Editing Green's theorem (section)

== Proof for rectifiable Jordan curves ==
We are going to prove the following

{{math theorem|math_statement= Let <math>\Gamma</math> be a rectifiable, positively oriented [[Jordan curve theorem|Jordan curve]] in <math>\R^2</math> and let <math>R</math> denote its inner region. Suppose that <math>A, B:\overline{R} \to \R</math> are continuous functions with the property that <math>A</math> has second partial derivative at every point of <math>R</math>, <math>B</math> has first partial derivative at every point of <math>R</math> and that the functions <math>D_1 B, D_2 A : R \to \R</math> are Riemann-integrable over <math>R</math>. Then
<math display="block">\int_{\Gamma}(A\,dx + B\,dy) = \int_R \left(D_1 B(x,y)-D_2 A(x,y)\right)\,d(x,y).</math>}}

We need the following lemmas whose proofs can be found in:<ref>{{Cite book |last=Apostol |first=Tom |author-link=Tom M. Apostol |title=Mathematical Analysis |date=1960 |publisher=[[Addison-Wesley]] |location=Reading, Massachusetts, U.S.A. |oclc=6699164}}</ref>

{{math theorem|name=Lemma 1 (Decomposition Lemma)|math_statement= Assume <math>\Gamma</math> is a rectifiable, positively oriented Jordan curve in the plane and let <math>R</math> be its inner region. For every positive real <math>\delta</math>, let <math>\mathcal{F}(\delta)</math> denote the collection of squares in the plane bounded by the lines <math>x=m\delta, y=m\delta</math>, where <math>m</math> runs through the set of integers. Then, for this <math>\delta</math>, there exists a decomposition of <math>\overline{R}</math> into a finite number of non-overlapping subregions in such a manner that
<ol type="i">
<li>Each one of the subregions contained in <math>R</math>, say <math>R_1, R_2,\ldots, R_k</math>, is a square from <math>\mathcal{F}(\delta)</math>.</li>
<li>Each one of the remaining subregions, say <math>R_{k+1},\ldots,R_s</math>, has as boundary a rectifiable Jordan curve formed by a finite number of arcs of <math>\Gamma</math> and parts of the sides of some square from <math>\mathcal{F}(\delta)</math>.</li>
<li>Each one of the border regions <math>R_{k+1},\ldots,R_s</math> can be enclosed in a square of edge-length <math>2\delta</math>.</li>
<li>If <math>\Gamma_i</math> is the positively oriented boundary curve of <math>R_i</math>, then <math>\Gamma=\Gamma_1+\Gamma_2+\cdots+\Gamma_s.</math></li>
<li>The number <math>s-k</math> of border regions is no greater than <math display="inline">4\!\left(\frac{\Lambda}{\delta} + 1\right)</math>, where <math>\Lambda</math> is the length of <math>\Gamma</math>.</li>
</ol>}}
{{math theorem|name=Lemma 2|math_statement= Let <math>\Gamma</math> be a rectifiable curve in the plane and let <math>\Delta_\Gamma(h)</math> be the set of points in the plane whose distance from (the range of) <math>\Gamma</math> is at most <math>h</math>. The outer Jordan content of this set satisfies <math>\overline{c}\,\,\Delta_\Gamma(h)\le2h\Lambda +\pi h^2</math>.}}

{{math theorem|name=Lemma 3|math_statement= Let <math>\Gamma</math> be a rectifiable curve in <math>\R^2</math> and let <math>f:\text{range of }\Gamma \to \R</math> be a continuous function. Then
<math display="block">\left\vert\int_\Gamma f(x,y)\,dy\right\vert \le\frac{1}{2} \Lambda\Omega_f,</math> and
<math display="block">\left\vert\int_\Gamma f(x,y)\,dx\right\vert \le\frac{1}{2} \Lambda\Omega_f,</math>
where <math>\Omega_f</math> is the oscillation of <math>f</math> on the range of <math>\Gamma</math>.}}

Now we are in position to prove the theorem:

'''Proof of Theorem.''' Let <math>\varepsilon</math> be an arbitrary positive real number. By continuity of <math>A</math>, <math>B</math> and compactness of <math>\overline{R}</math>, given <math>\varepsilon>0</math>, there exists <math>0<\delta<1</math> such that whenever two points of <math>\overline{R}</math> are less than <math>2\sqrt{2}\,\delta</math> apart, their images under <math>A, B</math> are less than <math>\varepsilon</math> apart. For this <math>\delta</math>, consider the decomposition given by the previous Lemma. We have
<math display="block">\int_\Gamma A\,dx+B\,dy=\sum_{i=1}^k \int_{\Gamma_i} A\,dx+B\,dy\quad +\sum_{i=k+1}^s \int_{\Gamma_i}A\,dx+B\,dy.</math>

Put <math>\varphi := D_1 B - D_2 A</math>.

For each <math>i\in\{1,\ldots,k\}</math>, the curve <math>\Gamma_i</math> is a positively oriented square, for which Green's formula holds. Hence
<math display="block">\sum_{i=1}^k \int_{\Gamma_i}A\,dx + B\,dy =\sum_{i=1}^k \int_{R_i} \varphi = \int_{\bigcup_{i=1}^k R_i}\,\varphi.</math>

Every point of a border region is at a distance no greater than <math>2\sqrt{2}\,\delta</math> from <math>\Gamma</math>. Thus, if <math>K</math> is the union of all border regions, then <math>K\subset \Delta_{\Gamma}(2\sqrt{2}\,\delta)</math>; hence <math>c(K)\le\overline{c}\,\Delta_{\Gamma}(2\sqrt{2}\,\delta)\le 4\sqrt{2}\,\delta+8\pi\delta^2</math>, by Lemma 2. Notice that
<math display="block">\int_R \varphi\,\,-\int_{\bigcup_{i=1}^k R_i} \varphi=\int_K \varphi.</math>
This yields
<math display="block">\left\vert\sum_{i=1}^k \int_{\Gamma_i} A\,dx+B\,dy\quad-\int_R\varphi \right\vert \le M \delta(1+\pi\sqrt{2}\,\delta) \text{ for some } M > 0.</math>

We may as well choose <math>\delta</math> so that the RHS of the last inequality is <math><\varepsilon.</math>

The remark in the beginning of this proof implies that the oscillations of <math>A</math> and <math>B</math> on every border region is at most <math>\varepsilon</math>. We have
<math display="block">\left\vert\sum_{i=k+1}^s \int_{\Gamma_i}A\,dx+B\,dy\right\vert\le\frac{1}{2} \varepsilon\sum_{i=k+1}^s \Lambda_i.</math>

By Lemma 1(iii),
<math display="block">\sum_{i=k+1}^s \Lambda_i \le\Lambda + (4\delta)\,4\!\left(\frac{\Lambda}{\delta}+1\right)\le17\Lambda+16.</math>

Combining these, we finally get
<math display="block">\left\vert\int_\Gamma A\,dx+B\,dy\quad-\int_R \varphi\right\vert< C \varepsilon,</math>
for some <math>C > 0</math>. Since this is true for every <math>\varepsilon > 0</math>, we are done.