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$$\iint\limits_D d\omega =\int_{Fr(D)} \omega $$ Where $\omega$ is differental form such as $\omega=Pdx+Qdy$ and $d\omega=\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}$
Let $Fr(D)$ be a positively oriented, piecewise smooth, simple closed curve in a plane.
Integral on the left side of Green theorem is double integral(Riemann integral), and right side is line integral (second type).
From Lebesgue's Criterion for Riemann integrability we know that $d\omega$ is is Riemann integrable if and only if it is continuous almost everywhere (the set of its points of discontinuity has measure zero.)
So my question is how we know that integral on the left side is well define.

josf
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3 Answers3

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The Green theorem states that given a plane region D bounded by a simple, closed, regular curve L, and given the functions P(x,y) and Q(x,y) and their derivatives, with P(x,y), Q(x,y) and their derivatives continuous on $D \cup L$ we have: \begin{equation} \iint\limits_D (\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})dD =\int_{L} (P dx + Q dy) \end{equation} where L=Fr(D) is traversed in the direction such that D appears to the left of an observer moving along L. Please, note that the integral on the left is over dD. Now when considering the Green theorem we can write that: \begin{equation} \iint\limits_D (\frac{\partial P}{\partial y})dD =\iint\limits_D (\frac{\partial P}{\partial y}) dx dy= \int\limits_a ^b dx \int\limits_{\psi_1(x)} ^{\psi_2(x)} \frac{\partial P}{\partial y} dy \end{equation} where $\psi_1(x)$ and $\psi_2(x)$ form the frontier Fr(D). Then the integrals on the right will be recognized as the line integrals of the function P(x,y) along L. You can complete the proof by repeating the same calculation for Q, and then subtracting the two expressions. When considering the relation above, it is clear that the integral on the left side is well defined.

Upax
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  • Unfortunately, you answer only exacerbates the confusion present in the OP's question. For Green's Theorem to be true, you need some regularity condition on the differential form $\omega$ which is supposed to be defined on the entire closed region. At the undergraduate level, the condition is continuous differentiability (and integrals are Riemann), which makes $d\omega$ continuous and, hence, automatically integrable, so no further proof is required. If you assume that $\omega$ is merely differentiable then Green's Theorem fails! – Moishe Kohan Jul 29 '17 at 15:45
  • See for instance https://math.stackexchange.com/questions/185303/does-the-everywhere-differentiability-of-f-imply-it-is-absolutely-continuous-o – Moishe Kohan Jul 29 '17 at 16:03
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Integrability is implicitly assumed in Green's theorem. You did not state the assumptions on the functions $P$ and $Q$ and the domain $D$. For example, if $D$ is bounded with $C^1$ boundary and $P,Q\in C^1(\Omega)$ for some $\Omega\subset\mathbb R^2$ so that $\bar D\subset\Omega$, then the formula of Green's theorem holds. It also follows from these assumptions that $d\omega$ is continuous in $\bar D$ and therefore integrable in $D$.

In general, formulas need assumptions to hold. You stated a formula without assumptions, so it's hard to judge what is exactly meant, in particular when it comes to regularity issues like integrability.

Joonas Ilmavirta
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Green's theorem is a special case of Stokes' theorem here .

Since Green's theorem falls out of Stokes' theorem in the 2-dimensional case, perhaps look to the more general theorem for your answer... To be sure the statement is (much) simpler.