measurable space structure on restricted product

Stream: maths

[Kevin Buzzard]

Reminder: if $\mathcal{F}$ is a filter on an index set $I$ and we have sets $X_i$ for $i\in I$ equipped with subsets $C_i$, then mathlib has the docs#RestrictedProduct of the $X_i$ with respect to the $C_i$, which as a set is the elements $(x_i)$ in $\prod_i X_i$ such that $\{i:x_i\in C_i\}\in\mathcal{F}$. For many choices of structure Foo, if $X_i$ is a Foo and the $C_i$ are sub-Foos then the restricted product is also a Foo. For example the restricted product of groups is a group, the restricted product of rings is a ring.

In mathlib we also have that the restricted product of topological spaces is a topological space (although the construction in the literature is perhaps not obvious: the topology is not the subspace topology of the product topology on $\prod_i X_i$). Anatole explains the construction in the module docstring -- if $\mathcal{F}$ is a principal filter $\mathcal{P}(S)$ then the restricted product is just a product $\prod_{i\in S}C_i\times\prod_{i\not\in S}X_i$ with the product topology, and a general restricted product is naturally a colimit (an inductive limit) of such things and gets the inductive limit topology.

But what about measure theory? Here are my questions.

1) Is a restricted product of measurable spaces wrt to measurable subspaces "naturally" a measurable space? Can one do an analogous construction as the topological case? Is a filtered colimit of measurable spaces naturally a measurable space?

2) If everything is second countable, is a restricted product of Borel spaces a Borel space?

3) Assuming the answer to the first two questions is "yes", are we likely to one day have this instance in mathlib? This is a slightly funny question because I know of no application of measure theory on restricted products unless everything is second countable and the measure is a Borel measure, and in this case one can just cheat and borelize the restricted product rather than having to do anything about inductive limit measures (if such a theory even exists and is useful). However if I borelize in FLT and then someone actually makes the general measurable space construction on a restricted product then this will cause diamonds. But before I have a clearer idea on the situation here I need to understand better the answers to Q1 and Q2.

[Aaron Liu]

Kevin Buzzard said:

1) Is a filtered colimit of measurable spaces naturally a measurable space?

You can give it a measurable space structure that makes it a colimit in the category of measurable spaces and measurable maps. This is the basically the same thing that you do with topological spaces.

[Yury G. Kudryashov]

What's known about the index type? Is it countable? Is the filter countably generated?

[Kevin Buzzard]

In the application I care about the index type is countably infinite and the filter is the cofinite filter. I know of no other applications of restricted products in mathematics other than "adelic" situations (there are variants on the adele ring, such as GL_n(adele ring) which is a restricted product of GL_n(completions) etc etc) but perhaps @Adam Topaz (who maybe knows something about higher class field theory) will know of more exotic examples (although I don't know if people do measure theory in more general situations).

[James E Hanson]

I believe I have an example where the colimit measurable $\sigma$-algebra is strictly finer than the Borel $\sigma$-algebra induced by the topology on the restricted product. I don't know whether the restricted product $\sigma$-algebra is a Borel $\sigma$-algebra for some topology on the restricted product. I can't see an easy way to ensure this. The filter in my example is not countably generated though. This seems essential to the example, and it's plausible to me that assuming $\mathcal{F}$ is countably generated might be enough to make the obvious nice statement work (i.e., that the restricted product $\sigma$-algebra is the Borel $\sigma$-algebra of the restricted product topology).

In the example, $I = \omega$, $X_i = 2 = \{0,1\}$ for each $i$, and $C_i = \{0\}$. (So the full product space is Cantor space.)

Given sets $A,B \subseteq \omega$, write $A \subseteq^\ast B$ to mean that $B \setminus A$ is finite. Let $A \subset^\ast B$ denote that $A \subseteq^\ast B$ and $B \not \subseteq^\ast A$. Let $A \equiv^\ast B$ denote that $A \subseteq^\ast B$ and $B \subseteq^\ast A$.

It's a nice set theory exercise to build a sequence $(S_\alpha)_{\alpha < \omega_1}$ of infinite subsets of $\omega$ such that for any $\alpha < \beta < \omega_1$, $S_\beta \subset^\ast S_\alpha$. Let $\mathcal{F}$ be the filter on $\omega$ generated by the $S_\alpha$'s. (This is something kind of ultrafilter-y, but it may not literally be an ultrafilter.) Note that for any $S \in \mathcal{F}$, there is an $\alpha < \omega_1$ such that $S_\alpha \subseteq^\ast S$. Let $R \subseteq 2^{\omega}$ be the (underlying set of the) restricted product with regards to this data.

For each $S \subseteq \omega$, write $X_S$ for the product $\prod_{i \in S} C_i \times \prod_{i \notin S} X_i$, regarded as a subset of Cantor space (i.e., $X_S$ is the set of elements of Cantor space that vanish on all elements of $S$). Obviously for any co-infinite $S$, $X_S$ is a non-empty perfect subset of Cantor space. Moreover, if $S \subset^\ast S'$, then $X_{S'} \cap X_{S}$ is nowhere dense in $X_S$. This means that for each $\alpha < \omega_1$, $X_{S_\alpha} \cap \bigcup_{\beta < \alpha,~n <\omega} X_{S_\beta \cap [n,\infty)}$ is meager in $X_{S_\alpha}$. By a minor variant of the Baire category theorem, this means we can find a non-empty perfect closed set $F_\alpha \subseteq X_{S_\alpha} \setminus \bigcup_{\beta < \alpha,~n <\omega} X_{S_\beta \cap [n,\infty)}$ for each $\alpha < \omega_1$. Note that by construction the $F_\alpha$'s are pairwise disjoint.

Unwinding the colimit definitions gives that a set $U \subseteq R$ is open in the restricted product topology iff $U \cap X_S$ is relatively open in $X_S$ for each $S \in \mathcal{F}$. Likewise $U \subseteq R$ is measurable in the restricted product measurable space structure iff $U \cap X_S$ is Borel in $X_S$ for each $S \in \mathcal{F}$.

Note that by this characterization of the topology, we know that the restricted product topology on $R$ is finer than the subspace topology on $R$ as a subset of $2^{\omega}$. In particular the restricted product topology is Hausdorff. Moreover, the inclusion maps of the $X_S$'s are continuous, so in particular we have that each $F_\alpha$ is homeomorphically embedded in the restricted product topology.

Now, for each $\alpha < \omega_1$, let $B_\alpha \subseteq F_\alpha$ be a Borel set of Borel rank $\alpha$. (Such a set always exists, since we chose the $F_\alpha$'s to be non-empty perfect closed sets.) I claim that $B = \bigcup_{\alpha < \omega_1} B_\alpha$ is in the restricted product measure space structure but is not Borel in the restricted product topology on $R$.

The fact that it is not Borel is easy. Since each $F_\alpha$ is homeomorphically embedded in $R$, we have that the Borel rank of $B$ is at least the Borel rank of $B \cap F_\alpha = B_\alpha$. Since these ranks are unbounded in $\omega_1$, $B$ cannot be Borel.

So we just need to establish that $B$ is measurable. We need to show that for any $S \in \mathcal{F}$, $B \cap X_S$ is Borel in $X_S$. By the above comment we know that there is an $\alpha$ such that $S_\alpha \subseteq^\ast S$. There is an $n$ such that $S_\alpha \cap [n,\infty) \subseteq S$. Therefore for all $\beta > \alpha$, we have that $F_\beta \cap X_S = \varnothing$, by construction. Therefore $X_S \cap B = X_S \cap \bigcup_{\gamma \leq \alpha} F_\gamma$, which is a Borel set, since it is a countable union of Borel sets.

[James E Hanson]

One thing to note is that while the filter $\mathcal{F}$ is clearly constructed just for this counterexample, this phenomenon can occur with naturally occurring filters. Assuming $\mathsf{CH}$, this will happen with any filter on the naturals that is not countably generated (such as the filter of sets $S$ such that the natural density of $\omega \setminus S$ is $0$). This might not even need $\mathsf{CH}$, but I can't quite see the argument right now.

[Sébastien Gouëzel]

Note that even for the product of two Borel spaces, the product sigma-algebra is in general strictly smaller than the Borel sigma-algebra of the product (because open sets in the product are arbitrary unions of open rectangles, while the Borel sigma-algebra will only contain the countable unions, and these can be different when the space is not second-countable). In Mathlib, we have the instance that a product of measurable spaces is a measurable space (with the product sigma-algebra) and the instance that, if the spaces are Borel and second-countable, then the product is Borel and second-countable. That's the route we should probably follow also in the restricted products, provided the answers to (1) and (2) are tue.

[James E Hanson]

Sébastien Gouëzel said:

That's the route we should probably follow also in the restricted products, provided the answers to (1) and (2) are tue.

Like Aaron said, (1) is true on the grounds of very general category theory. Specifically, the category of measurable spaces is complete and cocomplete and limits and colimits are computed analogously to the category of topological spaces (since they're both topologically concrete categories).

I think one can show that (2) is true in the sense of the Borel $\sigma$-algebra on the restricted product agreeing with the restricted product $\sigma$-algebra provided that

• the spaces are second-countable,
• the filter is countably generated, and
• each $C_i$ is a Borel subset of $X_i$.

I'm not sure that the third condition is necessary, but it's certainly general enough for most applications.

[Kevin Buzzard]

Just to clarify: in my situation the $C_i$ will be both open and compact so you could barely ask for more. I guess the cofinite filter on a countably infinite index set is countably generated (if "countably generated" means "it's the smallest filter containing this countable set of sets" then it seems to me that $I-{x}$ will work for $x$ running through $I$ as my $I$ is countable).

So this is then terrible news, because it means that one day someone in an attempt to be helpful to mathlib will put the colimit measurable space structure on a restricted product and break my FLT hack where I just use Anatole's topology and borelize :-) (the general instance will be equal but not defeq).

I don't know anything about geometric Langlands, but I wonder if they consider spaces like $\prod'_v(K_v,R_v)$ where $K=\mathbb{C}(t)$, $v$ runs through the points of $\mathbb{P}^1_{\mathbb{C}}$ (an uncountable set with the cofinite filter), $R_v$ is power series centred around $v$ (so $\mathbb{C}[[t-v]]$ if $v$ is finite and $\mathbb{C}[[1/t]]$ if $v=\infty$) and $K_v$ is the field of fractions of $R_v$. The fact that a rational function only has finitely many poles gives an injection $K\to\prod'_v(K_v,R_v)$, nothing is locally compact, but if geometric Langlands people attempt to do some geometric version of Tate's thesis then for all I know they will put a measure on this restricted product and it sounds like there might be some demons here.

[Sébastien Gouëzel]

I'm not sure it could be a problem: the day people construct the proper instance, you will just need to remove borelize, and that's it.

[Kevin Buzzard]

I tried this earlier in a situation with binary products and some convert proof broke :-)