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Stream: theory: mathematics

Topic: a lemma on the joins of measurable functions

Matteo Capucci (he/him) (Mar 26 2024 at 15:57):

The pointwise supremum of measurable function doesn't have to be measurable itself, but I found this MO answer claiming that the join, meant as a colimit in a poset, still exists but it's not given by pointwise supremum. They give a proof and also link to a reference with a very similar proof.
I'm only half confident it is a legit result, does anyone else want to weigh in?

Matteo Capucci (he/him) (Mar 26 2024 at 15:58):

the proof

Matteo Capucci (he/him) (Mar 26 2024 at 15:59):

The very last line "Obviously $\int_E v \, dx=s$. This easily implies that $v=\bigvee F$" is what confuses me

John Baez (Mar 26 2024 at 16:04):

Surely the sup of a countable family of measurable functions is measurable, and I feel that trying to go beyond the countable case would be going against the whole spirit of measure theory.

John Baez (Mar 26 2024 at 16:06):

So before thinking about this proof I have to wonder why you care.

Morgan Rogers (he/him) (Mar 26 2024 at 16:11):

Missing / implicit there is the argument that if there is $u\in \mathcal{F}$ which exceeds $v$ on any set of positive measure then the integral of $v$ must be less than $s$, a contradiction. Thus $v$ almost everywhere bounds all functions in $\mathcal{F}$, and since there is a sequence of functions converging to it from below, it must be a least upper bound.

John Baez (Mar 26 2024 at 16:11):

As for the last step I'd say maybe "obviously" $v = \bigvee \mathcal F$ a.e.. The argument needs more steps. But the best you can get from an equation between integrals is equality of their integrands almost everywhere. Are we working with equivalence classes of functions that are equal a.e.?

John Baez (Mar 26 2024 at 16:12):

(Morgan just added some more steps.)

Matteo Capucci (he/him) (Mar 26 2024 at 16:14):

John Baez said:

Surely the sup of a countable family of measurable functions is measurable, and I feel that trying to go beyond the countable case would be going against the whole spirit of measure theory.

Though I can buy one thing from that proof which is that you can always replace an uncountable sup of reals/real-valued functions with a countable one because of separability

Matteo Capucci (he/him) (Mar 26 2024 at 16:15):

John Baez said:

So before thinking about this proof I have to wonder why you care.

I'd like to say measurable functions $I \to [0,\infty]$ form a [[quantale]], thus in particular a [[suplattice]].

Matteo Capucci (he/him) (Mar 26 2024 at 16:16):

John Baez said:

As for the last step I'd say maybe "obviously" $v = \bigvee \mathcal F$ a.e.. The argument needs more steps. But the best you can get from an equation between integrals is equality of their integrands almost everywhere. Are we working with equivalence classes of functions that are equal a.e.?

Yes I forgot to mention that

John Baez (Mar 26 2024 at 16:16):

How about the sup of all functions on the real line that are characteristic functions of singletons?

Matteo Capucci (he/him) (Mar 26 2024 at 16:17):

Isn't it the same as the sup of all functions on the real line that are characteristic functions of the intervals $[n,n+1]$?

John Baez (Mar 26 2024 at 16:17):

The sup is 1; the sup of any countable subcollection is zero a.e..

Matteo Capucci (he/him) (Mar 26 2024 at 16:17):

Oh I see. No I'm not saying it has to be a subcollection!

John Baez (Mar 26 2024 at 16:20):

Well then the sup of any collection of functions is the sup of the one-element set of functions consisting of the sup.

Matteo Capucci (he/him) (Mar 26 2024 at 16:22):

uhm ok

Matteo Capucci (he/him) (Mar 26 2024 at 16:22):

image.png
also the theorem here says subcollection!

Matteo Capucci (he/him) (Mar 26 2024 at 16:23):

I guess your example doesn't apply since $1_{\{x\}} = 0$ almost everywhere

Tobias Fritz (Mar 26 2024 at 16:25):

Matteo Capucci (he/him) said:

The pointwise supremum of measurable function doesn't have to be measurable itself, but I found this MO answer claiming that the join, meant as a colimit in a poset, still exists but it's not given by pointwise supremum.

That claim is false. If it was true, then in particular the $\sigma$-algebra of measurable sets would have to be a complete Boolean algebra (just restrict to $\{0,1\}$-valued functions to see this), but this is not the case. For example, let $A \subseteq \mathbb{R}$ be any non-measurable set, and consider the family of its points considered as singletons. Let''s denote the claimed supremum of this family by $A'$. Then we must have $A' \subseteq A$, since for every $x \in \mathbb{R} \setminus A$, the set $\mathbb{R} \setminus \{x\}$ upper bounds the family. On the other hand, we must have $A \subseteq A'$, since every element of $A$ appears in some member of the family. Hence $A' = A$, which is a contradiction by the assumed non-measurability.

Tobias Fritz (Mar 26 2024 at 16:27):

Right? Please double-check, since saying that an old MO answer with a good number of upvotes is wrong seems like a steep claim :sweat_smile:

Tobias Fritz (Mar 26 2024 at 16:30):

What is true, and what seems to be the direction that @John Baez is hinting at, is that you do get a complete lattice if you identify measurable functions that coincide almost everywhere with respect to any fixed finite measure. (And also for certain infinite measures, probably s-finite is enough.)

Morgan Rogers (he/him) (Mar 26 2024 at 16:31):

I believe the ordering in the MO answer has $u \leq v$ iff that holds almost-everywhere (this is a preorder, and quotienting by almost-everywhere equality makes it a poset).

Morgan Rogers (he/him) (Mar 26 2024 at 16:31):

So your last message is the relevant one @Tobias Fritz

Tobias Fritz (Mar 26 2024 at 16:34):

Hmm okay, I see, I just skipped that part since I figured I know the definition of lattice supremum :sweat_smile: Perhaps that deserves to be made more explicit in that MO answer? In any case, that's a very old result, and I'm sure that it was known to von Neumann already

Tobias Fritz (Mar 26 2024 at 16:35):

Since algebras of bounded measurable functions modulo a.e. equality are the most basic examples of von Neumann algebras, and von Neumann algebras are well-known to have suprema of all bounded families (of self-adjoint elements). And as noted in the MO answer, lifting the boundedness assumption is easy

Matteo Capucci (he/him) (Mar 26 2024 at 16:54):

Thanks Tobias, your reply made me realize this is just a variation of something I know very well, i.e. that quotienting a $\sigma$-algebra by the ideal of null sets yields a complete poset.

Tobias Fritz (Mar 26 2024 at 16:56):

Yes, that's a nice way to put it!

Tobias Fritz (Mar 26 2024 at 16:58):

It may also be possible to give a clean proof based on that idea by looking at sublevel sets of the functions and noting that the sublevel sets of the supremum must be the suprema of the sublevel sets in the $\sigma$-algebra modulo null sets

Dmitri Pavlov (May 04 2024 at 03:38):

Matteo Capucci (he/him) said:

The pointwise supremum of measurable function doesn't have to be measurable itself, but I found this MO answer claiming that the join, meant as a colimit in a poset, still exists but it's not given by pointwise supremum. They give a proof and also link to a reference with a very similar proof.
I'm only half confident it is a legit result, does anyone else want to weigh in?

This is one of the many equivalent definitions of localizable measure spaces, a concept introduced by Irving Segal in 1950. In particular, measurable subsets of R^n are localizable. Perhaps the quickest introduction is a paper by Kelley “Decomposition and Representation Theorems in Measure Theory”. The nLab article [[categories of measure theory]] provides further examples and motivation, and also has additional references.