A description/proof of that from Euclidean normed topological space into equal or higher dimensional Euclidean normed topological space Lipschitz condition satisfying map image of measure 0 subset is measure 0
Topics
About: Euclidean normed topological space
About: measure
About: map
The table of contents of this article
Starting Context
- The reader knows a definition of Euclidean normed topological space.
- The reader knows a definition of Lipschitz condition.
- The reader knows a definition of measure.
Target Context
- The reader will have a description and a proof of 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.
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 topological spaces with the canonical metrics, \(\mathbb{R}^n\) and \(\mathbb{R}^m\) where \(n \leq m\), any open set, \(U \subseteq \mathbb{R}^n\), any Lipschitz condition satisfying map, \(f: U \rightarrow \mathbb{R}^m\), and any measure \(0\) subset, \(S \subseteq U\), the image of \(S\) under \(f\), \(f (S)\), is measure \(0\).
2: Proof
As \(f\) satisfies the Lipschitz condition, there is a constant, \(L\), such that for any points, \(p_1, p_2 \in U\), \(\Vert f (p_2) - f (p_1)\Vert \leq L \Vert p_2 - p_1 \Vert\). As \(S\) is measure \(0\), by the proposition that any area on any \(\mathbb{R}^n\) Euclidean metric space can be measured using only hypersquares, instead of hyperrectangles, for any real number, \(\epsilon \gt 0\), there are some countable disjoint half-open hypersquares, \(\{I_i\}\), \(S \subseteq \cup_i I_i, \sum_i m (I_i) \lt \epsilon\) where \(m (\bullet)\) is the measure of the argument. \(f (S) \subseteq f (\cup_i I_i) = \cup_i f (I_i)\) by the proposition that for any map, the map image of any union of sets is the union of the map images of the sets, so, \(m (f (S)) \leq \sum _i m (f (I_i))\). For any points, \(p_1, p_2 \in I_i\), \(\Vert p_2 - p_1 \Vert \lt \sqrt{n {l_i}^2}\) where \(l_i\) is the side length of the hypersquare, so, \(\Vert f (p_2) - f (p_1) \Vert \leq L \Vert p_2 - p_1 \Vert \lt L \sqrt{n {l_i}^2}\), so, \(f (I_i)\) is contained in a \(2 L \sqrt{n {l_i}^2}\) side-length hypersquare very safely (meaning that there is a smaller possible hypersquare), so, \(m (f (I_i)) \lt (2 L \sqrt{n {l_i}^2})^m\). \(\sum_i m (f (I_i)) \lt \sum_i (2 L \sqrt{n {l_i}^2})^m = (2 L \sqrt{n})^m \sum_i {l_i}^m\), but as \(m (I_i) = {l_i}^n\), \(\sum_i m (f (I_i)) \lt (2 L \sqrt{n})^m \sum_i (m (I_i))^{m / n}\). For any small enough \(\epsilon\), \(m (I_i) \lt 1\), so, as \(m / n \geq 1\), \((m (I_i))^{m / n} \leq m (I_i)\), so, \(\sum_i (m (I_i))^{m / n} \lt \epsilon\), so, \(\sum_i m (f (I_i)) \lt (2 L \sqrt{n})^m \epsilon\). For any real number, \(\epsilon' \gt 0\), \(\epsilon\) can be chosen such that \(\epsilon' \gt (2 L \sqrt{n})^m \epsilon\), then, \(\sum_i m (f (I_i)) \lt \epsilon'\), so, \(m (f (S)) \lt \epsilon'\).
3: Note
When \(m \lt n\), the proposition does not hold. As a counterexample, let \(n = 3, m = 2\) where \(\mathbb{R}^2\) is the \(z = 0\) plane in \(\mathbb{R}^3\) and \(U = \mathbb{R}^3\) and \(f\) be the projection. \(f\) satisfies the Lipschitz condition with \(L = 1\). A square on the \(z = 0\) plane is measure \(0\) on \(\mathbb{R}^3\), and its image under \(f\) is the same square on \(\mathbb{R}^2\), whose measure is not \(0\) on \(\mathbb{R}^2\).
