common dominating measure for a family of measures Given a family $\{\mu \}_{i\in I}$ on a Polish space (complete, separable metric space) $X$. When does there exist a measure $\lambda$ such that 
$$\mu_i=f_i \lambda$$
where the $f_i$ are densities (Radon-Nikodym) of $\mu_i$ with respect to $\lambda$. 
EDIT: What is a verifiable condition in the case $I$ is uncountable.
 A: At least for finite measures (the tags seem to suggest this question concerns mainly probability measures), this is equivalent to $\{\mu_i:i\in I\}$ being separable under the variation norm.
The variation norm on finite measures dominated by some finite measure $\nu$ agrees with the $L_1(\nu)$-norm of the corresponding Radon-Nikodym derivatives, so separability of $\{\mu_i:i\in I\}$ follows in this case from separability of $L_1(\nu)$.
Conversely, if the sequence of nonzero finite measures $\langle\lambda_n\rangle$ is dense in $\{\mu_i:i\in I\}$ under the variation norm, then whenever $\mu_i(A)>0$ for some $i\in I$, then there is some $n$ such that $\lambda_n(A)>0$. It follows that $\{\mu_i:i\in I\}$ is dominated by the probability measure $\kappa$ given by
$$\kappa(A)=\sum_{n=1}^\infty \frac{1}{2^n \lambda_n(X)}\lambda_n(A).$$
Relating this to the comment by  Jochen Wengenroth: If $I$ is compact and the function $i\mapsto\nu_i$ is continuous when the range is endowed with the variation distance, the range will be compact and be separable as a compact metrizable space, so in this case, a finite dominating measure will exist.
A: A countable family of sigma-finite measures, yes.  
Can we drop a condition?  
Drop sigma-finite: consider two measures on $\mathbb R$: Lebesgue measure and counting measure.
Drop countable: on $\mathbb R$ consider the family of measures, one of them is Lebesgue measure, and the rest are the unit point masses at all the points of $\mathbb R$.
