Isomorphism of Measure Spaces: Difference between revisions

Revision as of 07:55, 18 December 2020

Let $U$ be an open set of $\mathbb {R} ^{n}$ , and let

f\colon U\rightarrow \mathbb {R} ^{m}

be a smooth map. If

A\subset U

is of measure zero, then

f(A)

is of measure zero.

@@ Line 15: / Line 15: @@
 ===Smooth maps send sets of measure zero to sets of measure zero===
-Let <math> U </math> be an open set of
+Let <math> U </math> be an open set of <math> \mathbb{R}^n</math>, and let
-: <math> f\colon \mathbb{R}^n \rightarrow \mathbb{R}^m</math>
+: <math> f\colon U \rightarrow \mathbb{R}^m</math> be a smooth map. If <math> A\subset U</math> is of measure zero, then <math> f(A)</math> is of measure zero.
 ===Mini-Sards Theorem===
 ==Example==