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Though tangentially related to another post on MathOverflow (here), the questions below are mainly out of curiosity. They may be very-well known ones with very well-known answers, but...

Suppose $\Sigma$ is a sigma-algebra over a set, $X$. For any given topology, $\tau$, on $X$ denote by $\mathfrak{B}_X(\tau)$ the Borel algebra over $X$ generated by $\tau$.

Question 1. Does there exist a topology, $\tau$, on $X$ such that $\Sigma = \mathfrak{B}_X(\tau)$?

If the answer to the previous question is affirmative, it makes sense to ask for the following too:

Question 2. Denote by ${\frak{T}}_X(\Sigma)$ the family of all topologies $\tau$ on $X$ such that $\Sigma = \mathfrak{B}_X(\tau)$ and let $\tau_X(\Sigma) := \bigcap_{\tau \in {\frak{T}}_X(\Sigma)} \tau$. Is $\Sigma = \mathfrak{B}_X({\frak{T}}_X(\Sigma))$?

Updates. Q2 was answered in the negative by Mike (here).

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  • $\begingroup$ A comment on Question 1: If $\kappa_1<\kappa_2$ are uncountable cardinals, $X$ is a set with cardinality $\kappa_2$, and $\Sigma$ is the set of subsets $S$ of $X$ such that $S$ or $X\setminus S$ has cardinality at most $\kappa_1$, then can $\Sigma$ be the $\sigma$-algebra generated by a topology? $\endgroup$
    – Jonas Meyer
    Dec 7, 2011 at 4:02
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    $\begingroup$ @Jonas: Can’t you simply take $\tau$ to be $\{\varnothing\}\cup\{V\subseteq X:|X\setminus V|\le\kappa_1\}$? $\endgroup$
    – Brian M. Scott
    Dec 7, 2011 at 7:13
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    $\begingroup$ OK, elephant in the room: assuming choice, the Lebesgue sigma-algebra on $\mathbb{R}$ is probably not a Borel sigma-algebra. Of course this makes problem 1 into one that depends on your set theory since word on the street is there exist models of ZF in which all subsets of $\mathbb{R}$ are measureable. $\endgroup$
    – Mike F
    Dec 7, 2011 at 8:03
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    $\begingroup$ @Mike, I posted an answer arguing that the Lebesgue algebra is not a counterexample. $\endgroup$
    – JDH
    Dec 7, 2011 at 19:56
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    $\begingroup$ George, can you post a proof that your counterexample is a counterexample? I don't quite see it yet... $\endgroup$
    – JDH
    Dec 8, 2011 at 0:07

4 Answers 4

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This is a great question. I am posting not an answer to the question, but an answer to the comment posted by Mike about a proposed counterexample, the elephant in the room.

Theorem. The elephant in the room is not a counterexample. That is, the $\sigma$-algebra of Lebesgue measurable sets is the Borel algebra of the topology consisting of all sets of the form $O-N$, where $O$ is open in the usual topology and $N$ has measure $0$.

Proof. Note that the empty set and the whole of $\mathbb{R}$ have the desired form. Next, note that sets of that form are closed under finite intersection, since $(O-N)\cap(U-M)=(O\cap U)-(N\cup M)$. Next, I claim that they are closed under countable unions. This is because $\bigcup_i (O_i-N_i) = (\bigcup_i O_i)-N$, where $N$ is a certain subset of $\bigcup_iN_i$, which is still measure $0$ since this is a countable union. Next, note that it is closed under arbitrary unions in the case where the open set is the same, $\bigcup_i (O-N_i)$, since this is just the same as $O-(\bigcap_i N_i)$, which is $O$ minus a smaller null set. Finally, since we have a countable basis for the usual topology, using rational intervals, say, it follows now that my sets are closed under arbitrary unions, since for any sized union, we may rewrite it using basic open sets minus null sets, and then group together all the terms arising for each basic open set as a single term, thereby reducing the entire union to a countable union, which we've argued still has the desired form.

Thus, the sets of the form $O-N$ do indeed form a topology, and this topology is clearly contained within the Lebesgue algebra. Furthermore, the Borel algebra generated by my collection of sets includes all measure zero sets, as well as all open sets, and so it is the same as the algebra of all Lebesgue measurable sets. QED

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  • $\begingroup$ Thanks for the explaining this. I'm not sure how exactly this came about, but I've been walking around for a few years now thinking that the Lebesgue $\sigma$-algebra was somehow bigger than the completion of the Borel $\sigma$-algebra (wrt Lebesgue measure) - but I guess they are the same thing! $\endgroup$
    – Mike F
    Dec 8, 2011 at 6:42
  • $\begingroup$ ..yet somehow I was also aware that, when $S \subset \mathbb{R}$ is measureable, there exist $\mathscr{G}_\delta$ $G$ and $\mathscr{F}_\sigma$ $F$ such that $F \subset S \subset G$ and $G - F$ has measure zero. A couple of contradictory beliefs which somehow never bumped into each other. $\endgroup$
    – Mike F
    Dec 8, 2011 at 6:49
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I think that I can answer the second question. For each point $p \in \mathbb{R}$, let $\tau_p$ be the topology on $\mathbb{R}$ consisting of $\varnothing$ together with all the standard open neighbourhoods of $p$. Unless I've made some mistake, the Borel sigma-algebra generated by $\tau_p$ is the standard one. However, $\bigcap_{p \in \mathbb{R}} \tau_p$ is the indiscrete topology on $\mathbb{R}$.

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  • $\begingroup$ I can't see why the Borel sets would be the same as the standard one. $\endgroup$
    – Asaf Karagila ♦
    Dec 6, 2011 at 20:45
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    $\begingroup$ Asaf, it's because $\{p\}$ is Borel in $\tau_p$, arising from the intersection of nested intervals centered on $p$. Once you've got $\{p\}$, then you can get all the open sets not containing $p$ simply by subtracting, and so you get all the ordinary open sets. Thus, the Borel sets of $\tau_p$ include all the usual Borel sets. (And of course, it is included in this, so they are equal.) $\endgroup$
    – JDH
    Dec 6, 2011 at 21:13
  • $\begingroup$ @JDH: Ah, I figured it was something related to that. I did not keep in mind that we have the standard topology in the background; rather than just a continuum-sized set. $\endgroup$
    – Asaf Karagila ♦
    Dec 6, 2011 at 21:17
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    $\begingroup$ @JDH: Well $\{p\}$ isn't quite the whole story since subtracting that only gives you standard open sets having $p$ as a 2-sided limit point. But you can also get any closed interval containing $p$ by intersecting nested open intervals - and subtracting these should give the rest of the standard open sets (or at least the open intervals omitting $p$ which is enough). $\endgroup$
    – Mike F
    Dec 6, 2011 at 23:48
  • $\begingroup$ Mike, I agree with that. $\endgroup$
    – JDH
    Dec 7, 2011 at 2:14
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For Q1. How about this. $X = \{0,1\}^A$ for uncountable $A$, and $\Sigma$ is the product $\sigma$-algebra. So each element of $\Sigma$ depends on only countably many coordinates.

Now we just need a proof that this cannot be the Borel algebra of any topology.

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  • $\begingroup$ That is indeed true. I was about to post something along these lines and, in fact, I think you can convert it to a sigma algebra on A itself rather than on $2^A$. $\endgroup$
    – George Lowther
    Dec 7, 2011 at 21:38
  • $\begingroup$ I think you can show that there must exist an $x\in X$ such that $\{x\}$ is closed. But, $\{x\}\not\in\Sigma$. $\endgroup$
    – George Lowther
    Dec 7, 2011 at 21:49
  • $\begingroup$ @George: Of course this needs more work to show that... Our space is $T_0$ but not $T_1$. $\endgroup$
    – GEdgar
    Dec 7, 2011 at 22:19
  • $\begingroup$ Yes, I was too quick there. What I said is not true in the case where A is countably infinite, because $2^A$ is isomorphic (as a measurable space) to the reals with Borel sigma-algebra generated by the topology $\{(x,\infty):x\in R\}$, for which no $\{x\}$ is closed. $\endgroup$
    – George Lowther
    Dec 8, 2011 at 1:24
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EDIT: It's best just to ignore this nonsensical post.

The product space mentioned in the answer by Gerald Edgar does not provide a counterexample. The product $\sigma$-algebra is determined by countably many coordinate, the product topology by finitely many coordinates. A set determined by countably many coordiantes is a countable intersection of sets determined by finitely many coordinates.

Remark: This answer contained a flawed "proof" that the answer to the first question is yes. I've edited it.

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    $\begingroup$ If you still claim my space $X = \{0,1\}^A$ is not a counterexample, you need more explanation. Certainly that product sigma-algebra is not the Borel algebra for the product topology. But is it the Borel algebra for some other topology? That is the question! $\endgroup$
    – GEdgar
    Dec 24, 2011 at 17:16
  • $\begingroup$ You are right of course, the Borel sigma-algebra is too large. I'm sorry. Explaining what my confusion was would take more than the number of characters left... $\endgroup$
    – Michael Greinecker
    Dec 27, 2011 at 0:13

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