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.


  • 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.

  • $\begingroup$ 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? $\endgroup$ – Todd Trimble Feb 3 '16 at 5:54
  • $\begingroup$ 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? $\endgroup$ – Dmitri Pavlov Feb 3 '16 at 12:43
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    $\begingroup$ 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. $\endgroup$ – Taylor Friesen Feb 3 '16 at 17:06
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    $\begingroup$ 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? $\endgroup$ – Taylor Friesen Feb 3 '16 at 17:08
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    $\begingroup$ 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. $\endgroup$ – Taylor Friesen Feb 5 '16 at 17:36

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 RR 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.

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    $\begingroup$ 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). $\endgroup$ – Todd Trimble Feb 3 '16 at 12:34
  • $\begingroup$ @ToddTrimble: I guess the only way to find out is to ask the OP… $\endgroup$ – Dmitri Pavlov Feb 3 '16 at 12:42
  • $\begingroup$ Is the category of measurable locales accessible? Or else what do you mean? $\endgroup$ – Zhen Lin Feb 3 '16 at 17:20
  • $\begingroup$ @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. $\endgroup$ – Dmitri Pavlov Feb 3 '16 at 20:54
  • $\begingroup$ Err, the opposite of a locally presentable category is accessible if and only if it is a preorder... $\endgroup$ – Zhen Lin Feb 3 '16 at 22:06

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