu$-null set. We can think of $\mu(A)$ as measuring the distance $\mu$ is from being absolutely continuous wrt $\nu$, and I will write $d(\mu,\nu):=\mu(A)$.


Assuming $\mathcal{P}$ is nonempty then, using dependent choice, there exists a sequence $\mathbb{P}\\_1,\mathbb{P}\\_2,\ldots\in\mathcal{P}$ such that
$$
d(\mathbb{P}\\_n,\mathbb{P}\\_1+\cdots+\mathbb{P}\\_{n-1}) \ge \frac12\sup\left\lbrace d(\mathbb{P},\mathbb{P}\\_1+\cdots+\mathbb{P}\\_{n-1})\colon\mathbb{P}\in\mathcal{P}\right\rbrace.
$$
We can show that $\mathbb{Q}:=\sum\_{n=1}^\infty2^{-n}\mathbb{P}\\_n$ is a dominating measure for $\mathcal{P}$.


Let me start by showing that $\alpha\_n:=d(\mathbb{P}\\_n,\mathbb{P}\\_1+\cdots+\mathbb{P}\\_{n-1})$ tends to zero. By definition, there exists $A\_n\in\mathcal{F}$ such that $\mathbb{P}\\_n(A\_n)=\alpha\_n$ and $\mathbb{P}\\_m(A\_n)=0$ for $m < n$. Replacing $A\_n$ by $A\_n\setminus\bigcup\_{m > n}A\_m$ if necessary, we can further suppose that $A\_n$ are disjoint. Now, set $B\_n=\bigcup\_{m\ge n}A\_m$ so that $\mathbb{P}\\_m(B\_n)\ge\alpha\_m$ for $m\ge n$ and $B\_n\downarrow\emptyset$ as $n\to\infty$. If $\alpha\_n$ does not tend to zero then, by passing to a subsequence if necessary, it can be assumed that $\alpha\_n\rightarrow\alpha > 0$. Now, by compactness, the sequence $\mathbb{P}\\_m$ has a limit point $\mathbb{P}$ in the weak-\* topology. By the definition of the topology given in the question, this means that $\mathbb{P}(B\_n)$ is a limit point of $\mathbb{P}\\_m(B\_n)\ge\alpha\_m$ as $m\to\infty$, so $\mathbb{P}(B\_n)\ge\alpha$. This contradicts the fact that $\mathbb{P}(B\_n)\to0$ as $n\to\infty$ (by countable additivity). So, $\alpha\_n\to0$ as required.


Finally, lets show that $\mathbb{Q}$ is a dominating measure. Choosing any $\mathbb{P}\in\mathcal{P}$ set $\alpha:=d(\mathbb{P},\mathbb{Q})$. If $\alpha > 0$ then we would have $\alpha\_n < \alpha/2$ for large $n$, in which case
$$
\begin{align}
d(\mathbb{P}\\_n,\mathbb{P}\\_1+\cdots+\mathbb{P}\\_{n-1})&=\alpha\_n < \alpha/2 =\frac12d(\mathbb{P},\mathbb{Q})\cr
&\le\frac12d(\mathbb{P},\mathbb{P}\\_1+\cdots+\mathbb{P}\\_{n-1}).
\end{align}
$$
The last inequality uses the fact that $\mathbb{P}\\_1+\cdots+\mathbb{P}\\_{n-1}$ is absolutely continuous wrt $\mathbb{Q}$. This inequality contradicts the choice of $\mathbb{P}\\_n$, so we have $\alpha=0$ and $\mathbb{P}$ is absolutely continuous wrt $\mathbb{Q}$.


