136: Area on Euclidean Metric Space Can Be Measured Using Only Hypersquares, Instead of Hyperrectangles

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A description/proof of that area on Euclidean metric space can be measured using only hypersquares, instead of hyperrectangles

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Topics

About: metric space

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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 admits the proposition that the area of any hyperrectangle can be approximated by the area of covering finite number hypersquares to any precision.

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Target Context

• The reader will have a description and a proof of the proposition that any area on any ${\mathbb{R}}^n$ Euclidean metric space can be measured using only hypersquares, instead of hyperrectangles.

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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 area, $a$, on any ${\mathbb{R}}^n$ Euclidean metric space, $a$ can be measured using only hypersquares.

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2: Proof

By the usual definition of area by the measure theory, for any real number, $ϵ>0$, there are some countable number disjoint half-open intervals (hyperrectangles) that cover the area, $\{R_i\}$, such that $m(R_i)-a<ϵ$ where $m(\bullet )$ means the measure of the argument.

By the proposition that the area of any hyperrectangle can be approximated by the area of covering finite number hypersquares to any precision, a covering finite number hypersquares, $\{S_{i_j}\}$, can be chosen such that $\sum _{j}m(S_{i_j})-m(R_i)<\frac{1}{k}m(R_i)$ where $k$ is any natural number.

Then, $\sum _i(\sum _{j}m(S_{i_j}))-a<\sum _i(m(R_i)+\frac{1}{k}m(R_i))-a=\sum _i\frac{k+1}{k}m(R_i)-a=\frac{k+1}{k}\sum _{i}m(R_i)-a<\frac{k+1}{k}(a+ϵ)-a$. We want to choose $k$ and $ϵ$ such that for any ${ϵ}^{\prime }$, $\frac{k+1}{k}(a+ϵ)-a\le {ϵ}^{\prime }$. $\frac{1}{k}a+\frac{k+1}{k}ϵ\le {ϵ}^{\prime }$; we choose $\frac{1}{k}a\le \frac{{ϵ}^{\prime }}{2}$ and $\frac{k+1}{k}ϵ\le \frac{{ϵ}^{\prime }}{2}$, which is possible by $k\ge \frac{2a}{{ϵ}^{\prime }}$ and $ϵ\le \frac{k}{k+1}\frac{{ϵ}^{\prime }}{2}$.

The hypersquares are countable, because we can count them, for example, by the order of $i+j$ and then $i$ for $S_{i_j}$, like $S_{1_1},S_{1_2},S_{2_1},S_{1_3},S_{2_2},S_{3_1},...$.

So, for any real number ${ϵ}^{\prime }>0$, there are some countable number hypersquares, $S_{i_j}$, such that $\sum _i\sum _j{S}_{i_j}-a<{ϵ}^{\prime }$.

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References

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