# Sigma algebra without atoms ?

I'm looking for an example of a set S, and a sigma algebra on it, which has no atoms.

Motivation: It seems to me that a lot of definitions in probability and stochastic processes - conditional probability, filtrations, adapted processes - become a lot simpler if phrased in terms of a sample space partitioned into atoms.

In the book I'm reading, this is done for finite sample spaces. But the only problem I find with extending the definitions to general spaces seems to be that, in general, you might have a sigma algebra without any atoms. (Note: All this definitions can be made without introducing a measure, so objections on grounds of uncountability don't apply.)

Does anyone have an example ?

2) Related question: What if you replace sigma algebra by algebra - closed under finite, rather than countable unions ?

I suspect that the algebra of subsets of R generated by open intervals has no atoms, but can't prove it rigorously.

• I know what an atom is in a measure space. What is the definition of an atom in a sigma algebra? Apr 25, 2010 at 3:20
• I think it's a nonempty element of the sigma algebra whose only proper subset in the sigma algebra is $\emptyset$. Apr 25, 2010 at 3:25
• The existence of such algebras follows directly from the Stone Representation Theorem; for example the clopen subsets of Cantor space form a free Boolean algebra with countably many generators. However, the case of sigma-algebras is more complicated... Apr 25, 2010 at 4:03
• @Cosmonut: ℝ[(ℝ(-1,1))∪(-1,0)∪(0,1)] = {0} is an atom in your example. Apr 25, 2010 at 4:38
• A theory of measurable partitions and its applications to stochastic processes indeed exists. It was developed by Rokhlin and it can be found in Sinai's books on ergodic theory. Apr 26, 2010 at 1:06

In your second question, you are asking merely for an atomless Boolean algebra, of which there are numerous examples. One easy example related to the one given on the Wikipedia page is the collection of periodic subsets of the natural numbers N. That is, the subsets $A\subset N$ such that there is a finite set $a\subset k$ for some $k$ and $kn+m\in A$ if and only if $m\in a$ for $m<k$. This is closed under finite intersections, unions and complements, but has no atoms, since any nonempty periodic set can be made smaller by reducing it to a set with a larger period.

For your main question, a similar idea works, when generalized to the transfinite, to produce the desired atomless $\sigma$-algebra. Namely, consider the collection of periodic subsets of $\omega_1$, the first uncountable ordinal. This ordinal is bijective with the set of reals, if the Continuum Hypothesis holds, but in any case (under AC), it is bijective with a subset of the reals, so one can take the underlying set of points here to be a set of real numbers. By periodic here I mean the collection of sets $A\subset \omega_1$, such that there is a set $a\subset\alpha$ for some countable ordinal $\alpha$, such that $\alpha\beta+\xi\in A$ if and only if $\xi\in a$, where $\xi<\alpha$. Note that if $\alpha$ is fixed, every ordinal has a unique representation as $\alpha\beta+\xi$ for $\xi<\alpha$, since this is just dividing the ordinals into blocks of length $\alpha$. This means that $A$ consists of the pattern in $a$ repeated $\omega_1$ many times. This collection of sets is easily seen to be a Boolean algebra and atomless, but it is also $\sigma$-closed, since for any countably many such $A$, I can find a common countable period since the collection of multiples of any fixed $\alpha$ form a club subset of $\omega_1$. Thus, the intersection (or union) is again periodic, as desired. There are no atoms, since any nonempty periodic set can be made smaller by reducing it to have a larger period.

There are numerous atomless Boolean algebras arising in the forcing technique of set theory, used by Cohen to prove the independence of the Continuum Hypothesis and many others subsequently.

• That's an incredibly simple example! Apr 25, 2010 at 4:56
• @Joel Where can I find more about the properties of periodic subsets of natural numbers? Dec 28, 2022 at 22:21
• @MadHatter I don't have any specific references, sorry, but I find that topic to be suitable for any determined student fruitfully to explore. For example, the periodic subsets of $\mathbb{N}$ form an atomless Boolean algebra. Can you compute the meet $a\wedge b$ and join $a\vee b$ of any two periodic sets? And the complement $\neg a$? How many such periodic sets are there? Does this algebra have a computable presentation? Is the theory decidable? Are all countable atomless Boolean algebras isomorphic? Dec 28, 2022 at 22:33
• @JoelDavidHamkins OK. Just to make sure that I understand the definition correctly. Every periodic set can be uniquely determined by a pair $(n,F)$, where $F$ is a subset of $n$. Thus, for example, $(1,\{0\})$ is the set of natural numbers, and $(2^3,\{0,1\})=\{0,1,8,9,16,17,\ldots\}$. Dec 28, 2022 at 22:43
• Yes, that's right. Dec 28, 2022 at 22:44

Space $\{0,1\}^A$ where $A$ is uncountable, with the standard (product) $\sigma$-algebra $\cal F$. Single-point sets are not measurable, but $\cal F$ separates points. Single-point sets are not in $\cal F$, since any set in $\cal F$ depends on countably-many coordinates. Alternate description: $\{0,1\}^A$ is a compact Hausdorff space, and $\cal F$ is the $\sigma$-algebra of Baire sets.

• Very nice, Gerald. Apr 25, 2010 at 20:53
• I suppose that your space is rather large, with size $2^{\omega_1}$ (which I admit may have size only continuum, but often is larger), but can you cut it down by reducing to a "dense" set of size continuum? Perhaps it works if you restrict $2^A$ just to the periodic sequences? Apr 25, 2010 at 22:25
• Gerald's example does a little more: $\mathcal{F}$ is the free $\sigma$-algebra with $|A|$ generators. Apr 26, 2010 at 1:22
• @Joel: I think cutting down is going backwards. Plus putting structure on the set $A$ is making the example more complicated. If anything, $A$ should be $\mathbb{R}$ and not $\omega_1$. Apr 26, 2010 at 11:57
• Perhaps only a logician would think $\omega_1$ is simpler than the reals. In the topological space $\{0,1\}^A$, no point is extraneous: if you maliciously remove one, any topologist would just put it back in! Apr 27, 2010 at 17:38

A good source for results on atomless $\sigma$-algberas is chapter 3 of Borel Spaces (1981) by Rao and Rao (Diss. Math.). Here are two remarks, taken from this text:

1. There is no atomless countably generated $\sigma$-algebra.

Let $\mathcal{C}$ be a countable set of generators for a $\sigma$-algebra on $X$. W.l.o.g., we can assume that $\mathcal{C}$ is closed under complements. Then for every $x\in X$, the set $A(x)=\bigcap \{C:x\in C,C\in\mathcal{C}\}$ is measurable in $\sigma(\mathcal{C})$ as a countable intersection of measurable sets. Since one cannot separate points by $\sigma(\mathcal{C})$ that one cannot separate by $\mathcal{C}$, for each $x$, $A(x)$ is an atom of $\sigma(\mathcal{C})$

2. On every uncountable set, there exists an atomless $\sigma$-algebra that separates points.

Let $X$ be an uncountable set. Clearly, it suffices to show that an atomless $\sigma$-algebra that separates points exists on a set with the same cardinality as $X$. The set $Y$ of all finite subsets of $X$ has the same cardinality as $X$, so we work with $Y$. We verify that the $\sigma$-algebra on $Y$ generated by elements of the form $G_x=\{F:x\in F\}$ for some $x\in X$ does the job.

It is obvious that this $\sigma$-algebra separates the elements of $Y$. So if there were any atom, it would be a singleton $\{F\}$ and it would be generated by countably many of the $G_x$. So let $C$ be a countable subset of $X$ such that $\{F\}\in\sigma\{G_c:c\in C\}$. For the reason pointed out in 1., $\{F\}$ would be the intersection of all elements in $\{G_c:c\in C\}\cup\{G_c^C:c\in C\}$ that contain $F$. Now $F\in G_c$ only if $c\in F$. So let $x\notin F\cup C$. Then $F\cup\{x\}\in G_c$ for all $c\in F$ and $F\cup\{x\}\in G_c^C$ for $c\notin F$. So the intersection doesn't contain only $F$, which is a contradiction.

• I have one question and would be grateful if you could explain: "it would be generated by countably many of the Gx" I understand everything except that statement. How do I know it's not finitly or uncountably generated?
– Meow
Oct 17, 2013 at 21:19
• @Mare Let $\mathcal{F}$ be any family of subsets and $A\in\sigma(\mathcal{F})$. Then there exists a countable family $\mathcal{C}\subseteq\mathcal{F}$ with $A\in\mathcal{F}$. To prove this, you have to just verify that the sets generated by countable subfamilies of $\mathcal{F}$ form a $\sigma$-algebra containing every element of $\mathcal{F}$. Oct 17, 2013 at 22:50
• Can I just use this argument to skip that: {F} must be interection of countably many elements from $\{G_c:c \in X\}\cup\{G_c^c: c \in X\}$ since it's from sigma-algebra. (and now I say I can take countable $C \subseteq X$ etc.
– Meow
Oct 18, 2013 at 6:48
• @Mare Every atom is indeed such a countable intersection. Oct 18, 2013 at 8:20