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Is every sigma-algebra the Borel algebra of a topology?

inspires the present question which asks for less.

Question: Given a $\sigma$-algebra $\mathcal A$ on a set $X$, does there exist a topology $\mathcal T$ on perhaps some other set $Y$ such that $\mathcal A$ is isomorphic to the Borel sets determined by $\mathcal T$?

Examples contained in the answers to the quoted question indicate an answer of "not necessarily" if one also requires $X=Y$. I may be wrong, but it seems to me that a negative answer here (if appropriate) will require a new idea.


I've changed the title of my question to account for Gerald Edgar's comment. One could still ask to represent any abstract $\sigma$-algebra as a Borel field, but this isn't possible, as noted by Loomis in On the representation of $\sigma$-complete Boolean algebras. That said, the theorem Loomis proves indeed realizes abstract $\sigma$-algebras as Borel fields modulo a $\sigma$-ideal. I don't believe this settles my intended question though.

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    $\begingroup$ What is an isomorphism of $\sigma$-algebras? An isomorphism of Boolean algebras that preserves countable sups? $\endgroup$ Oct 14, 2012 at 8:33
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    $\begingroup$ An isomorphism $f$ between any two posets obviously preserves sups (i.e. $I$ has a sup iff $f(I)$ has a sup in which case $f(\sup(I))=\sup(f(I))$. So an isomorphism between $\sigma$-algebras just means an isomorphism of the underlying Boolean algebras. $\endgroup$
    – YCor
    Oct 14, 2012 at 9:24
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    $\begingroup$ Plus you added the problem of starting with an abstract sigma-algebra, and realizing it as a sigma-algebra of sets: where the countable sups correspond to countable unions. $\endgroup$ Oct 14, 2012 at 12:19

1 Answer 1

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Edit: My previous answer does not work as is. I edited it so that it becomes more of a long comment. It discusses a few aspects of the case where $X$ has a topology that makes it compact totally separated, and $\mathcal A$ is generated by the algebra of clopen sets.

Let $(X,\mathcal A)$ be a measurable space. We assume $\mathcal A$ is isomorphic to the Borel algebra $\mathcal B(Y)$ of some topological space $(Y,\mathcal T)$ (as posets). We denote by $\phi:\mathcal A\to\mathcal B(Y)$ the isomorphism of posets.

Since the Borel poset of a topological space does not change upon taking the Kolmogorov quotient, we can assume $Y$ is $T_0$. We also do the equivalent construction on $X$ (take the quotient by the relation “the measurable sets containing $x$ and $x'$ are the same”), so $\mathcal A$ separates points of $X$ and $\mathcal B(Y)$ separates points of $Y$

Define $\omega\mathfrak U(\mathcal A)$ (ultrafilters stable under countable intersections) the collection of all $U\subset\mathcal A$ such that

  • $\varnothing\notin U$;
  • $U$ is stable under countable intersections;
  • if $A\subset A'$ are elements of $\mathcal A$ and $A\in U$, then $A' \in U$;
  • for all $A\in\mathcal A$, either $A\in U$ or $A^\complement\in U$.

There are natural maps from $X$ and $Y$ to $\omega\mathfrak U(\mathcal A)$, namely $$\begin{aligned}x&\mapsto\{A\in \mathcal A:x\in A\}=:U_x, &y&\mapsto\{A\in \mathcal A:y\in\phi(A)\}=:V_y.\end{aligned}$$

What remains of my answer is the following, which we shall prove later.

Lemma.

  • The maps $x\mapsto U_x$ and $y\mapsto V_y$ are injective.
  • If moreover $X$ is a Stone space and $\mathcal A$ is the $\sigma$-algebra generated by its clopen sets, then
    • the map $x\mapsto U_x$ is bijective;
    • the induced map $i:Y\to X$ is such that $\phi(A)=i^{-1}A$.

In the latter case, since the preimage has much nicer properties than $\phi$, this gives a lot more information about how the latter behaves. For instance, it shows that if a (not necessarily countable) union of elements of $\mathcal A$ is itself in $\mathcal A$, then $\phi$ commutes with the union. It is also true for arbitrary intersections. However, this property need not hold for the inverse $\phi^{-1}$ over $\mathcal B(Y)$: to be sure to have $$\phi^{-1}\big(\bigcup_{j\in J}B_j\big) = \bigcup_{j\in J}\phi^{-1}(B_j), $$ one needs of course all the $B_j$ and their union to be in $\mathcal B(Y)$ (so that the equation even makes sense) but also the union of the $\phi^{-1}(B_j)$ to be in $\mathcal A$. For this reason, the set of all $\phi^{-1}(B)$ for $B\in\mathcal T$ is not clearly a topology. However, if we can show it is by other means, then it must generate the $\sigma$-algebra.

Fact. If $\mathcal S:=\{A\in\mathcal A:\phi(A)\in\mathcal T\}$ is a topology over $X$, then it generates $\mathcal A$ as a $\sigma$-algebra.

In Jochen Wengenroth's answer to Is every sigma-algebra the Borel algebra of a topology?, it is shown that this property would not hold for the Stone space $\{0,1\}^{\mathbb R}$, so to get a contradiction it would suffice to show that $\mathcal S$ is a topology.

Proof of the fact. Define $\mathcal B'=\{ B\in\mathcal B(Y):\phi^{-1}B\in\sigma(\mathcal S)\}$. It is a $\sigma$-algebra (because $\phi^{-1}$ preserves countable unions and intersections) and it contains $\mathcal T$, so $\mathcal B(Y)\subset\mathcal B'$, and by definition it means $\mathcal A\subset\sigma(\mathcal S)\subset\mathcal A$.

It remains to prove the lemma.

Proof of the lemma: injectivity. It should be clear that the map $X\to\omega\mathfrak U(\mathcal A)$ is well-defined, i.e. $U_x\in\omega\mathfrak U(\mathcal A)$. The fact that it is injective comes from the fact that $\mathcal A$ separates points. The same reasoning holds for $y\mapsto V_y$.

Proof of the lemma: surjectivity. For the surjective part, write $\mathcal T_X$ for the topology on $X$ which makes it a compact totally separated space. Let $U$ be an element of $\omega\mathfrak U(\mathcal A)$. Note that by the finite intersection property for $U$, the collection of open sets $O\in(\mathcal A\setminus U)\cap\mathcal T_X$ cannot be an open cover. Choose some $x\in X$ not covered by this collection; we will show that $U=U_x$.

The fact that $U=U_x$ is equivalent to $U_x\subset U$ (by the third and fourth properties of elements of $\omega\mathfrak U(\mathcal A)$), which is equivalent to all elements $A$ of $\mathcal A$ satisfying the following property: “exactly one of $A$ or $A^\complement$ contains $x$, and the one that does belongs to $U$”. In other words, setting $\mathcal A'$ those elements, we want to have $\mathcal A'=\mathcal A$.

In fact it is not too difficult to see that $\mathcal A'$ is a $\sigma$-algebra, and it contains the clopen sets because $x$ is not in the open cover discussed above. This concludes the proof that $U=U_x$.

Proof of the lemma: the map $i$. To be clear, this map sends $y\in Y$ to the only $x\in X$ such that $V_y=U_x$. It is clearly well-defined and injective by the first two points of the lemma. The fact that $\phi(A)$ is always the inverse image of $A$ under $i$ is a matter of unfolding the definitions: $$ y\in\phi(A) \Leftrightarrow A\in V_y \Leftrightarrow A\in U_{i(y)} \Leftrightarrow i(y)\in A \Leftrightarrow y\in i^{-1}A. $$


¹ For every pair $x,x'\in X$ of distinct elements, there is some $A\in\mathcal A$ containing precisely one of the two.

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  • $\begingroup$ It sounds right to me. $\endgroup$ Jan 19, 2023 at 10:42
  • $\begingroup$ Still trying to figure out whether the answer is positive or negative... $\endgroup$ Mar 8, 2023 at 10:30
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    $\begingroup$ @AndrejBauer The current answer is neither, as I explain in the introduction. The previous (flawed) version was negative. All of this is trying to prove that the "Stone $\sigma$-algebra" of $[0,1]^{\mathbb R}$ is not an abstract Borel $\sigma$-algebra. What I am saying is that Stone algebras must be Borel algebras on a subset through the expected isomorphism. This almost reduces the question to the linked one, since some Stone algebras cannot come from Borel algebras on the full set. $\endgroup$
    – Pierre PC
    Mar 8, 2023 at 11:52

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