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Is there a decent characterization of measure on an infinite countable set? At page 7 of "Introduction to Measure Theory and Integration" (Ambrosio, Da Prato, Mennucci), example 1.10 I found that "clearly" any measure on a finite or countable set has to be atomic, where atomic means linear composition of Dirac distributions. As far as I know nobody tells me that singletons have to be in my $\sigma$-algebra. I can easily invent a measure on integers where singletons do not have a measure, for example assigning measure $0.5$ to the set of even numbers and $0.5$ to the set of the odds. Am I wrong?

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    $\begingroup$ I think you are right. You can even take $\sigma$-algebra $\{\varnothing,\Omega\}$. But actually you can recognize in your example a measure on a set of two "elements" (even,odd) in disguise. If you only distinguish even and odd then why not doing it with a two-element set? $\endgroup$ – drhab Jan 8 '16 at 8:59
  • $\begingroup$ I might whant to study the behaviour of a function on the set on the integers with respect to that measure... Is it so trivial? $\endgroup$ – ThePunisher Jan 8 '16 at 9:12
  • $\begingroup$ I have to point out that I didn't notice that the $\sigma$-algebra was $P(\Omega)$... It was my fault... $\endgroup$ – ThePunisher Jan 8 '16 at 18:17
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You have to specify a $\sigma$-algebra of measurable sets. Almost all $\sigma$-algebras will just be the power set (if we have all the singletons in it), or a power set of "classes" of the countable set, where the classes are atomic (they're the $\sigma$-algebra and no subset of them is), I think. This measure is still called atomic, usually. So your example is just a trivial $\sigma$-algebra on a two "point" set. I think modulo such identifications the statement is in fact true.

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  • $\begingroup$ Ok, but do we just believe in that or is there a proof? Because the one given on the book relies on Dirac measure. $\endgroup$ – ThePunisher Jan 8 '16 at 9:14
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    $\begingroup$ There is a proof. Try to give it yourself! I might have time later. $\endgroup$ – Henno Brandsma Jan 8 '16 at 9:19
  • $\begingroup$ Ok, I will try to answer my own question. $\endgroup$ – ThePunisher Jan 8 '16 at 9:31
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Given our countable set $\Omega$ and a $\sigma$-algebra $\mathfrak{F}$ on it we will introduce an equivalence relation. We will say that two elements $x,y \in \Omega$ are equivalent ($x \sim y$) if $$ \forall F \in \mathfrak{F} \quad x \in F \iff y \in F $$ The set of equivalence classes has to be countable since it can't be "bigger" than $\Omega$. We can show that every equivalence classes is an element of $\mathfrak{F}$ since it is a countable intersection (the class of $x$ can be seen as the intersection of all the sets $F$ containing $x$) and in the same way every element of $\mathfrak{F}$ is the union of equivalence classes. For this reason a function on the equivalence classes to $[0,+ \infty]$ can be uniquely extended to a measure on $\mathfrak{F}$. The atoms are in fact the equivalence classes, not the singletons.

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