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In my book, "Probability and Stochastics" by Cinlar, it's stated that for some measurable space $(E,\scr E)$, and fixed $x\in E$, the Dirac measure $\delta_x(A)=\left\{ \begin{array}{lcc} 1, x\in A \\ \\ 0, &\mbox{ otherwsie} \\ \\ \end{array} \right.$

is purely atomic (i.e. The set $D$ of atoms of $\delta_x$ is countable and $\delta_x(E\setminus D)=0$, where an atom is a singleton $A$ such that $\delta_x(A)>0$).

I guess $\delta_x$ would have only one atom, that being $\{x\}$, but how can you say that $\{x\}\in\scr E$ for any $\sigma$-algebra $\scr E$? What if $\scr E$ is the trivial $\sigma$-algebra $\{E,\varnothing\}$?

user153582
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    Isn't an atom defined to be a minimal member of $\mathscr{E}$ of positive measure? So an atom is not necessarily a singleton. – angryavian Aug 07 '15 at 06:43
  • I'm not sure of other definitions. My book reads, "A point $x$ is said to be an atom (of a measure) $\mu$ if $\mu ({x})>0.$" My book doesn't mention non-singleton sets being atoms, but maybe you are right, and your definition is more standard. – user153582 Aug 07 '15 at 06:48
  • @angryavian Wikipedia uses the same definition you gave, so maybe my book is making some assumption about $E$ or $\scr E$ – user153582 Aug 07 '15 at 06:51
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    I see the part of the book you are reading. At the beginning of the paragraph there is an assumption that every singleton set belongs to $\mathcal{E}$. I don't know if Cinlar later gives the more general definition of atom that Wikipedia uses. Make sure you note however that given this assumption "$\mathcal{E}$ contains the singleton sets," Cinlar's definition and Wikipedia's definition are equivalent. – angryavian Aug 07 '15 at 06:59
  • Oh I completely missed that haha. I was looking for such an assumption earlier in the section. Thank you very much! :) – user153582 Aug 07 '15 at 07:03
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    Very relevant question: http://math.stackexchange.com/questions/41142/questions-on-atoms-of-a-measure. See in particular the answer by Michael Greinecker. He provides an example where all singletons are measurable and have measure zero, but the measure still has atoms in the sense of Wikipedia. @angryavian: This means that the two definitions are not equivalent in general. – PhoemueX Aug 07 '15 at 07:45
  • @PhoemueX Thank you, I stand corrected! – angryavian Aug 07 '15 at 15:02

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