9
$\begingroup$

$\newcommand{\A}{\mathcal{A}}\newcommand{\T}{\mathcal{T}}$The notions of continuous map of topological spaces and measurable function of measurable spaces are very similar:

  • A map of topological spaces from $(X,\T_X)$ to $(Y,\T_Y)$ is continuous if for each $V\in\T_Y$, we have $f^{-1}(V)\in\T_X$.
  • A map of measureable spaces from $(X,\A_X)$ to $(Y,\A_Y)$ is measurable if for each $V\in\A_Y$, we have $f^{-1}(V)\in\A_X$.

Similarly, a map from $(X,\T_X)$ to $(Y,\T_Y)$ is open if for each $U\in\T_X$, we have $f(U)\in\T_Y$.

Does the analogue of openness for measurable spaces have a name? Is there some place where I can read more about it and its properties?

$\endgroup$
2
  • 3
    $\begingroup$ $\newcommand{\A}{\mathcal{A}}$(Some miscellaneous remarks: the analogue of a closed map for a measurable space agrees with that of an open map, as $\sigma$-algebras are closed under complements. Further, we could also try to replace the left adjoint $f_*$ of $f^{-1}$ with its right adjoint $f_!$, as in my question from some time back here. However, the condition $U\in\A_X\implies f_!(U)\in\A_Y$ ends up being equivalent to $U\in\A_X\implies f(U)\in\A_Y$ by მამუკა's argument there. $\endgroup$ Commented Feb 13, 2023 at 14:49
  • 1
    $\begingroup$ Lastly, the assignment sending a topological space to its Borel measurable space defines a functor from $\mathsf{Top}$ to $\mathsf{Meas}$, as continuous functions give rise to measurable functions, but this is not the case for open maps (or even maps which are both open and closed), as these need not satisfy $U\in\mathcal{A}_X\implies f(U)\in\mathcal{A}_Y$) $\endgroup$ Commented Feb 13, 2023 at 14:54

2 Answers 2

5
$\begingroup$

There are at least three different answers that can be given to this question, and in all three interpretations the answer essentially states that all maps are “open”, for the appropriate analogue of “open”.

In the first interpretation, we may ask: under what conditions is the image of a measurable set measurable? Of course, in measure theory we typically work with equivalence classes of measurable maps modulo equivalence almost everywhere, i.e., up to a set of measure 0 (alias negligible set), so first we need to give a more precise definition of an image that interacts nicely with negligible sets. Below, I reuse the notation of the main post.

Definition. The measurable image of a mesurable map $\def\cA{{\cal A}}f\colon X→Y$ is a subset $V∈\cA_Y$ such that $f^{-1}(Y∖V)$ is negligible and, furthermore, for any $W∈\cA_Y$ such that $W⊂V$ and $f^{-1}(W)$ is negligible, the set $W$ itself is negligible. The measurable image of $U∈\cA_X$ is defined as the measurable image of the map $f|_U\colon U→Y$, where $U$ is equipped with the induced structure of an enhanced measurable space.

Of course, this definition requires the additional data of σ-ideals of negligible sets $\def\cN{{\cal N}}\cN_X⊂\cA_X$ and $\cN_Y⊂\cA_Y$, which we therefore include in the given data. Adopting the terminology of arXiv:2005.05284, we refer to a triple $(X,\cA_X,\cN_X)$ as an enhanced measurable space. If we want composition of measurable maps to respect the equivalence relation of equality almost everywhere, we must also require that preimages of negligible sets are negligible, in complete analogy to the existing requirement that preimages of measurable sets are measurable. Below, a morphism of enhanced measurable spaces is an equivalence class of such maps modulo equality almost everywhere; in the uncountably separated case some care must be exercised when defining equality almost everywhere, see Definition 4.13 in arXiv:2005.05284.

The following theorem is due to Fremlin (Lemma 451Q in his Measure Theory). An exposition that matches the presentation of this post can be found in Proposition 4.65 of arXiv:2005.05284.

Theorem (Fremlin, 2003). The measurable image exists whenever $(X,\cA_X,\cN_X)$ is Marczewski-compact and $(Y,\cA_Y,\cN_Y)$ is strictly localizable. The equivalence class (up to a negligible set) of the measurable image is uniquely defined.

Marczewski-compactness is essentially the abstract reformulation of a Radon measure (not relying on the structure of a topological space) and strict localizability amounts to being isomorphic to a disjoint union of σ-finite spaces. In particular, with the exception of trivial counterexamples constructed for a specific purpose, all measure spaces used in standard textbooks are compact and strictly localizable, since Radon measure spaces are automatically compact and all σ-finite measure spaces are strictly localizable. See the sources cited above for more details.

In the second interpretation we may ask for a more direct relation between measurable maps of measurable spaces and open maps of topological spaces. Here is a theorem that establishes such a relation.

Theorem (Theorem 1.1 in arXiv:2005.05284). The category of compact strictly localizable enhanced measurable spaces is equivalent to the category of hyperstonean topological spaces and open continuous maps. In one direction, the equivalence sends an enhanced measurable space $(X,\cA_X,\cN_X)$ to the Gelfand spectrum of the von Neumann algebra of bounded measurable functions on $X$, or, equivalently, the Stone spectrum of the Boolean algebra $\cA_X/\cN_X$. In the other direction, the equivalence sends a hyperstonean topological space $\def\cT{{\cal T}}(X,\cT_X)$ to the enhanced measurable space $(X,\cA_X,\cN_X)$, where $\cN_X$ is the set of all meager subsets of $X$ (alias sets of first category) and $\cA_X$ is the set of all subsets of $X$ with the Baire property (i.e., symmetric differences of open subsets and meager subsets).

Thus, morphisms of enhanced measurable spaces correspond precisely to open maps of hyperstonean topological spaces.

Finally, if you are willing to go a little bit beyond point-set topological spaces in the setting of locales (which are a rather mild adjustment of the usual notion of a topological space), then a third answer is possible.

Theorem (Theorem 1.1 in arXiv:2005.05284). The category of compact strictly localizable enhanced measurable spaces is equivalent to the category of measurable locales.

Here the category of measurable locales is a full subcategory of locales whose underlying frames are Boolean algebras admitting a completely additive measure. It is in this precise sense that we can consider measure theory to be a part of general topology (understood here as including the study of locales). Having nonspatial locales (i.e., locales that do not come from a point-set topological space) is essential, since almost all measurable locales are not spatial. The notion of an open map makes sense for locales, and can be defined in a formally similar way: the image of an open sublocale must be open. With this definition, it is easy to prove (Lemma 2.44 in arXiv:2005.05284) that any morphism of measurable locales is an open map of locales.

$\endgroup$
7
  • $\begingroup$ Thank you very much, Dmitri! I'm trying to better understand the notion of measurable image you defined, but I'm struggling a bit; could it be that there's a typo in it? It seems the set $U\in\mathcal{A}_X$ doesn't enter the definition in its current form $\endgroup$ Commented Feb 14, 2023 at 17:50
  • $\begingroup$ @Emily: Yes, for an arbitrary U, we must first restrict f to U. (Fixed now.) $\endgroup$ Commented Feb 14, 2023 at 21:34
  • $\begingroup$ This is really great! Now that I understand that definition I see that Fremlin's result (exposed in the way you did) was exactly what I was looking for. Again, thank you so much, Dmitri =) $\endgroup$ Commented Feb 14, 2023 at 23:37
  • $\begingroup$ I am confused. For example, what is the locale corresponds to a measurable space? $\endgroup$ Commented Feb 15, 2023 at 12:06
  • $\begingroup$ @Z.M: The locale corresponding to $(X,A_X,N_X)$ is precisely the Boolean algebra $A_X/N_X$. This is explained in detail in arXiv:2005.05284. $\endgroup$ Commented Feb 15, 2023 at 16:14
11
$\begingroup$

A classical theorem about open mappings is the invariance of domain. This theorem asserts that an injective continuous map from ${\bf R}^n$ to ${\bf R}^n$ is open: the image of an open set is an open set. The space ${\bf R}^n$ can be replaced by any manifold. This is proven in most algebraic topology books, such as Hatcher.

There is an analogous result in measure theory that asserts that for an injective Borel map from $[0,1]$ to $[0,1]$, the image of a Borel set is a Borel set. The interval $[0,1]$ can be replaced by any standard Borel space. A reference is the book of Kechris, Classical Descriptive Set Theory.

$\endgroup$
1
  • $\begingroup$ That's a very nice result, thank you! $\endgroup$ Commented Feb 14, 2023 at 17:51

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.