Is a function discrete or continuous w.r.t. to the Lebesgue measure? I just read a paper where it said that

However, if the base measure $G_0$ is absolutely continuous with respect
  to Lebesgue measure, parameters will not be shared between groups.
  Only in the case where the base measure G0 is discrete will there be a
  strictly positive probability of the group specific distributions
  having overlapping support (i.e. sharing parameters between groups.)

What does the part in bold mean? I cannot seem to find a answer on Google that I understand, nor any useful papers.
My understanding is that the Lebesgue measure can tell us if a function is discrete or continuous? The wiki is not very helpful either...
 A: There are two important characterizations of this notion. In both, I use $\lambda$ to refer to the Lebesgue measure and $\mu$ to refer to some other measure.


*

*$\mu$ is absolutely continuous with respect to Lebesgue measure if for all Lebesgue measurable sets $A$ with $\lambda(A)=0$, $\mu(A)=0$.

*$\mu$ is absolutely continuous with respect to Lebesgue measure $\lambda$ if there exists a Lebesgue measurable function $f$ such that $\mu(A) = \int_A f d \lambda$ for all Lebesgue measurable sets $A$.
It is not hard to see that 2 implies 1. The statement "1 implies 2" is an instance of the Radon-Nikodym theorem. Accordingly, in analysis the function $f$ in the second characterization is referred to as the Radon-Nikodym derivative of $\mu$ with respect to $\lambda$, and is sometimes denoted by $\frac{d \mu}{d \lambda}$.
A different place that you may have encountered this notion under a different name is in probability theory. The probability distribution of a (one-dimensional) continuous random variable is a probability measure on $\mathbb{R}$ which is absolutely continuous with respect to Lebesgue measure. In this context the function $f$ in my second characterization is called the density of the random variable.
