1
$\begingroup$

Let $(\Omega, B, \mu)$ be a measure space. How can we characterize the space $(\Omega, B, \mu)$ so that

  1. the counting measure on $B$ is absolutely continuous with respect to $\mu$?
  2. the Dirac measure on $B$ is absolutely continuous with respect to $\mu$?

Edit: What I mean by the Dirac measure on $B$ is the following: Let $x\in \Omega$. Then the Dirac measure at $x$ assigns $1$ to a set in $B$ that contains $x$ and $0$ to a set that does not contain $x$.

  • 0
    When you say "the dirac measure on $B$" what exactly do you mean?2012-04-12
  • 0
    @AsafKaragila Isn't the [dirac measure](http://en.wikipedia.org/wiki/Dirac_measure) defined with respect to a point $x$? $$ \delta_x (A) := \begin{cases} 1 & x \in A \\ 0 & x \notin A \end{cases}$$2012-04-12
  • 1
    @MattN. That would be **a** Dirac measure, not *the* Dirac measure.2012-04-12
  • 0
    I suppose @Michael Greinecker's answer is what you had in mind, but you should word you question more precisely because "What should $\Omega$, $B$, and $\mu$ be..." is too vague. That could mean that you just want any example, in which case you could take $\Omega$ to be any set, $B$ the full power set of $\Omega$, and in 1. $\mu$ the counting measure, and in 2. $\mu$ the Dirac measure concentrated at any $x \in \Omega$. The point is that every measure is absolutely continuous with respect to itself.2012-04-12
  • 0
    Okay, so this is the Dirac measure *at $x$*, not "the Dirac measure on $B$", the $x$ matters! :-)2012-04-12

1 Answers 1

1
  1. Counting measure is usually defined on the $\sigma$-algebra of all subsets. So let $\mu$ be another measure defined on all subsets. Counting measure is absolutely continuous with respect to $\mu$ if no $\mu$-zero set has positive counting measure. Every nonempty set has positive counting measure. So $\mu$ needs to put positve measure on every nonempty-set, in particular on every singleton. The measure on all singletons determines the measure on all countable sets and since uncountable sums of positve real numbers are always infinite, every such measure is determined by having a positive value at each singleton. So for $\mu$ there exists a function $f:\Omega\to\mathbb{R}\cup\{\infty\}$ with strictly positive values such that $\mu(A)=\sum_{\omega\in A}f(\omega)$ and this property characterizes the measures you are looking for.

  2. Dirac measures are bit more delicate since they can be defined on any $\sigma$-algebra. $\delta_x$ is absolutely continuous with respect to $\mu$ if $\mu(A)>0$ for every measurable set containing $x$. If the $\sigma$-algebra contains a smallest set $S$ containing $x$, which is true if it is countably generated or the powerset, then we just need $\mu(S)>0$.

  • 0
    I'm a bit confused. what does $\mu$-zero set mean?2012-04-12
  • 0
    A measurable set $S$ with $\mu(S)=0$.2012-04-12
  • 0
    Oh ok. Thanks. one more question. Could you please explain why $\delta_x \ll \mu$ if $\mu(A) \gt 0$ for every measurable set containing $x$?2012-04-12
  • 0
    Because these are exactly the ones that have positive measure under the dirac measure at $x$.2012-04-12
  • 0
    Great. Thanks again.2012-04-12
  • 0
    If you can define the counting measure on "all subsets", then you can define it on any $\sigma$-algebra! I don't see how this makes the $\delta_x$ more delicate. In fact, isn't the $\delta_x$ a counting measure itself?2012-04-13
  • 0
    @AndréCaldas: Of course. The issue is that there might exist measures $\mu$ on a smaller $\sigma$-algebra such that counting measure on that smaller $\sigma$-algebra is absolutely continuous with respect to $\mu$, but $\mu$ might have no extension to all subsets or no extension such that counting measure is absolutely continuous with respect to that extension. So one has to consider fewer measures when looking at measures defined on all subsets. But I agree, solving the more general problem is not that difficult either.2012-04-13
  • 0
    I might be missing something, but I guess that if you talk about measures being absolutely continuous one with respect to another, they have to be defined on the same $\sigma$-algebra. That is, $B$ is fixed. I also do not see what $\mu$ having or not having an extension has anything to do with this problem... :-(2012-04-17
  • 0
    @AndréCaldas: My point is simply that in principle, characterizing measures with a certain property for one single $\sigma$-algebra is easier than doing it for all $\sigma$-algebras. The present problem is easy enough that I could have done the more general problem.2012-04-17