This question arises from the excellent question posed on math.SE by Salvo Tringali, namely, Correspondence between Borel algebras and topology. Since the question was not answered there after some time, I am bringing it up here on mathoverflow in the hopes that it may find an answer here.

For any topological space, one may consider the Borel sets of the space, the $\sigma$-algebra generated by the open sets of that topology. The question is whether every $\sigma$-algebra arises in this way.

Question. Is every $\sigma$-algebra the Borel algebra of a topology?

In other words, does every $\sigma$-algebra $\Sigma$ on a set $X$ contain a topology $\tau$ on $X$ such that $\Sigma$ is the $\sigma$ algebra generated by the sets in $\tau$?

Some candidate counterexamples were proposed on the math.SE question, but ultimately shown not to be counterexamples. For example, my answer there shows that the collection of Lebesgue measurable sets, which seemed at first as though it might be a counterexample, is nevertheless the Borel algebra of the topology consisting of sets of the form $O-N$, where $O$ is open in the usual topology and $N$ is measure zero. A proposed counterexample of Gerald Edgar's there, however, remains unresolved. And I'm not clear on the status of a related proposed counterexample of George Lowther's.

Meanwhile, allow me to propose here a few additional candidate counterexamples:

  • Consider the collection $\Sigma_0$ of eventually periodic subsets of $\omega_1$. A set $S\subset\omega_1$ is eventually periodic if above some countable ordinal $\gamma$, there is a countable length pattern which is simply repeated up to $\omega_1$ to form $S$. This is a $\sigma$-algebra, since it is closed under complements and countable intersections (one may find a common period among countably many eventually periodic sets by intersecting the club sets consisting of starting points of the repeated pattern).

  • Consider the collection $\Sigma_1$ of eventually-agreeing subsets of the disjoint union $\omega_1\sqcup\omega_1$ of two copies of $\omega_1$. That is, sets $S\subset \omega_1\sqcup\omega_1$, such that except for countably many exceptions, $S$ looks the same on the first copy as it does on the second. Another way to say it is that the symmetric difference of $S$ on the first copy with $S$ on the second copy is bounded. This is a $\sigma$-algebra, since it is closed under complement and also under countable intersection, as the countably many exceptional sets will union up to a countable set.

Please enlighten me by showing either that these are not actually counterexamples or that they are, or by giving another counterexample or a proof that there is no counterexample.

If the answer to the question should prove to be affirmative, but only via strange or unattractive topologies, then consider it to go without saying that we also want to know how good a topology can be found (Hausdorff, compact and so on) to generate the given $\sigma$-algebra.

  • $\begingroup$ Consider a set base B for a sigma algebra, and let $\Sigma$ be the sigma-algebra generated by B. It will contain countable unions of members of B, but not arbitrary unions. Won't something like a discrete set of singletons for B work? Or is the issue that we can't tell if a member if $\Sigma$ is open? In any case, if B is large enough, shouldn't there be cardinalities which may be achieved in the topology generated by part of $\Sigma$ that won't be achieved by any member of $\Sigma$? (Assume Choice first, to satisfy my intuition.) Gerhard "Ask Me About System Design" Paseman, 2012.02.07 $\endgroup$ Feb 7, 2012 at 21:03
  • $\begingroup$ Gerhard, the topology generated by the singletons would include all sets, and this is not contained within the sigma algebra you mention. Your cardinality idea is a good one---perhaps one can get a consistency result that way---but I don't quite see my way through the full details. $\endgroup$ Feb 7, 2012 at 21:06
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    $\begingroup$ Yes. I was going to say that $\Sigma$, if it contained any topology at all, had to contain the topology generated by B, to get a contradiction, but stopped myself in time. Even so, I am struck by the idea that there is an algebraic resolution to this: there must be a large (but not too large) $\Sigma$ which contains a lot of small sets, and that enough of these small sets need to be open. If I knew how to finish this idea, I would be using the answer box instead. Gerhard "Algebra Solves Every Problem, Right?" Paseman, 2012.02.07 $\endgroup$ Feb 7, 2012 at 21:30
  • $\begingroup$ Obviously, the topology suggested by Joel David Hamkins having the class of Lebesgue measurable sets as its Borel $\sigma$−algebra is neither locally compact nor first countable, hence not metrizable. $\endgroup$
    – TaQ
    Feb 9, 2012 at 16:09
  • $\begingroup$ I have accepted Jochen Wengenroth's answer, since it provides a counterexample answering the main question. I do remain interested, however, in the resolution of the two proposed counterexamples in the question. Please post further answers with information about that, or with additional interesting counterexamples to the main question. $\endgroup$ Feb 9, 2012 at 18:31

3 Answers 3


Unfortunately, I can only provide a reference but no ideas since I don't have the paper. In "On the problem of generating sigma-algebras by topologies", Statist. Decisions 2 (1984), 377-388, Albert Ascherl shows (at least according to the summary to be found on MathSciNet) that there are $\sigma$-algebras which can't be generated by a topology.

Robert Lang (same journal 4 (1986), 97-98) claims to give a shorter proof.

As suggested by Joel, I add the ideas of Lang's example. The underlying space is $\Omega= 2^{\mathbb R}$, that is the space of all indicator functions, and the $\sigma$-algebra is $\mathcal A = \bigotimes_{\mathbb R} \mathcal P$ where $\mathcal P$ is the power set of the two element set. It is generated by the system $\mathcal E$ of the "basic open sets" of the product topology (prescribed values in a finite number of points). This generator has the cardinality $c$ of the continuum and since the generated $\sigma$-algebra can be obtained in $\omega_1$ (transfinite) induction steps the cardinality of $\mathcal A$ is also $c$. On the other hand, if $\mathcal T$ is a topology with $\mathcal A=\sigma(\mathcal T)$ then $\mathcal T$ separates points (this should follow from the "good sets principle"), in particular, for two distinct points of $\Omega$ the closures of the corresponding singletons are distinct. Hence $\mathcal T$ has at least $|\Omega|=2^c$ elements.

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    $\begingroup$ I had a short look at Lang's paper. The example is the $\mathbb R$-fold product of $\{0,1\}$ (the latter endowed with the power set). The underlying space consists, thus, of all indicator functions on $\mathbb R$. The argument goes by showing that the $\sigma$-algebra has cardinality $| \mathbb R |$ and any generating topology would be point separating and thus had cardinality $\ge |2^{\mathbb R}|$. $\endgroup$ Feb 8, 2012 at 13:59
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    $\begingroup$ Jochen, to answer my own question, I guess you mean that $\Sigma$ is the $\sigma$-algebra generated by the basic open sets of the product topology. And then the point is that this $\sigma$-algebra has size continuum, since there are only continuum many such basic open sets, and closing under complements and countable unions remains size continuum since $\omega_1\cdot|\mathbb{R}|^\omega=|\mathbb{R}|$. $\endgroup$ Feb 8, 2012 at 14:39
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    $\begingroup$ Of course, I didn't have a proof... the cardinality argument is something I didn't think of. So: a $T_0$ topology on a set of cardinal $2^c$ must have at least $2^c$ open sets... but this sigma-algebra has only $c$ elements, total. $\endgroup$ Feb 8, 2012 at 14:46
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    $\begingroup$ Jochen, could I kindly ask you to edit your answer to include a summary of the argument, following the details now given in the comments? Since the question has been popular, it seems best to provide a thorough answer to it. $\endgroup$ Feb 8, 2012 at 19:03
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    $\begingroup$ Here is your first article booksc.org/s/… $\endgroup$ Sep 20, 2016 at 6:50

If $\Sigma$ is countably generated -- and any sigma-algebra of practical interest will be [Edit: or will be the completion of such an algebra] -- then I believe the answer is yes. Here is the argument.

Let $G$ be a countable generating set for $\Sigma$. Let $\tau$ be the topology generated by $G$: that is,

  1. Define $B$ to be the set of all finite intersections of elements of $G$ (including the intersection of 0 sets, i.e. $X$ itself).
  2. Define $\tau$ to the set of all unions of subsets of $B$.

$\tau$ is the topology on $X$ for which $\Sigma$ is its Borel algebra.

To see this, first note that $B$ is a base and $G$ is a subbase for $\tau$. Since $G$ is countable, so is $B$, and hence all elements of $\tau$ are countable unions of subsets of $B$. Since $B\subseteq \Sigma$, this means that $\tau \subseteq \Sigma$ also.

Recall that the Borel algebra on $\tau$ is just the sigma-algebra generated by $\tau$, $\sigma(\tau)$. Since $\tau \subseteq \Sigma$ we have $\sigma(\tau) \subseteq \Sigma$. But $\Sigma = \sigma(G)$ where $G\subseteq \tau$, so $\Sigma \subseteq \sigma(\tau)$. That is, $\Sigma = \sigma(\tau)$.

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    $\begingroup$ The algebra of Lebesgue measurable sets is not of practical interest? $\endgroup$ yesterday
  • $\begingroup$ OK, but it's the completion of a countably generated sigma algebra. $\endgroup$ 18 hours ago
  • $\begingroup$ But your construction doesn't apply to completions of countably generated $\sigma$-algebras, does it? $\endgroup$ 17 hours ago
  • $\begingroup$ No. But completion serves only to satisfy an intuition that any subset of a negligible set should be measurable with measure 0, so I don't consider it an issue of practical interest. $\endgroup$ 16 hours ago

The silly error has been found! Sorry for the hassle, and please disregard this nonsense.

I'm sure this is suboptimal, but it seems to work without much fuss. Of course, that probably means I made some silly error.

Take the space to be $\omega_2 \times 2$, and the $\sigma$-algebra to consist of those sets $A$ such that $A\cap (\omega_2 \times \{0\})$ is countable or cocountable and $A\cap (\omega_2 \times \{1\})$ is countable or cocountable. Suppose towards a contradiction that you've found a compatible topology. The intuition is that open sets really want to be cocountable, but they can't succeed on both halves.

[This paragraph has the error!] More precisely, given any cocountable set $X \subseteq \omega_2$, there is an open set $U$ in the topology and some $i$ such that $U \cap (X \times \{i\})$ is countable and nonempty. Otherwise there couldn't be any $A$ in the $\sigma$-algebra with $A \cap (X \times \{0\})$ countable nonempty and $A \cap (X \times \{1\})$ cocountable (but of course there are such $A$).

Now inductively build an increasing $\omega_1$-length sequence of countable sets $X_\alpha$ and an $\omega_1$-length sequence of open sets $U_\alpha$ such that (a) for some $i$, $U_\alpha \cap (\omega_2 \times \{i\}) \subseteq X_\alpha \times \{i\}$ and (b) for some $i$, $U_\alpha \cap (\omega_2 \times \{i\})$ is a countable, nonempty subset of $(\omega_2 \setminus \bigcup_{\beta < \alpha} X_\beta) \times \{i\}$. Without loss of generality, we may thin down to an $\omega_1$-length sequence where $i$ may always be chosen to be $0$. But then $\bigcup_{\alpha<\omega_1} U_\alpha$ would have an $\aleph_1$-sized intersection with $\omega_2 \times \{0\}$, contradicting the definition of the $\sigma$-algebra.

  • $\begingroup$ Unless I have misunderstood your example, I don't see that it works. Consider the topology $\tau$ consisting of the sets that are either co-countable or empty in each slice. These sets are in your $\sigma$-algebra, and an arbitrary union of sets of this form is also of this form; and intersections of finitely many sets of this form is also of this form. So it is a topology. And it generates your $\sigma$-algebra. Right? $\endgroup$ Feb 8, 2012 at 0:09
  • $\begingroup$ Your example is the sum of two copies of the countable/co-countable $\sigma$-algebra, which is the Borel algebra of the co-countable topology. So each slice separately is the Borel algebra of the co-countable topology on that slice, and then we put them together with a disjoint sum topology. $\endgroup$ Feb 8, 2012 at 0:14
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    $\begingroup$ Joel, you are of course absolutely right, and the error is clearly in the "More precisely" paragraph. Anyway I'll leave this embarrassment up as a cautionary tale for others. $\endgroup$ Feb 8, 2012 at 0:17
  • $\begingroup$ Clinton, yes, I agree. But thanks for thinking about it! I had also tried some examples like this at first, and it led me to the eventual-agreement candidate example that I mention in my question, which is similar, and for which I don't know the answer. $\endgroup$ Feb 8, 2012 at 0:20

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