# Stone representation theorem and $\sigma$-isomorphism

By Stone representation theorem we know that every Boolean algebra $$\mathcal{B}$$ is (Boolean) isomoprhic to the Boolean algebra of the clopen-sets of its associated Stone space. If $$\mathcal{B}$$ is a $$\sigma$$-algebra (closed under countable union and intersection), to my knowledge, the abovementioned isomorphism may not be a $$\sigma$$-isomorphism (that is, it may not preserve countable union and intersection). I am curious if there is any result on when the Boolean isomorphism in the Stone representation theorem is in fact a $$\sigma$$-isomorphism. (To give more context to my question, I am working with tail-$$\sigma$$-algebras in probability theory, and I am trying to verify if or when a given tail-$$\sigma$$-algebra is $$\sigma$$-isomorphic to the Borel algebra of its associated Stone space). Many thanks in advance!

• A boolean isomorphism is also an order isomorphism and so it preserves everything under sight : any join, any meet, anything you can define order-theoretically – Max Jul 6 at 17:56

First, a note on terminology: your use of the term "$$\sigma$$-isomorphism" doesn't really make sense. An isomorphism between two Boolean algebras always preserves all joins and meets of all sizes (it's an isomorphism!). What you mean is really whether the embedding of $$\mathcal{B}$$ into the power set of its Stone space is a $$\sigma$$-homomorphism (not a $$\sigma$$-isomorphism, because it's almost never going to be surjective!).

Now, here's the answer: if $$\mathcal{B}$$ is an infinite $$\sigma$$-algebra, then the embedding into the power set of its Stone space is never a $$\sigma$$-homomorphism. Indeed, let $$(b_n)$$ be any sequence of disjoint nonzero elements of $$\mathcal{B}$$ (such a sequence exists as long as $$\mathcal{B}$$ is infinite). Then the union of the clopen sets corresponding to the $$b_n$$ cannot be clopen (it is not compact, since by definition it is covered by infinitely many disjoint open sets!), and thus cannot be in the image of $$\mathcal{B}$$ under the embedding. In particular, it cannot be equal to the image of $$\bigvee b_n$$.

What is true is that the embedding preserves countable joins and meets (indeed, all joins and meets) modulo meager sets. That is, let $$X$$ be the Stone space of $$\mathcal{B}$$ and let $$I$$ be the $$\sigma$$-ideal in $$\mathcal{P}(X)$$ consisting of all meager sets. Then the composition of the embedding $$\mathcal{B}\to\mathcal{P}(X)$$ with the quotient map $$\mathcal{P}(X)\to\mathcal{P}(X)/I$$ preserves all countable joins and meets that exist in $$\mathcal{B}$$. Here's a sketch of the proof: identifying $$\mathcal{B}$$ with the clopen subsets of $$X$$, the join of elements $$b_n\in \mathcal{B}$$, if it exists, must be the closure of the union $$\bigcup b_n$$. But $$\overline{\bigcup b_n}\setminus\bigcup b_n$$ is meager (it is closed and has empty interior), so the join is the same as the union, modulo meager sets.

In particular, when $$\mathcal{B}$$ is a $$\sigma$$-algebra, this gives an isomorphism of $$\mathcal{B}$$ with the Baire $$\sigma$$-algebra of $$X$$ modulo the ideal of meager Baire sets.

• I see. Thank you so much! Just a quick clarification on the terminology: I was reading Terence Tao's notes on Loomis-Sikorski representation theorem (terrytao.wordpress.com/2009/01/12/…) and he says Applying Stone’s representation theorem, we can find a Stone space X such that there is a Boolean algebra isomorphism $\phi: \mathcal B \to Cl(X)$ from $\mathcal B$ (viewed now only as a Boolean algebra rather than a sigma-algebra to the clopen algebra of X... – discretizer Jul 6 at 20:08
• The map $\phi$ need not be a \sigma-algebra isomorphism, being merely a Boolean algebra isomorphism one instead; it preserves finite unions and intersections, but need not preserve countable ones.'' Is this in tension with your remark that isomorphisms between Boolean algebras always preserve all joints and meets? – discretizer Jul 6 at 20:09
• Tao is abusing the term "$\sigma$-algebra isomorphism" the same way you did, using it to refer to a map that sends countable joins to countable unions (rather than countable joins, which might not be the same). – Eric Wofsey Jul 6 at 21:23
• Ah I see! Many thanks again! – discretizer Jul 6 at 22:00
• @discretizer In the clopen algebra, countable joins (if they exist) are not just countable unions. These are open but not always closed, so are not always clopen. – Henno Brandsma Jul 7 at 6:10