Although in an abstract category the morphisms are not explicitly defined, in a concrete example (model theory?), morphisms are (always/usually?) mappings that preserve some properties.

In the category of topological spaces, the morphisms are continuous mappings. One property that is preserved under a continuous mapping is for example convergence of a filter/net/sequence, as Arturo Magidin noted in a comment

In order to view topological maps (i.e., continuous maps) as "preserving a structure", you really need to think of them in terms of preserving the notion of "closeness", not the notion of "open sets" (by the "inverse" of the mapping). It just so happens that the right way to say "f sends points that are close-to-one-another to points that are close-to-one-another" is via inverse images when you consider open sets. To define it in terms of direct images, you consider instead the filter of neighborhoods of a point. – Arturo Magidin

In the category of measurable spaces with measurable mappings being the morphisms, I was wondering what properties are preserved by a measurable mapping? Quickly browsing through Wikipedia and some other sources doesn't provide me the answer.

I used to think that a measurable mapping preserving structures is in the sense that the inverse of the mapping maps a measurable subset to a measurable subset. But as Arturo said, preserving structure should be done by the mapping in the forward direction not its inverse in the backward direction.

Thanks and regards!

  • $\begingroup$ This article proves that a pushforward measure induced by a measurable map is a measure: proofwiki.org/wiki/Pushforward_Measure_is_Measure. By the way, reflection is the dual of preservation. $\endgroup$ – alancalvitti Jan 24 '13 at 15:32
  • $\begingroup$ @alancalvitti: THanks! Does "reflection" mean that some property on the codomain is preserved when being mapped to the domain by the inverse mapping? $\endgroup$ – Tim Jan 24 '13 at 15:36
  • $\begingroup$ yes exactly, and I wish someone would put together a database of properties preserved and reflected across categories, by arrows, by functors, etc. For example in $\bf Set$, functions preserve (existence of) elements and reflect distinctions. One can do this for various classes of arrows like monos and epis, or full and faithful functors. $\endgroup$ – alancalvitti Jan 24 '13 at 15:39
  • $\begingroup$ @alancalvitti: Are there some references/sources for the concepts of reflection and preservation? $\endgroup$ – Tim Jan 24 '13 at 15:43
  • $\begingroup$ @alancalvitti: I just realized that measure is an additional set function that is not determined by the sigma algebras. This is unlike convergence in a topological space. Given a topological space, which filter/net/sequence converges is completely determined. $\endgroup$ – Tim Jan 24 '13 at 15:51

Great question! Here's one possible answer. Let a measurable space be a set $X$ together with a Boolean algebra $\mathcal{M}$ of subsets of $X$, called the "measurable" sets (you can require $\mathcal{M}$ to be a $\sigma$-algebra if you want, but this is irrelevant to the discussion below). As usual, if $X$ and $Y$ are measurable spaces and $f:X\to Y$, say $f$ is measurable if $f^{-1}(A)$ is measurable for every measurable set $A\subseteq Y$.

Given a filter $F$ on a measurable space $X$, say that $F$ is convergent if for all measurable $A\subseteq X$, either $A\in F$ or $X\setminus A\in F$ (that is, $F$ restricts to an ultrafilter on $\mathcal{M}$). Note that given any filter $F$ on $X$, we can extend $F$ to a convergent filter by adjoining only measurable sets to $F$. (You might be wondering, what are these "convergent" filters converging to? The answer is a point in the Stone space of $\mathcal{M}$; see the discussion after the break below.)

The answer to your question is then that measurable maps are exactly the maps that preserve convergent filters. More precisely:

Theorem: Let $X$ and $Y$ be measurable spaces. Then a map $f:X\to Y$ is measurable iff for any convergent filter $F$ on $X$, the pushforward $f_*F=\{A\subseteq Y:f^{-1}(A)\in F\}$ is convergent on $Y$.

(If you find it strange that $f$ is only required to preserve whether filters converge, rather than what they converge to, you may find it reassuring that actually the same thing happens for $T_1$ topological spaces. That is, a map $f$ between $T_1$ topological spaces is continuous iff whenever a filter $F$ converges, so does $f_*F$, without requiring that $f$ preserve the limit! In the case of sequences, you prove this as follows. Suppose $(x_n)$ converges to $x$, and that $f$ of a convergent sequence is convergent. Then the sequence $(x_1,x,x_2,x,x_3,x,\dots)$ converges, so $(f(x_1),f(x),f(x_2),f(x),\dots)$ converges, which means $(f(x_n))$ must converge to $f(x)$.)

Proof of Theorem: First, suppose $f$ is measurable and let $F$ be a convergent filter on $X$. Then for any measurable $A\subseteq Y$, either $f^{-1}(A)\in F$ or $f^{-1}(Y\setminus A)=X\setminus f^{-1}(A)\in F$. Thus either $A\in f_*F$ or $Y\setminus A\in f_*F$, so $f_*F$ is convergent.

Conversely, suppose $f$ is not measurable; say $f^{-1}(A)$ is not measurable for some measurable $A\subseteq Y$. Let $F$ be the filter on $X$ generated by all the measurable subsets of $X$ that contain either $f^{-1}(A)$ or $X\setminus f^{-1}(A)$ (these sets have the finite intersection property because $f^{-1}(A)$ is not measurable). Extend $F$ to a convergent filter $G$ by adjoining only measurable sets. Then I claim $f_*G$ is not convergent. Indeed, suppose $A\in f_*G$. Since $G$ is generated by measurable sets, there is then some measurable $B\in G$ such that $B\subset f^{-1}(A)$. But then $X\setminus B$ contains $X\setminus f^{-1}(A)$, so by construction $X\setminus B\in F\subseteq G$. This is a contradiction, so $A\not\in f_*G$. By a symmetric argument, $Y\setminus A\not\in f_*G$ as well. Thus $f_*G$ is not convergent.

(If you like, you can also define convergent nets in a measurable space and prove a version of the theorem above with nets in place of filters.)

Let me now sketch another proof of this theorem, which is more complicated but also more explanatory (at least to me). Note that if $(X,\mathcal{M})$ is a measurable space, then each point of $X$ determines an ultrafilter on $\mathcal{M}$, giving a map $i:X\to S(\mathcal{M})$ from $X$ to the Stone space of $\mathcal{M}$ with dense image. To slightly simplify the discussion, let us assume that this map $i$ is injective (this is equivalent to saying that $\mathcal{M}$ separates points of $X$), so we identify $X$ with a dense subspace of $S(\mathcal{M})$. Now note that $f:X\to Y$ is measurable iff it induces a Boolean algebra homomorphism $\mathcal{N}\to\mathcal{M}$ (where $\mathcal{N}$ is the measurable subsets of $Y$). Through Stone duality, this is equivalent to saying that the map $f:X\to Y$ extends to a continuous map $S(\mathcal{M})\to S(\mathcal{N})$, identifying $X$ with a dense subspace of $S(\mathcal{M})$ and $Y$ with a dense subspace of $S(\mathcal{N})$. By some general theory of compact Hausdorff spaces, such a continuous extension exists iff whenever a filter $F$ on $X$ converges to a point of $S(\mathcal{M})$, the image $f_*F$ converges to a point of $S(\mathcal{N})$. Filters on $X$ which converge in $S(\mathcal{M})$ are exactly what we called "convergent filters" above. Thus we have recovered the theorem above.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.