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There are many questions on this site about the following two topics:

  • In the definition of a topology, why do we require it to be closed under arbitrary unions but only under finite intersections?

  • In measure theory, why do we require a $\sigma$-algebra to be closed under countable unions, rather than finite unions or arbitrary unions?

For the first of these I feel like I more or less understand the reasons. I particularly like Thomas Andrews' answer here, interpreting topology in terms of intuitionistic logic, but I also understand the more familiar justifications in terms of 'closeness' or by generalising from real analysis.

However, the second question, about $\sigma$-algebras, still seems a bit opaque to me. The answers I understand are along the lines of "because finite unions are too weak to give the right results, and arbitrary unions are too strong." This is fine as far as it goes, but it doesn't give me the feeling I understand why.

Now we get to my question. The two definitions seem like they should be related, both because Borel measures are in some way canonical and are defined in terms of open sets, and because the concept of a valuation (see Wikipedia, nlab) is closely related to the concept of measure, but is defined on a topology instead of a $\sigma$-algebra.

So my question is, how exactly are the two definitions above related, particularly in regards to the distinction between arbitrary, countable and finite unions and intersections? And where exactly do valuations come in to the story?

I'm interested in understanding probability theory as an extension of logic, so I would be particularly keen on an answer along the lines of Thomas Andrews' answer about topologies, explaining why $\sigma$-algebras are the appropriate object from a logic point of view. But any answer is welcome.

N. Virgo
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    Regarding $\sigma$-algebras, here's how I used to motivate the use of countable unions and countable intersections. We want closure under complements, so has to be the same (finitely many, countably many, arbitrarily many) for both unions and intersections. The reason for "countably many" is to be robust with countability features of analysis, such as: (1) convergence of sequences (of numbers, of functions, of sets); (2) separability/second-countability of spaces such as ${\mathbb R}^n$ and ${\mathcal L}^p$; (3) countable additivity of measure and problems with arbitrary additivity. – Dave L. Renfro Mar 02 '20 at 17:59
  • @DaveL.Renfro that does make sense, but it's still quite close to the "finite is too weak, arbitrary is too strong" argument I mentioned in the question - it still makes me want to ask "yes, but why?" (In particular, why do we need that particular set of features for probability? If we chose countable union instead we would lose those features but in return we'd get a uniform measure on the natural numbers and a translation invariant measure on the reals, which also seem desirable.) I'm sorry to be such a fussy student! – N. Virgo Mar 02 '20 at 18:27
  • Actually, I didn't intend for my comment to be much of an answer to your question (character limitations prevented me from saying this), since you're asking for a modern "high level" (i.e. use modern logic and category theory, if possible) explanation, whereas what I gave was pretty much forced on the early giants of the field during the development of measure theory (Borel, Lebesgue, etc.) and its abstraction to more general settings in the 1910s (I think Frechet originated the $\sigma$-algebra approach) through the 1930s. For probability, I'm guessing Kolmogorov, but I assume you know this. – Dave L. Renfro Mar 02 '20 at 18:39

1 Answers1

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I’d say that the two are different things.

There are simple logical reasons for the topological thing, whereas measure turns out more elusive.

Essentially, we have an intuition for measure, and it turns out, there is no way to define measure on Euclidean space that works for all sets. The definition of Borel then is a work-around because measure is tricky. You can certainly have measures that are uncountably additive, but they tend to be trivial (each point has a measure, and the measure of the set is the sum of the measures of the points - a discrete measure.)

Uncountably additive is, of course, also not much use, because any sum of uncountably many positive reals is $+\infty.$ So we define countably additive as “enough,” because we can’t always get a measure on all sets.

I suppose, if you think of topologies as metric spaces, then the topology condition can be thought of as:

The infimum of finitely many positive real numbers is a positive real number, but the infimum of infinitely many positive real numbers might be zero.

The measure theory reason to only care about countable sums is due to:

A sum of positive reals cannot be a positive real unless the sum is countable.

These seem logically very different to me. They are similar, as statements about sets of positive real numbers, but they act in different directions.

But the real reason, to me, is that the definition for measure theory is entirely motivated by the failures of our intuition to be fully realized in our most basic case, the real line. The solution always felt like a hack, to me. “Oh well, best we can do!”

I suppose there is one similarity - if we allowed arbitrary intersections in topology (at least for Hausdorff topologies) we get only discrete topologies.

If we allow arbitrary unions in measure theory, we get stuck with just discrete metrics.

Interestingly, in constructive math, all sets of real numbers are measurable. And all functions are continuous.

So, on some level, the problem in measure theory is “solved” by removing the things in set theory and logic that allow for the “sick” counterexamples.

But in intuitionist logic, any non-finite process can only be constructive, so there are no uncountable unions there, anyway.

It might also be true that in constructive math, all subsets of the real line are open. Not sure. However, in constructive math, non-finite intersections are not allowed because membership of an infinite intersection has no constructive test, in general.

So, on the real line, at least, constructive math only allows countable unions and finite intersections. Perhaps this is the ultimate common reason - we are defining these things these ways to emulate constructive math.

Thomas Andrews
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