3

In the book Lectures and Exercises on Functional Analysis by Helemskii I have stumbled upon the following note:

The Rohlin theorem and similar results (see e.g., [19],[20]) show that the structure of a measure space is the coarsest among all the substantial structures on a set (see the discussion in [21, pp. 46-47]).

By the Rohlin theorem (I believe) the author means the theorem classifying standard measure spaces in the category $\mathsf{Meas}$, whose objects are measure spaces and whose arrows are equivlance classes of bimeasurable mappings such that inverse images of null-sets are null-sets.

My question is what exactly does this mean? Does this statement mean that topological structure for instance is more informative than "measurable structure"? If so, in what sense? If it's a very deep statement which cannot be briefly explicated, can you please point me to the things I need to learn to understand it?

The relevant bibliography:

  • [19] J von Neumann Uber Funktionen von Funktionaloperatoren. Ann. of Math. 32 (1931), 191-226.
  • [20] I. E. Segal, Equivalences of measure spaces. Amer. J. Math. 73 (1951 ), 275-313.
  • [21] A. Connes, N oncommutative geometry. Academic Press, Orlando, FL, 1990.

-Thanks in advance!

1 Answers1

1

Connes argument is the following: Thanks to these isomorphism results, essentially all measure spaces one encounters are isomorphic. It follows that there exists no invariants of a measure space that is preserved under isomorphism and makes it possible to differentiate between different measure spaces. In this sense, measure spaces (in the class) have no structure.

Compare that with invariants such as the number of connected components, dimension, etc. that are available in geometry and analysis, which have therefore much more structure.

Michael Greinecker
  • 32,841
  • 6
  • 80
  • 137