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Is there some notion of "nice" measurable spaces and "nice" maps between them which satisfies the following properties?

  • The real line equipped with the Lebesgue $\sigma$-algebra is nice.
  • Any translation $\mathbb R \to \mathbb R$ is nice. (Ideally, any continuous map $\mathbb R \to \mathbb R$ is nice, but right now I'm not picky.)
  • Any finite or countable set equipped with the discrete $\sigma$-algebra is nice.
  • Nice measurable spaces and nice maps form an elementary topos.

Observations:

  • If such a topos exists, it is Boolean but does not satisfy the axiom of choice. To prove the latter: Construct the unit circle as a quotient of $\mathbb R$, and then consider the quotient of the unit circle by an irrational rotation. This quotient map cannot split, because if it did the image of its section would be a Vitali set.
  • If such a topos does not exist, any proof of its nonexistence must rely on the axiom of choice, since in the Solovay model the category of discrete measurable spaces satisfies the conditions.

Edited to add: I would also be interested in the answer to the same question with the Lebesgue $\sigma$-algebra replaced by the Borel $\sigma$-algebra.

Taylor Friesen
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  • We are told so little about the morphisms that it seems hard to deduce anything about any such topos. The only function that can equalize two distinct translations of $\mathbb{R}$ is the one with empty domain, so the initial object could only be the empty measurable space. But I'm honestly having trouble seeing that the terminal could only be the one-point space $1$ (e.g., we don't know that there are any nice maps to $1$, except the identity on $1$). How do you see the topos would have to be Boolean? – Todd Trimble Feb 03 '16 at 05:54
  • Do you want your nice category to be a subcategory of measurable spaces, or do you want it to contain measurable spaces as a full subcategory? – Dmitri Pavlov Feb 03 '16 at 12:43
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    I was looking for a subcategory of measurable spaces, although as multiple people have pointed out I didn't put enough restrictions on the morphisms involved. I also meant "measurable space" to mean a set equipped with a $\sigma$-algebra, not one additionally equipped with an ideal of null sets. I'm trying to rewrite my question to better capture what I'm looking for, but it's possible that I won't manage to make it rigorous and I'll need to make this into a soft question. – Taylor Friesen Feb 03 '16 at 17:06
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    In the meantime, here's the intuition: There are various functors which take some kind of space $X$ to a space of measures on $X$. (The latter space has some linear or linear-like structure, depending on the type of measure.) If we have anything which looks like "the uniform measure on the unit interval", Vitali's theorem suggests that the source category can't be a topos with choice. My question is: Can we do the next best thing, and have the source category be a topos, while maintaining the ability to study at least the more common measur(abl)e spaces coming from other fields of mathematics? – Taylor Friesen Feb 03 '16 at 17:08
  • Considering that virtually all nontrivial theorems in measure theory require a σ-ideal of null sets either explicitly or implicitly, it would be really surprising if one could do without it in this context. Also, it seems to me that the quotient of a circle by an irrational rotation should be a point. Indeed, functions on such a quotient are functions on a circle invariant under irrational rotations, and the only such functions are constant functions. In fact, the real line can be replaced by any target, and the Yoneda lemma tells us that the quotient is isomorphic to the point. – Dmitri Pavlov Feb 04 '16 at 16:32
  • Generally speaking, one cannot expect to have good quotients in the context of 1-toposes anyway, and passage to ∞-toposes is usually necessary. Indeed, the stacky quotient of a circle by an irrational rotation is a well-known (elementary) example in noncommutative geometry, and it is studied there by means of its groupoid convolution algebra. – Dmitri Pavlov Feb 04 '16 at 16:36
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    Unless I'm badly misunderstanding something, though, the quotient of the unit circle by an irrational rotation exists and is not a point in the classical theory: The points are orbits of the rotation and the measurable sets are sets of orbits whose union is Lebesgue measurable. It's not a space which is amenable to study by its $L^1$ functions, but in the category of sets equipped with $\sigma$-algebras, it's not isomorphic to a point. – Taylor Friesen Feb 05 '16 at 17:36
  • This second point implies that there is a sub-object classifier given by the algebra $\mathbb{C} \oplus \mathbb{C}$, or two points locales, and hence that this category is itself a boolean elementary topos (which seems extremely weird, hence the disclaimer in the beginning) in fact it also have arbitrary co-products and a generator (the standard measure space) hence it is a Grothendieck topos (which I found even weirder).

    Finally this category can be described as a category of nice measure spaces, with measurable map modulo equality almost everywhere.

     – Simon Henry Feb 03 '16 at 13:52
  • How do you show that this category is cartesian closed? (Recall that an elementary topos is a finitely complete cartesian closed category with a subobject classifier.) I remember reading in one of Andre Kornell's papers that it isn't, which is why I'm confused now. – Dmitri Pavlov Feb 03 '16 at 16:17
  • In fact, Theorem 5.8 in Kornell's “Quantum collections” shows that the opposite category of (noncommutative) von Neumann algebras is not cartesian closed. – Dmitri Pavlov Feb 03 '16 at 16:24
  • ... I knew I was missing something obvious ! This being said, a quick google search gave me this note http://www.math.ru.nl/~landsman/Uijlen.pdf (I think a master thesis of a student of K.Landsman) With inside a proof that the opposite of the category of VN algebras is monoidal closed for the spatial tensor product... This might be a start... – Simon Henry Feb 03 '16 at 16:42
  • The main theorem of Kornell's paper is to show that the opposite of VNA is closed monoidal for the spatial tensor product, and his paper appeared earlier. – Dmitri Pavlov Feb 03 '16 at 16:54
  • Indeed, I just found it. Kornell's paper also prove that there is no co-exponential in the categories of commutative VN algebras ( corrolary 6.7). So this shows I was wrong. But I have the impression that his argument say something one the initial question. – Simon Henry Feb 03 '16 at 17:01
  • So I guess we could say that measurable locales form a closed symmetric monoidal Boolean pretopos. – Dmitri Pavlov Feb 03 '16 at 17:20
  • Instead of deleting this post with the comments, you can flag it and ask the moderators to convert it to comments. In fact, this is what I will do right now. – Dmitri Pavlov Feb 03 '16 at 17:21
  • Simon, let's roll back so that people can see what the comments were about. (It will also help others discern some of the issues involved.) You could also convert to CW so that any downvotes carry no penalty. Or, I could roll back and convert everything to comments as Dmitri suggested, which will have the effect of deleting this as an answer – Todd Trimble Feb 03 '16 at 18:59
  • I would simply change “elementary topos” to “pretopos” (which will fix the only mistake), then convert to comments. – Dmitri Pavlov Feb 03 '16 at 20:55
  • By the way, another problem with measurable locales not being a Grothendieck topos is that they don't admit a set of generators. Indeed, consider the cartesian power W_κ of arbitrary cardinality κ of the two-point discrete locale. A map W_κ→W_λ exists only if λ≤κ, so in order to “see” all measurable locales one needs a proper class of objects of the form W_κ. (Any measurable locale can be presented as a coproduct of W_κ, which gives an exhaustive classification of measurable locales.) – Dmitri Pavlov Feb 05 '16 at 17:00
  • Disclaimer: I have the impression that there is something very wrong with my answer... I'm hoping someone will tell me if it is wrong.) Let $C$ be the opposite category of the category of commutative von Neumann algebras, equivalently the category of measurable boolean locale Dimitri was talking about in his answer. Then: 1) C has all finite limits (fiber products are given by tensor product of Von Neumann algebras) 2) Monomorphisms are maps $A \twoheadrightarrow pA$ for $p$ a symmetric idempotent, or open inclusion in terms of locales. (continued) – Simon Henry Feb 03 '16 at 13:52

1 Answers1

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Take the category of measurable locales, equip it with its natural Grothendieck topology, and take the topos of sheaves of sets on the resulting site. (Apply standard disclaimers about universes, coaccessibility, or cosmall sheaves to avoid size issues.)

The resulting Grothendieck topos contains the category of measurable locales (or, equivalently, the category of measurable spaces, see https://mathoverflow.net/a/20820 for a detailed description of the latter, and https://mathoverflow.net/a/49542 for more information). Therefore it satisfies all of your properties, except for the part about arbitrary continuous maps: the preimage of a measure 0 set under a continuous map R→R is not necessarily of measure 0, so you cannot hope to get such maps. But you do get all continuous maps whose preimage preserves measure 0 sets.

Community
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Dmitri Pavlov
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    I think I read the question a little differently. It seems to me OP wants a subcategory of measurable spaces and measurable maps to form a topos (where "subcategory" is in the set-theoretic sense: subclass of objects, and homs are subsets -- which is of course not a notion that is invariant under equivalence). Here you've instead expanded the category (which might be the more sensible thing to consider, but still). – Todd Trimble Feb 03 '16 at 12:34
  • @ToddTrimble: I guess the only way to find out is to ask the OP… – Dmitri Pavlov Feb 03 '16 at 12:42
  • Is the category of measurable locales accessible? Or else what do you mean? – Zhen Lin Feb 03 '16 at 17:20
  • @ZhenLin: The category of measurable locales is accessible because it is the opposite of the category of commutative von Neumann algebras, and the latter is locally presentable. – Dmitri Pavlov Feb 03 '16 at 20:54
  • Err, the opposite of a locally presentable category is accessible if and only if it is a preorder... – Zhen Lin Feb 03 '16 at 22:06
  • @ZhenLin: You are right, of course, measurable locales are coaccessible, not accessible. (In my mind I had the example of the category of Hilbert spaces, which is accessible and coaccessible—but it's not complete or cocomplete either.) – Dmitri Pavlov Feb 03 '16 at 22:17
  • I suppose coaccessibility is what we want, though. Small presheaves on an coaccessible categories are the same as (covariant) accessible functors, if I recall correctly. On the other hand, restricting to small sheaves is not going to give you a Grothendieck topos... – Zhen Lin Feb 04 '16 at 00:00
  • @ZhenLin: You're right, we need some additional work to show that we have (or don't have) a Grothendieck topos, or something close to it. We do seem to have a problem with a set of small generators: taking cartesian powers of arbitrary high cardinality of the discrete two-point measurable locale will eventually give us measurable locales that cannot be probed by any fixed set of measurable locales. Perhaps we have here a class-locally presentable elementary topos (in the sense of class-locally presentable categories of Chorny and Rosický)? – Dmitri Pavlov Feb 04 '16 at 01:01
  • @ZhenLin: In fact, taking sheaves on the full subcategory of measurable locales consisting of two objects, the point and the real line, gives us an honest Grothendieck topos satisfying the desired properties. – Dmitri Pavlov Feb 04 '16 at 02:49
  • In situations like this we should a class-locally presentable pretopos. I'm not so sure whether we get a cartesian closed category, or even whether we get a subobject classifier. Restricting to a small "subsite" (not standard terminology) would give a Grothendieck topos but it seems like an arbitrary thing to do. – Zhen Lin Feb 04 '16 at 07:23