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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asked Feb 21, 2014 at 17:51
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$\begingroup$ 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. $\endgroup$– Jochen WengenrothCommented Feb 21, 2014 at 19:17
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3$\begingroup$ @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. $\endgroup$– Todd TrimbleCommented Feb 21, 2014 at 19:24
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$\begingroup$ I did not claim to answer the question, I just can't imagine a Boolean isomorphism with some $\sigma$-algebra. $\endgroup$– Jochen WengenrothCommented Feb 21, 2014 at 19:32
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1$\begingroup$ Every complete Boolean algebra is the regular open sets in some topological space, so Jochen's suggestion doesn't actually say much. $\endgroup$– Eric WofseyCommented Feb 22, 2014 at 5:49
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1$\begingroup$ 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. $\endgroup$– alphaCommented Feb 25, 2014 at 8:54
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4 Answers
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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 by 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.
answered Feb 21, 2014 at 19:02
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$\begingroup$ Can we assume $X\subseteq [0,1]$? $\endgroup$ Commented Feb 21, 2014 at 19:10
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2$\begingroup$ 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 $[a/k,(a+1)/k]$ " $\endgroup$ Commented Feb 21, 2014 at 20:01
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2$\begingroup$ @IoannisSouldatos : I stand by what I said : $\Sigma$ is isomorphic as an ordered set to the subset of the algebra $L^{\infty}([0,1],\mu)$ of equivalences class of essentially bounded function up to equality almost everywhere. It is the subset of function with value (almost everywhere) in $\{0,1\}$ and this algebra has (as every von Neuman algebra) all supremum and this subset is closed under supremum (because it is characterized by $x^2=x$. The point is that uncountable supremum of element of $\Sigma$ might have nothing to do with union of sets(...) $\endgroup$ Commented Jan 10, 2018 at 22:03
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2$\begingroup$ For exemple an uncountable supremum of singleton is the still the empty set. If you want a more explicit proof that it is complete, the point is that if you have an ascending chain indexed by an ordinal, if the ordinal is countable then you just take the union, if the ordinal is uncountable then the measure of each terms is an uncountable increasing sequence of real number smaller than $1$, and such a sequence is constant after some rank, and so the sequence became constant up to the equivalence relation after some rank, so it has a supremum. $\endgroup$ Commented Jan 10, 2018 at 22:05
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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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1$\begingroup$ Thanks, it's nice to have an answer that says that "while false, it's almost true". $\endgroup$ Commented Feb 21, 2014 at 21:01
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3$\begingroup$ 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. $\endgroup$ Commented Feb 22, 2014 at 0:59
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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...
answered Feb 22, 2014 at 1:27
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$\begingroup$ 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. $\endgroup$ Commented Feb 22, 2014 at 1:54
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$\begingroup$ 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... $\endgroup$ Commented Feb 22, 2014 at 19:19
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1$\begingroup$ 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 $$\bigvee\{\bigwedge_{i\in I}x_{i}|x_{i}\in R_{i}\,\textrm{for}\,i\in I\}=1.$$ $\endgroup$ Commented Feb 22, 2014 at 19:35
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1$\begingroup$ 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. $\endgroup$ Commented Feb 22, 2014 at 19:45
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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
answered Feb 21, 2014 at 19:21
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$\begingroup$ I added more details. The completion of an atomless Boolean algebra has no atoms, since the original algebra is dense in it. $\endgroup$ Commented Feb 21, 2014 at 19:38
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$\begingroup$ 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. $\endgroup$ Commented Feb 21, 2014 at 19:46
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$\begingroup$ That was a long time ago, but I think that the example I gave in my answer is actually not the Cohen algebra (but I might be wrong on the terminology). From my understanding, the Cohen algebra corresponds to the double negation topology on $[0,1]$ and it admit no absolutely continuous valuation, while the example I mentioned might not be countably generated, but admit a completely additive valuation given by the Lebesgue measure. Through the frame <-> locales duality, the Cohen algebra is the space of "Bair generic reals" while my example is the space of "Lebesgue generic reals". $\endgroup$ Commented Jan 10, 2018 at 22:16
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$\begingroup$ @JoelDavidHamkins : My above comment, is actually a question on whether my terminology is correct or if "cohen algebra" really means the thing I'm talking about in my answer as you said. $\endgroup$ Commented Jan 10, 2018 at 22:19
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1$\begingroup$ Simon, your algebra is what is what is known as the random algebra, corresponding to the forcing to add a random real. This is different from the Cohen algebra, which corresponds to the forcing to add a Cohen real. The Cohen algebra is the unique atomless complete Boolean algebra with a countable dense set. Your algebra does not have a countable dense set. $\endgroup$ Commented Jan 10, 2018 at 23:28