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Stream: theory: categorical probability

Topic: Does the category of standard Borel spaces have pullbacks?

Nathaniel Virgo (Sep 07 2021 at 06:04):

Consider the category whose objects are standard Borel spaces and whose morphisms are measurable functions. (So the deterministic subcategory of BorelStoch.) Is it known whether this category has pullbacks?

(I hope it does but I don't have a great intuition for it, and my measure theory isn't good enough to attempt to work it out.)

As a separate but related question, does the same category have coproducts, or finite coproducts at least?

Matteo Capucci (he/him) (Sep 07 2021 at 10:32):

Re pullbacks, I'd say they do, even more in general. A pullback is product followed by an equalizer. Products of measurable spaces exists (they're constructed like product topologies). Now taking the equalizer here means restricting your space to a subspace, hence you can equip this with the subspace $\sigma$-field (again, works like topology).

Btw, in my thesis, I reference Theorem 10.6.2 in "Cohn, Donald L. (2013). Measure Theory: Second Edition" saying that this proves Meas doesn't have all limits (I don't remember why I claimed that...). Notably, the hypothesis of the theorem (Kolmogorov Consistency Theorem) are that the involved measure spaces are standard.

Matteo Capucci (he/him) (Sep 07 2021 at 10:33):

Coproducts should be even easier

John Baez (Sep 07 2021 at 12:24):

I'd imagine disjoint unions of Borel spaces provide countable coproducts. I don't think the category you mention has uncountable coproducts: an ucountable coproduct of copies of the point with its unique $\sigma$-algebra is not a standard Borel space. But if you're doing measure theory, you probably never wanted uncountable coproducts in the first place!

John Baez (Sep 07 2021 at 12:26):

I'd guess the category of standard Borel spaces has countable products too - and pullbacks, built in the sort of obvious way using a binary product and an equalizer.

John Baez (Sep 07 2021 at 12:28):

But I haven't carefully checked any of these things. It feels like a routine exercise, but it's good to actually prove such things, since sometimes one is surprised.

Jens Hemelaer (Sep 07 2021 at 12:49):

For the equalizers this result could help:
if $(X, \mathcal{B})$ is a standard Borel space and $S \subseteq X$ is a subset, then $(S, \mathcal{B}\cap S)$ is standard Borel if and only if $S$ is a Borel subset of $X$. This appears as Corollary 3.2.8 in S. K. Berberian's thesis.

The equalizer of two maps $f,g : X \to Y$ is given by the subset $A = \{ (u,v) : f(u) = g(u) \}$. To show that this is a Borel set, you can show that it is the preimage of the diagonal along the map $(f,g) : X \to Y \times Y$.

Matteo Capucci (he/him) (Sep 07 2021 at 13:37):

John Baez said:

I don't think the category you mention has uncountable coproducts: an ucountable coproduct of copies of the point with its unique $\sigma$-algebra is not a standard Borel space. But if you're doing measure theory, you probably never wanted uncountable coproducts in the first place!

Uncountable measure theory can be very interesting, Tao and @Asgar Jamneshan have been looking into it in the context of ergodic theory iirc

Dmitri Pavlov (Oct 11 2021 at 01:16):

It does admit pullbacks as long as you identify measurable maps that are equal almost everywhere. (Which is what you do normally in measure theory anyway.) See this paper: https://arxiv.org/abs/2005.05284

John Baez (Oct 11 2021 at 16:15):

Nice! Now I know what paper to read in case I ever want to do measure theory in a category-theoretic way again. In Infinite-Dimensional Representations of 2-Groups we studied representations of 2-groups whose the set of objects and set of morphisms were both standard Borel spaces (with some extra conditions, see Appendix A). It took us forever to write this book, but if I had superhuman energy I'd rewrite it using more von Neumann algebras.

John Baez (Oct 11 2021 at 16:16):

I find the intersection of higher categories and hard-core measure theory to be one of those areas that doesn't get enough love... for more or less obvious reasons. Someday it'll become important.

David Michael Roberts (Oct 12 2021 at 05:06):

John Baez said:

I find the intersection of higher categories and hard-core measure theory to be one of those areas that doesn't get enough love... for more or less obvious reasons. Someday it'll become important.

I agree. Higher representation theory is not topological enough, for my liking.

Nathaniel Virgo (Oct 12 2021 at 05:19):

Thank you very much Dmitri Pavlov!

Does anybody know how this category (with almost-everywhere-equal maps identified) relates to BorelStoch as Fritz uses it? I think that the deterministic subcategory of BorelStoch is the category of standard Borel spaces and all measurable maps, without identifying almost-everywhere-equal ones, so this answer seems like a negative one - is that correct?

If so, might there be any way to fix it, e.g. by defining a probability monad on the category Dmitri Pavlov described? My hope was to have all the nice known properties of BorelStoch (e.g. that it's representable and has conditionals) while also having pullbacks in its deterministic subcategory, but it could be that that's just wishful thinking.

Sam Staton (Oct 12 2021 at 06:34):

Hi, I thought that Jens' argument (here) was good for pullbacks in Borel spaces and measurable maps.

As to the category that Dmitri described, it is a nice category and I agree that it fits some things you might often do in probability theory. But iirc the diagonal map R -> R(x)R is not a morphism in that category, because the diagonal is negligible. (Correct me if I am mixing up some definitions!) If so, well, the diagonal map is very handy in various arguments about BorelStoch etc.. We discussed this a bit at the CPS workshop last year, and it should still be somewhere on this zulip.

Nathaniel Virgo (Oct 12 2021 at 07:16):

Jens' argument seemed good to me too, but I took Dmitri's "as long as" to mean "only if", so I thought I might have missed some important subtlety - I'm not sure.

Jens Hemelaer (Oct 12 2021 at 09:32):

In the category BorelStoch by Fritz, there is no identification of maps when they are almost everywhere equal. Fritz discusses alternatives to BorelStoch (featuring negligible sets) in this paper, in Examples 4.3 and 4.4 and the discussion afterwards.

Sam Staton said:

As to the category that Dmitri described, it is a nice category and I agree that it fits some things you might often do in probability theory. But iirc the diagonal map R -> R(x)R is not a morphism in that category, because the diagonal is negligible. (Correct me if I am mixing up some definitions!) If so, well, the diagonal map is very handy in various arguments about BorelStoch etc.. We discussed this a bit at the CPS workshop last year, and it should still be somewhere on this zulip.

I think the notion of almost-everywhere-equal in the paper by @Dmitri Pavlov is as follows: maps $f,g : X \to Y$ are almost everywhere equal if and only if $\{ x \in X : f(x) \neq g(x) \}$ is a negligible set. In particular, the diagonal map is not almost everywhere equal to a map that just sends everything to a point (because $\mathbb{R}-\{0\}$ is not negligible in the domain).

There are different ways to equip a standard Borel space with a notion of "negligible set", but I believe that there is a unique good choice up to isomorphism (where isomorphism means that $f \circ g$ and $g\circ f$ are almost everywhere equal to the identity map).

Dmitri Pavlov (Oct 12 2021 at 13:41):

Sam Staton said:

Hi, I thought that Jens' argument (here) was good for pullbacks in Borel spaces and measurable maps.

As to the category that Dmitri described, it is a nice category and I agree that it fits some things you might often do in probability theory. But iirc the diagonal map R -> R(x)R is not a morphism in that category, because the diagonal is negligible. (Correct me if I am mixing up some definitions!) If so, well, the diagonal map is very handy in various arguments about BorelStoch etc.. We discussed this a bit at the CPS workshop last year, and it should still be somewhere on this zulip.

The diagonal is not negligible in the categorical product. It is negligible in the measure-theoretic product, which is a nontrivial subobject of the categorical product. This is explained in detail here: https://mathoverflow.net/questions/49426/is-there-a-category-structure-one-can-place-on-measure-spaces-so-that-category-t/49542#49542

Dmitri Pavlov (Oct 12 2021 at 13:47):

Nathaniel Virgo said:

Jens' argument seemed good to me too, but I took Dmitri's "as long as" to mean "only if", so I thought I might have missed some important subtlety - I'm not sure.

While you may be able to squeeze out some categorical limits or colimits, depending on how you arrange the definitions, my experience show that the resulting objects either coincide with what you get after identifying almost everywhere equal morphisms, and if they don't, they tend to be “wrong”.

Here is a fun fact about the category of measurable spaces I mentioned above (calles CSLEMS in the paper).
It is complete and cocomplete. Its opposite category is locally presentable.

It has a nontrivial monoidal structure (given by the measure-theoretic product) and this monoidal structure is closed (!). This last point is not addressed at all without sets of measure 0 and I do not believe it can be (there are obstructions discovered by Aumann).

Another fun fact: this category is (equivalent to) a full subcategory of the category of locales. Think about it: measure theory is, quite literally, a part of (pointfree) general topology!

Sam Staton (Oct 15 2021 at 08:44):

Thanks! Has anyone tried to look at internal languages for this category CSLEMS?
It could be really great if it would work. I have tried a bit over the years, but I haven't managed yet.

For example, if I understand correctly, addition is not a morphism R^2 -> R from the categorical product, only from the other monoidal product. (I’m thinking about the inverse image of {0}.) So we cannot easily interpret an expression like "x+x", which would ordinarily be interpreted as a composite R -diag-> R^2 -add-> R. If you see what I mean, the two morphisms are not ok together, even though the composite function R -> R is fine as a map.

Sam Staton (Oct 15 2021 at 08:45):

[I’m asking partly because the internal language of Markov categories can be regarded as a sort-of probabilisitic programming language, of the kind that are already used in stats and ML. Tobias’s work gives a great way to understand negligibility and almost sureness in that context. But it would also be very interesting to see different perspectives and different internal languages.]

Dmitri Pavlov (Oct 19 2021 at 17:47):

Sam Staton said:

Thanks! Has anyone tried to look at internal languages for this category CSLEMS?
It could be really great if it would work. I have tried a bit over the years, but I haven't managed yet.

For example, if I understand correctly, addition is not a morphism R^2 -> R from the categorical product, only from the other monoidal product. (I’m thinking about the inverse image of {0}.) So we cannot easily interpret an expression like "x+x", which would ordinarily be interpreted as a composite R -diag-> R^2 -add-> R. If you see what I mean, the two morphisms are not ok together, even though the composite function R -> R is fine as a map.

Addition actually works perfectly fine with both products. It is easy to map out of the measure-theoretic product, and the diagonal does not pose a problem when mapping out.

Sam Staton (Oct 22 2021 at 08:47):

Dmitri Pavlov said:

Addition actually works perfectly fine with both products. It is easy to map out of the measure-theoretic product, and the diagonal does not pose a problem when mapping out.

Thanks @Dmitri Pavlov. Can I check I understand you correctly? I can see that addition is a morphism out of the measure-theoretic tensor in CSLEMS. But are you saying that it's also a morphism out of the categorical product? I’ll write $\mathbb{R}^2$ for the categorical product.

I’m pretty sure negation is a morphism $\mathbb{R} \to \mathbb{R}$. So if addition $\mathbb{R}^2 \to \mathbb{R}$ is a morphism, then
subtraction $\mathbb{R}^2 \to \mathbb{R}$ is also a morphism. I think $\{0\}$ is negligible in $\mathbb{R}$. But $subtraction^{-1}(\{0\})=\{(x,x)|x \in \mathbb{R}\}$, the diagonal in $\mathbb{R}^2$, which you had said is _not_ negligible in $\mathbb{R}^2$. A contradiction? Am I making a silly mistake?

Dmitri Pavlov (Oct 22 2021 at 15:04):

Sam Staton said:

Dmitri Pavlov said:

Addition actually works perfectly fine with both products. It is easy to map out of the measure-theoretic product, and the diagonal does not pose a problem when mapping out.

Thanks Dmitri Pavlov. Can I check I understand you correctly? I can see that addition is a morphism out of the measure-theoretic tensor in CSLEMS. But are you saying that it's also a morphism out of the categorical product? I’ll write $\mathbb{R}^2$ for the categorical product.

I’m pretty sure negation is a morphism $\mathbb{R} \to \mathbb{R}$. So if addition $\mathbb{R}^2 \to \mathbb{R}$ is a morphism, then
subtraction $\mathbb{R}^2 \to \mathbb{R}$ is also a morphism. I think $\{0\}$ is negligible in $\mathbb{R}$. But $subtraction^{-1}(\{0\})=\{(x,x)|x \in \mathbb{R}\}$, the diagonal in $\mathbb{R}^2$, which you had said is _not_ negligible in $\mathbb{R}^2$. A contradiction? Am I making a silly mistake?

The notation R is ambiguous here. For spatial (i.e., measure-theoretic) products, R refers to the Lebesgue real line, with the Borel σ-algebra of measurable sets, and Borel σ-ideal of negligible sets. For categorical products, R is not the Lebesgue real line, but rather the real line constructed here: https://mathoverflow.net/questions/31603/why-do-probabilists-take-random-variables-to-be-borel-and-not-lebesgue-measura/31724#31724

In fact, by the representable functor theorem, we have a functor from locally compact locales to measurable locales (given a locally compact locale L, consider the restricted Yoneda embedding of L, given by the functor M↦hom(M,L) on measurable locales, and take the representing object of this functor), and I am pretty sure (but I have not checked all the details) that the subtraction morphism for the categorical product is obtained by applying this morphism to the subtraction map on R as a locally compact locale.

The nLab article [[measurable locales]] also mentions this construction.

Sam Staton (Oct 25 2021 at 16:31):

Ah thanks very much, that's clear now. So the internal language of CSLEMS might be a bit complicated, with (at least) two types of real numbers, two tensors, and conversions between them. But it might still be informative.