Isomorphism of Measure Spaces: Difference between revisions

From Optimal Transport Wiki
Jump to navigation Jump to search
Line 25: Line 25:


===Mini-Sards Theorem===
===Mini-Sards Theorem===
Let <math> U </math> be an open set of <math> \mathbb{R}^n</math>, and let <math> f\colon U \rightarrow \mathbb{R}^m</math> be a smooth map. Then if <math> m > n </math>, <math>f(U)</math> has measure zero in <math> \mathbb{R}^m</math>.  
Let <math> U </math> be an open set of <math> \mathbb{R}^n</math>, and let <math> f\colon U \rightarrow \mathbb{R}^m</math> be a smooth map. Then if <math> m > n </math>, <math>f(U)</math> has measure zero in <math> \mathbb{R}^m</math>.


==Example==
==Example==

Revision as of 00:37, 19 December 2020

Motivation

Definition

Let be a measurable space and a sigma algebra on . Similary, Let be a measurable space and a sigma algebra on . Let and be measurable spaces.

  • A map is called measurable if for every .
  • These two measurable spaces are called isomorphic if there exists a bijection such that and are measurable (such is called an isomorphism).

Basic Theorem

Let and be Borel subsets of complete separable metric spaces. For the measurable spaces and to be isomoprhuic, it is necessary and sufficient that the sets and be of the same cardinality.


Properties

Smooth maps send sets of measure zero to sets of measure zero

Let be an open set of , and let be a smooth map. If is of measure zero, then is of measure zero.

  • Note that we can loosen the restriction of our map being smooth. It is enough to consider absolutely continuous functions and the statement still holds.

Mini-Sards Theorem

Let be an open set of , and let be a smooth map. Then if , has measure zero in .

Example

Consider where . is easily seen to be a smooth map since it has partial derivatives of all order. Let . Pick . Consider the cover . Then is covered by the union of all . Now, . Since was aribtary .

Reference