# 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 Commented Jul 6, 2019 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...
– Y.Z.
Commented Jul 6, 2019 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?
– Y.Z.
Commented Jul 6, 2019 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). Commented Jul 6, 2019 at 21:23
• Ah I see! Many thanks again!
– Y.Z.
Commented Jul 6, 2019 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. Commented Jul 7, 2019 at 6:10