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Does there exist a complete Boolean algebra that is not isomorphic to any $\sigma$-algebra? If so, what is an easy or canonical example or construction?

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A candidate would be the complete Boolean algebra of regular open sets ($A$ is the interior of its closure) in $\mathbb R$ or some more exotic topological space. As open intervals are regular open the class itself is clearly not a $\sigma$-algebra. – Jochen Wengenroth Feb 21 '14 at 19:17
@JochenWengenroth Is that actually enough? The question is whether there isn't a possibly exotic embedding of that Boolean algebra $B$ onto a $\sigma$-algebra supported on some other set $X$ (one preserving Boolean algebra structure and countable intersections), not necessarily the underlying set of the topological space you started with. – Todd Trimble Feb 21 '14 at 19:24
I did not claim to answer the question, I just can't imagine a Boolean isomorphism with some $\sigma$-algebra. – Jochen Wengenroth Feb 21 '14 at 19:32
Actually, your proposal is what occurred to me too, and I think with an extra nudge it can be made into an answer as well. – Todd Trimble Feb 21 '14 at 19:49
It would be a pity not to mention the "Richtungsstreit" amongst measure theoreticians due to the distinction between the concepts of a $\sigma$ algebra or a complete Boolean algebra. The two approaches are measure spaces with points as the basic elements and measure algebras with events. We are using the terminology of David Fremlin who discusses the two approaches in his book "Topological Riesz Spaces" and his monumental five volume tome on measure theory (in particular, chapter 32) where he makes a lucid case for the second, less popular approach. – alpha Feb 25 '14 at 8:54
up vote 14 down vote accepted

Let $\Sigma_0$ be the $\sigma$-algebra of Lebesgue measurable sets on the real interval $[0,1]$. Define $\Sigma$ to be $\Sigma_0$ quotiented by the relation $U \simeq V$ if $U$ and $V$ differ from a Lebesgue negligible set.

$\Sigma$ is a complete boolean algebra, but $\Sigma$ is not a $\sigma$-algebra.

Indeed, assume $\Sigma$ is identified with a $\sigma$-algebra in $\mathcal{P}(X)$ for some set $X$, and let $x \in X$. Then for each integer $k$, $x$ has to belong to a set of the form $[a/k,(a+1)/k]$ but the countable intersection of a family of such set is always empty in $\Sigma$ (it has zero measure), hence $x$ belong to the empty set, which yields a contradiction.

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Can we assume $X\subseteq [0,1]$? – Bjørn Kjos-Hanssen Feb 21 '14 at 19:10
+ 1. Very nice. – Joel David Hamkins Feb 21 '14 at 19:21
No, I do not assume that $X$ is a subset of [0,1], when I said that $x$ belongs to $[a/k,(a+1)/k]$ I meant " x belong to the subset of $X$ which is identified with the object of $\Sigma$ corresponding to $[a/k,(a+1)/k]$ " – Simon Henry Feb 21 '14 at 20:01

Every $\sigma$-complete Boolean algebra is isomorphic to a quotient $\mathcal{M}/I$ where $(X,\mathcal{M})$ is a $\sigma$-algebra and $I$ is a $\sigma$-complete ideal (See the Handbook of Boolean Algebras for a proof or look here). Here by a $\sigma$-complete Boolean algebra, we mean a Boolean algebra where every countable subset has a least upper bound. A $\sigma$-complete Boolean algebra $B$ is isomorphic to an algebra of sets if and only if each element in $B\setminus\{0\}$ is contained in some $\sigma$-complete ultrafilter.

$\textbf{Proposition}$ Suppose that $B$ is an atomless $\sigma$-complete Boolean algebra that satisfies the countable chain condition. Then $B$ has no $\sigma$-complete ultrafilters. In particular, $B$ is not representable as an algebra of sets.

$\textbf{Proof}$ Suppose to the contrary that $\mathcal{U}$ is a $\sigma$-complete ultrafilter. Then since $B$ is atomless, $\mathcal{U}$ is non-principal. Let $c\subseteq B$ be a maximal family under inclusion such that $c\cap\mathcal{U}=\emptyset$ and $a\wedge b=0$ for each $a,b\in c,a\neq b$. Such a set $c$ exists by Zorn's lemma. By maximality, $c$ is in fact a partition of $B$. By the countable chain condition, the set $c$ must be countable, so since $c\cap\mathcal{U}=\emptyset$, we have $1=\bigvee c\not\in\mathcal{U}$ since $\mathcal{U}$ is $\sigma$-complete. $\mathbf{QED}$

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Thanks, it's nice to have an answer that says that "while false, it's almost true". – Bjørn Kjos-Hanssen Feb 21 '14 at 21:01
By the way, the representation theorem in the first paragraph is known as the Loomis-Sikorski Theorem but often as just the Sikorski Representation Theorem. – François G. Dorais Feb 22 '14 at 0:59

Let me elaborate on Simon Henry's nice answer.

What we prove is that the Cohen algebra, the completion of the unique countable atomless Boolean algebra, is not isomorphic to any $\sigma$-algebra. Since the Cohen algebra has a countable dense set, this follows immediately from the following:

Lemma. If a $\sigma$-algebra $\Sigma$ has a countable dense set, then it must have atoms.

Proof. Enumerate the countable dense set $b_0,b_1,\dots$, where dense here means that every nonempty set in $\Sigma$ must contain some $b_k$ as a subset. Now fix any point $x$ in the underlying set, and let $a_n$ be either $b_n$ or the complement, chosen to ensure $x\in a_n$. Thus, $x\in a^*$, where $a^*=\bigcap_n a_n$ is the intersection. Notice that $a^*$ decides every $b_n$, in the sense that it is contained either in $b_n$ or in the complement. Since it is not empty, it must therefore be an atom. QED

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I added more details. The completion of an atomless Boolean algebra has no atoms, since the original algebra is dense in it. – Joel David Hamkins Feb 21 '14 at 19:38
Oh okay, so if we think in terms of propositional logic with variables $p_1,p_2,\ldots$ then an atom would correspond to a complete truth assignment on all the variables, but (1) in the Cohen algebra a complete truth assignment gets identified with 0, and (2) if $\mathbb P(p_n=\text{True})=1/2$ then any particular truth assignment has probability 0, i.e., is Lebesgue negligible, and that's the connection with @Simon Henry's answer. Thanks. – Bjørn Kjos-Hanssen Feb 21 '14 at 19:46

Here is yet another way to see this. In a $\sigma$-algebra of sets, countable intersections an unions are computed setwise, which means that we always have $$\bigwedge_{n<\omega} \bigvee_{m < \omega} A_{m,n} = \bigcap_{m<\omega} \bigcup_{n < \omega} A_{m,n} = \bigcup_{f:\omega\to\omega} \bigcap_{m<\omega} A_{m,f(m)},$$ where the right hand side is to be considered in a purely set theoretic basis. In any complete boolean algebra, we have $$\bigwedge_{n<\omega} \bigvee_{m < \omega} A_{m,n} \geq \bigvee_{f:\omega\to\omega} \bigwedge_{m<\omega} A_{m,f(m)}.$$ In the case of a $\sigma$-algebra of sets, we have $$\bigvee_{m<\omega} A_{m,f(m)} = \bigcap_{m<\omega} A_{m,f(m)}$$ for every $f:\omega\to\omega$, hence the set theoretic identity above shows that in a complete $\sigma$-algebra we must have $$\bigwedge_{n<\omega} \bigvee_{m < \omega} A_{m,n} = \bigvee_{f:\omega\to\omega} \bigwedge_{m<\omega} A_{m,f(m)}.$$ There are plenty of complete Boolean algebras that do not satisfy this distributive identity. In fact, from the perspective of forcing, this distributive property for a complete atomless Boolean algebra is equivalent to not adding new reals, so any complete atomless Boolean algebra that adds a new real is not isomorphic to any $\sigma$-algebra of sets. This includes Cohen forcing as Joel pointed out, Random forcing as Simon pointed out, and a whole lot more...

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Note that it is consistent with ZFC that there is a ccc algebra that doesn't add reals (a Souslin algebra) and so this answer does not subsume all of Joseph's answer. – François G. Dorais Feb 22 '14 at 1:54
Here is the usual recursion theoretical comment: I gather that if $A_{\langle\sigma,n\rangle,\tau}$ says $\sigma$ is a prefix of $\tau$ and $\tau$ belongs to or strongly avoids a certain dense set $D_n$, then François' last equation says there is an $f$ that given $\sigma$ and $n$ finds such a $\tau$; if the sequence $D_n$ is the collection of $\Sigma^0_1$ sets then $f$ computes a 1-generic $G$. Interesting... – Bjørn Kjos-Hanssen Feb 22 '14 at 19:19
The distributivity law mentioned in this problem can be generalized to necessary and sufficient conditions. A $\sigma$-complete Boolean algebra $B$ is representable as a $\sigma$-algebra if and only if whenever $I$ is an index set, $R_{i}\subseteq B$ is a countable set with $\bigvee R_{i}=1$ for $i\in I$, and whenever $x_{i}\in R_{i}$ for $i\in I$ there is a countable $J\subseteq I$ with $\bigwedge_{i\in I}x_{i}=\bigwedge_{j\in J}x_{j}$ (i.e. you have some compactness), then $$\bigvee\{\bigwedge_{i\in I}x_{i}|x_{i}\in R_{i}\,\textrm{for}\,i\in I\}=1.$$ – Joseph Van Name Feb 22 '14 at 19:35
One can strengthen these distributivity conditions to obtain necessary and sufficient conditions for whether every quotient $B/I$ of a $\sigma$-complete Boolean algebra $B$ by a $\sigma$-complete ultrafilter $I$ can be represented as an algebra of sets, and these algebras can be put into a one-to-one correspondence with the Lindelof $P$-spaces. These results generalize to any Boolean algebra which is endowed with a notion of which least upper bounds are important and which least upper bounds are unimportant. – Joseph Van Name Feb 22 '14 at 19:45

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