# General nonatomic measure that cannot be expressed as an integral

I read in a paper (Kingman — Poisson Processes, 2005) that:

In most cases the mean [of an inhomogenous Poisson process on a set $A$] is given in terms of the rate function $\lambda(x)$ on $S$ by

\begin{align} \mu(A) &= \int_A \lambda(x)dx\qquad\textrm{for }A > \subseteq S \end{align}

but it can be a general (nonatomic) measure on $S$.

I know this is a silly question, but what kinds of nonatomic measures cannot be expressed as in the equation?

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## 3 Answers

The Cantor set has Lebesgue measure zero, so it is enough to find an atomless measure that puts positive measure on the Cantor set and nowhere else. Now one can identify the Cantor set with the space $\{0,1\}^\mathbb{N}$ in a measurably isomorphic way and the latter space can be endowed with the coin flipping probability measure.

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Thanks! (Now, I wonder if there are any real-world Poisson processes whose intensity varies as the cantor set ;) ) –  Neil G Jul 13 '12 at 9:46

Here's a "terminological" answer with no actual content. See also this question, whose answers subsume both Michael Greinecker's and mine.

A measure $\tilde{\nu}$ is absolutely continuous with respect to another measure $\nu$ if it doesn't assign positive measure to any set which $\nu$ says has measure zero.

The Radon-Nikodym theorem says that if $\tilde{\nu}$ is absolutely continuous with respect to $\nu$, and both have finite total measure, then $\tilde{\nu}$ can be expressed in terms of $\nu$ in the way you describe: $$\tilde{\nu}(A) = \int_A \lambda\;d\nu$$ for some measurable function $\lambda$. On the other hand, if $\tilde{\nu}$ isn't absolutely continuous with respect to $\nu$, it's clear that $\tilde{\nu}$ can't be expressed in this way.

So, the measures which can't be expressed in terms in the way you describe are precisely the ones which are not absolutely continuous with respect to whatever measure $dx$ means.

Michael Greinecker's answer gives an excellent example of a non-atomic measure on $[0, 1]$ which is not absolutely continuous with respect to the Lebesgue measure.

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I think you read his book "Poisson Processes"………..he mentions that the Poisson Process cannot arise from an atomic mean measure….

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