Is there a natural way to interpret the category of measure spaces as an internal category in the category of topological spaces? any references would be helpful.
17
-
3$\begingroup$ It seems to me that this could be a sensible question, if only it were more clear. I guess it's asking whether there's a natural way to interpret the category of measure spaces as an internal category in Top. Presumably the answer is yes (disregarding niggles over set-theoretic size); perhaps there are several ways to do it. $\endgroup$– Tom LeinsterCommented Dec 25, 2012 at 15:11
-
3$\begingroup$ My impression is that this question is somehow the wrong way round, and should be more of the form: is there a category of measured spaces? I mention this because G.W. Mackey did work on measured groupoids and I have always assumed these were groupoids internal to the appropriate category! See G.W. Mackey, ‘Ergodic theory and virtual groups’, Math. Ann. 166 (1966) 187-207. He asked the question: if a transitive action of a group $G$ corresponds to a subgroup of $G$ (or conjugacy class thereof) what then corresponds to an erodic action, which is "almost transitive"! $\endgroup$– Ronnie BrownCommented Dec 25, 2012 at 18:54
-
3$\begingroup$ Ronnie, you may be right; all the same, the question does seem reasonable to me. For example, in my own work I've used the fact that the category of measures spaces with finite underlying set can naturally be regarded as an internal category in Top. This question removes the restriction of finiteness, which makes it harder. $\endgroup$– Tom LeinsterCommented Dec 25, 2012 at 19:41
-
3$\begingroup$ @Walker: you should check out Dmitri Pavlov's posts on this site dating back over the past three(!) years, as well as the nLab entry on measurable spaces. In short, the philosophy is that the category of measurable spaces is nicer to work with than that of measured spaces. * mathoverflow.net/questions/49426/… * mathoverflow.net/questions/20740/… * ncatlab.org/nlab/show/measurable+space $\endgroup$– Tom LaGattaCommented Dec 26, 2012 at 2:48
-
2$\begingroup$ Clearly, the category of measurable spaces cannot be interpreted as an internal category in spaces, because measurable spaces (or even their isomorphism classes) form a proper class, whereas a space can only have a set of points. I presume that the question actually means enriched categories. $\endgroup$– Dmitri PavlovCommented Dec 26, 2012 at 9:18
|
Show 12 more comments
1 Answer
$\begingroup$
$\endgroup$
3
As demonstrated by Andre Kornell in http://arxiv.org/abs/1202.2994, the category of measurable spaces is closed with respect to the monoidal structure given by the spatial tensor product (Theorem 9.5). Points (atoms) of the resulting internal hom are precisely morphisms of measurable spaces (Corollary 9.12, ibid.). In particular, we obtain an enrichment of the category of measurable spaces over itself. There is a fully faithful forgetful functor from the category of measurable spaces (= measurable locales) to the category of locales (e.g., see the nLab article measurable locale), which gives you the desired enrichment over (not necessarily spatial) topological spaces.
answered Dec 26, 2012 at 8:26
-
3$\begingroup$ It's an interesting answer, but not exactly an answer to the question asked. I'd still like to know about ways of making the category of measure spaces (not measurable spaces) internal to (not enriched in) the category of topological spaces. As both Dmitri and I have pointed out, there's a set-theoretic niggle to deal with, but that shouldn't be a serious obstacle. $\endgroup$ Commented Dec 26, 2012 at 15:40
-
$\begingroup$ I am not expert; but as I have understood the answer,if it is impossible for measurable spaces category ,so it is impossible for measure spaces category $\endgroup$– TWalkerCommented Dec 26, 2012 at 16:39
-
$\begingroup$ @Tom Leinster and Walker: I answered in the main comment thread. $\endgroup$ Commented Dec 26, 2012 at 19:18