A description/proof of that from convex open set whose closure is bounded on Euclidean normed \(C^\infty\) manifold into equal or higher dimensional Euclidean normed \(C^\infty\) manifold polynomial map image of measure 0 subset is measure 0
Topics
About: Euclidean normed \(C^\infty\) manifold
About: measure
About: map
The table of contents of this article
Starting Context
- The reader knows a definition of Euclidean normed \(C^\infty\) manifold.
- The reader knows a definition of measure.
- The reader admits the proposition that any \(C^1\) map from any open set on any Euclidean normed \(C^\infty\) manifold to any Euclidean normed \(C^\infty\) manifold satisfies the Lipschitz condition in any convex open set whose closure is bounded and contained in the original open set.
- The reader admits the proposition that the image of any measure 0 subset under any Lipschitz condition satisfying map from any Euclidean normed topological space to any equal or higher dimensional Euclidean normed topological space is measure 0.
Target Context
- The reader will have a description and a proof of the proposition that for any polynomial map from any convex open set whose closure is bounded on any Euclidean normed \(C^\infty\) manifold, into any equal or higher dimensional Euclidean normd \(C^\infty\) manifold, the map image of any measure 0 subset is measure 0.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Description
For any Euclidean normed \(C^\infty\) manifolds, \(\mathbb{R}^n\) and \(\mathbb{R}^m\) where \(n \leq m\), any convex open set whose closure is bounded, \(U \subseteq \mathbb{R}^n\), any polynomial map, \(f: U \rightarrow \mathbb{R}^m\), and any measure \(0\) subset, \(S \subseteq U, m (S) = 0\) where \(m (\bullet)\) is the measure of the argument, the image of \(S\) under \(f\) is measure \(0\), which is \(m (f (S)) = 0\).
2: Proof
The polynomial map can be regarded to be defined on \(\mathbb{R}^n\), canonically extended from \(U\) to \(\mathbb{R}^n\). As \(f\) is \(C^1\), by the proposition that any \(C^1\) map from any open set on any Euclidean normed \(C^\infty\) manifold to any Euclidean normed \(C^\infty\) manifold satisfies the Lipschitz condition in any convex open set whose closure is bounded and contained in the original open set, \(f\) satisfies the Lipschitz condition on \(U\).
By the proposition that the image of any measure 0 subset under any Lipschitz condition satisfying map from any Euclidean normed topological space to any equal or higher dimensional Euclidean normed topological space is measure 0, \(m (f (S)) = 0\).
