Meaning of a probability distribution being dominated by a measure The following comes from Ghosh & Ramamoorthi (2003) Bayesian Nonparametrics.



In terms of notations, $\Theta$ is a parameter space with Borel $\sigma$-algebra $\mathcal B(\Theta)$.
For each $\theta\in\Theta$, $P_\theta$ is a probability distribution on a measurable space $(\mathbf{X},\mathcal A)$ such that, for each $A\in\mathcal A$, $\theta\mapsto P_\theta(A)$ is $\mathcal B(\Theta)$ measurable.
$X_1,X_2,\dots$ is a sequence of $\mathbf X$-valued random variables independently and identically distributed as $P_\theta$.
Lastly, $\Pi$ is a prior, i.e. a probability measure on $(\Theta,\mathcal B(\Theta))$.
Question
What does it mean when the authors say that
"[$P_\theta$ is] dominated by a $\sigma$-finite measure $\mu$"? More generally, how is dominance defined in probability theory?
 A: A measure $\xi$ is absolutely continuous with respect to $\mu$ or dominated by $\mu$ if:


*

*$\mu(A) = 0$ implies $\xi(A) = 0$ for every measurable set $A$


Verbally, $\xi$ is never non-zero if $\mu$ is zero.
This is also written as $\xi \ll \mu$.
There is another equivalence called the Radon-Nikodym theorem:
If a $\sigma$-finite measure $\xi$ is dominated by $\mu$ - both measures on $(X,\Sigma)$ - then there is a measurable function $f: X \rightarrow [0,\infty)$ such that for any $X \subset A$:


*

*$\xi(A) = \int_A f d\mu$


The function $f$ is the Radon–Nikodym derivative $\frac{d\xi}{d\mu}$ or in your case $\frac{dP_\theta}{d\mu}$. It is similar to an ordinary derivative but defined over sets (which is what is required in measure theory). An ordinary derivative does also not always exist and similar care (or even more care: continuous is not yet differentiable) is required. 
If someone is defining a conditional probability - or as in this case a posterior distribution - you run the risk to run into them.
A: Addition to the answer of "Davide Giraudo", I want to add some informal intuition on the domination of measure. 
"If one measure of a certain set is zero,then another particular measure is always zero". That means, if the former device can't measure some sets, the latter device measures them, neither. 
Les's think a telescope having two lenses of which one is implemented on the front(to the eyes) and the other one is at the end(to the sky), if the latter one can't catch some subject in the sky, the former one is also impossible to catch that subject.
A: A measure $\nu$ is dominated by the measure $\mu$ if $\nu\ll\mu$, which means that if $\mu(A)=0$  for some measurable $A$, then $\nu(A)=0$. 
A family of probability measures $(P_\theta)_{\theta\in\Theta}$ is dominated by a probability measure $\mu$ if and only if for each $\theta\in\Theta$, the measure $P_\theta$ is dominated by $\mu$.  
