356: 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

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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

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Topics

About: Euclidean normed topological space
About: measure
About: map

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof
• 3: Note

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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.

• The reader admits the proposition that any area on any ${\mathbb{R}}^n$ Euclidean metric space can be measured using only hypersquares, instead of hyperrectangles.
• The reader admits the proposition that for any map, the map image of any union of sets is the union of the map images of the sets.

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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.

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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.

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Main Body

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1: Description

For any Euclidean normed topological spaces with the canonical metrics, ${\mathbb{R}}^n$ and ${\mathbb{R}}^m$ where $n\le m$, any open set, $U\subseteq {\mathbb{R}}^n$, any Lipschitz condition satisfying map, $f:U\to {\mathbb{R}}^m$, and any measure $0$ subset, $S\subseteq U$, the image of $S$ under $f$, $f(S)$, is measure $0$.

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2: Proof

As $f$ satisfies the Lipschitz condition, there is a constant, $L$, such that for any points, $p_1,p_2\in U$, $\Vert f(p_2)-f(p_1)\Vert \le L\Vert p_2-p_1\Vert$. 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, $ϵ>0$, there are some countable disjoint half-open hypersquares, $\{I_i\}$, $S\subseteq {\cup }_i{I}_i,\sum _{i}m(I_i)<ϵ$ 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, $m(f(S))\le \sum _{i}m(f(I_i))$. For any points, $p_1,p_2\in I_i$, $\Vert p_2-p_1\Vert <\sqrt{n{l_i}^2}$ where $l_i$ is the side length of the hypersquare, so, $\Vert f(p_2)-f(p_1)\Vert \le L\Vert p_2-p_1\Vert <L\sqrt{n{l_i}^2}$, so, $f(I_i)$ is contained in a $2L\sqrt{n{l_i}^2}$ side-length hypersquare very safely (meaning that there is a smaller possible hypersquare), so, $m(f(I_i))<(2L\sqrt{n{l_i}^2}{)}^m$. $\sum _{i}m(f(I_i))<\sum _i(2L\sqrt{n{l_i}^2}{)}^m=(2L\sqrt{n}{)}^m\sum _i{l_i}^m$, but as $m(I_i)={l_i}^n$, $\sum _{i}m(f(I_i))<(2L\sqrt{n}{)}^m\sum _i(m(I_i){)}^{m/n}$. For any small enough $ϵ$, $m(I_i)<1$, so, as $m/n\ge 1$, $(m(I_i){)}^{m/n}\le m(I_i)$, so, $\sum _i(m(I_i){)}^{m/n}<ϵ$, so, $\sum _{i}m(f(I_i))<(2L\sqrt{n}{)}^mϵ$. For any real number, ${ϵ}^{\prime }>0$, $ϵ$ can be chosen such that ${ϵ}^{\prime }>(2L\sqrt{n}{)}^mϵ$, then, $\sum _{i}m(f(I_i))<{ϵ}^{\prime }$, so, $m(f(S))<{ϵ}^{\prime }$.

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3: Note

When $m<n$, 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 ${\mathbb{R}}^2$.

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References

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