What is the meaning of writing the differential inside of a function? I am reading through Resnick's "Extreme Values, Regular Variation and Point Processes" and have come across some notation that I am not familiar with. In talking about moving a Poisson point process into higher dimensions, we are introduced to the mean measure function:
\begin{align*}
\mu^*(dx, dy)=\mu(dx) K(x, dy),
\end{align*}
my question here is strictly about notation like: $\mu(dx)$. I know the following notation
\begin{align*}
\int_\Omega f(x)\mu(dx)=\int_\Omega f(x)d\mu(x)=\int_\Omega fd\mu
\end{align*}
and I know that when we write something like
\begin{align*}
\int_\Omega f(X, y)K(X, dy)
\end{align*}
we are freezing $X$ and integrating with respect to $K$, viewed now as a function only of $y$. My question here is, what is meant when we write the differential inside a function, like $\mu(dx)$, outside of the integral?
 A: This can be a quick way of defining a measure.  For example, if $\nu$ is the measure defined by $\nu(A) = \int_{A} f(x) \, \mu(dx)$ (where $\mu$ is some other measure and $f$ is some nice enough function), I could instead simply say "Define $\nu$ by $\nu(dx) = f(x) \, \mu(dx)$."  You might also see this written as "Define $\nu$ by $d \nu = f\, d \mu$."
You might ask: is this really quicker than writing $\nu(A) = \cdots$?  I don't think so, but what can I say...  I guess in geometric measure theory it's quicker to write $\mu = f \, d\mathcal{H}^{d-1} \restriction_{\partial \Omega}$, for instance, than "Define $\mu$ by $\mu(A) = \int_{A \cap \partial \Omega} f \, d\mathcal{H}^{d-1}$."  Also, if I wanted to describe the Lebesgue decomposition of a measure, it's very quick to write $\nu = f \, d \mu + \nu^{s}$ rather than some alternative.
The same idea seems to apply to product measures.  The measure $\mu^{*}$ you mentioned I assume is understood to be the one given by
\begin{equation*}
\mu^{*}(A \times B) = \int_{A} K(x,B) \mu(dx).
\end{equation*}
You can see how writing $\mu^{*}(dx, dy) = \mu(dx) K(x,dy)$ is quicker way to convey the same information in this case.  (I would write "$\mu^{*}(dx \otimes dy) = K(x,dy) \mu(dx)$" at the risk of watching my audiences' eyes roll out of their sockets...)  Of course, the simplest example of how to do this in two variables would be $\nu(dx \otimes dy) = \mu_{1}(dx) \mu_{2}(dy)$ for the product $\nu = \mu_{1} \otimes \mu_{2}$ (or $\nu = \mu_{1} \times \mu_{2}$).
