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!

  • $\begingroup$ 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 $\endgroup$ Jul 6 '19 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.

  • $\begingroup$ 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... $\endgroup$ Jul 6 '19 at 20:08
  • $\begingroup$ ``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? $\endgroup$ Jul 6 '19 at 20:09
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    $\begingroup$ 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). $\endgroup$ Jul 6 '19 at 21:23
  • $\begingroup$ Ah I see! Many thanks again! $\endgroup$ Jul 6 '19 at 22:00
  • $\begingroup$ @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. $\endgroup$ Jul 7 '19 at 6:10

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