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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 $x$.

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    Okay, so this is the Dirac measure *at $x$*, not "the Dirac measure on $B$", the $x$ matters! :-)2012-04-12

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  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 $\mu(S)>0$.

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    @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