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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?

The Mad Scientist
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  • Measures inherently make sense with respect to integration so to me, it's really just shorthand for defining it as its integral, i.e. $\mu(dx) = d\mu(x)$ is really shorthand for $$\int_X f(x),\mu(dx) = \int_X f(x),d\mu(x).$$ – Cameron L. Williams Jan 05 '21 at 18:00

1 Answers1

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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 $\nu = \mu_{1} \times \mu_{2}$).