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Reading an article in probability theory I faced with phrase atomic $\sigma$-field. I tried to search for the definition, but google doesn't give any meaningful result. As a result I'm looking for the definition here.

Mihran Hovsepyan
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    can you link the article or quote the relevant portion? definition of atom(ic) for a measure: http://en.wikipedia.org/wiki/Atom_%28measure_theory%29 my guess is that the $\sigma$-algebra must have at least one element that is not a null set and whose proper subsets are all null sets – suissidle Apr 07 '13 at 15:56
  • I found that definition of atom, but here I'm looking for the definition of atomic $sigma-field$. – Mihran Hovsepyan Apr 07 '13 at 15:59
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    in the absence of a measure you could take the maximal ideal as the collection of null sets; or maybe they include the measure in the definition of $\sigma$-field... that's why you should quote the passage as it appears to be a nonstandard definition – suissidle Apr 07 '13 at 16:03

2 Answers2

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Absent the notion of a $\sigma$-ideal of null sets an atomic $\sigma$-algebra $\Sigma$ has the property that for every $E\in \Sigma$ there is $E'\in \Sigma$ satisfying $E'\neq \emptyset$, $E'\subseteq E$, and if $E''\subseteq E'$, $E''\in \Sigma$, then $E'=\emptyset$.

If $\Sigma$ admits a strictly positive measure, then it contains a countable many atoms.

But without a strictly positive measure the algebra of Borel sets on $[0,1]$ is atomic but restricted to $(0,1]$ it is not.

Now suppose that $\mathcal{N}$ is a $\sigma$-ideal of $\Sigma$, then taking the quotient algebra $\mathcal{F}=\Sigma|_{\mathcal{N}}$ and ordering it by set inclusion. We say that $\mathcal{F}$ is atomic if every non-trivial chain in $\mathcal{F}$ has a second smallest element.

Rabee Tourky
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To expand on Rabee's answer, an atom in a measurable space is a nonempty measurable set that has no nonempty measurable proper subsets. The measurable space is atomic if every point lies in an atom.

Every countably generated measurable space is atomic. An example of a measurable space that is not atomic is $\{0,1\}^\kappa$ endowed with the product $\sigma$-algebra for $\kappa$ uncountable.

For countably generated measurable spaces endowed with a probability measure $\mu$, the measure $\mu$ is atomless if and only if $\mu(A)$ for every atom $A$. The assumption that the space is countably generated can not be weakened to it being atomic. To see this, endow $\mathbb{R}$ with the countable-cocountable $\sigma$-algebra and the probability measure $\mu$ that assigns probability $1$ to uncountable sets and probability $0$ to countable sets. Then $\mu$ is atomic but the atoms of the $\sigma$-algebra, the singletons, have all measure zero.

Proofs of these statements can be found in this answer of mine. A great reference for these issues is the booklet Borel Spaces by Rao and Rao.

Community
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Michael Greinecker
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    You said the measurable space is atomic if every point lies in an atom. But by defining for every $\omega$ the set $\mathfrak a(\omega)$ as the set of all $\omega'$ which cannot be separated from $\omega$ by a measurable set, you will end up that every $\omega$ is contained in an atom, not? So every measurable space is atomic?? – Andy Teich Jan 25 '14 at 01:22
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    @AndyTeich No., $\mathfrak a(\omega)$ may not be measurable. – Michael Greinecker Jan 25 '14 at 01:25