357: 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

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A description/proof of that from convex open set whose closure is bounded on Euclidean normed $C^{\mathrm{\infty }}$ manifold into equal or higher dimensional Euclidean normed $C^{\mathrm{\infty }}$ manifold polynomial map image of measure 0 subset is measure 0

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Topics

About: Euclidean normed $C^{\mathrm{\infty }}$ manifold
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

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Starting Context

• The reader knows a definition of Euclidean normed $C^{\mathrm{\infty }}$ 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 $C^{\mathrm{\infty }}$ manifold to any Euclidean normed $C^{\mathrm{\infty }}$ 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.

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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 $C^{\mathrm{\infty }}$ manifold, into any equal or higher dimensional Euclidean normd $C^{\mathrm{\infty }}$ manifold, the map image of any measure 0 subset 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 $C^{\mathrm{\infty }}$ manifolds, ${\mathbb{R}}^n$ and ${\mathbb{R}}^m$ where $n\le m$, any convex open set whose closure is bounded, $U\subseteq {\mathbb{R}}^n$, any polynomial map, $f:U\to {\mathbb{R}}^m$, 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 $m(f(S))=0$.

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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 $C^{\mathrm{\infty }}$ manifold to any Euclidean normed $C^{\mathrm{\infty }}$ 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, $m(f(S))=0$.

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References

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