Green’s theorem is one of the most profound theorems in mathematics. The version here appeared in 1951 and involves a rectifiable simple closed curve. It will always be assumed that the x,y axes are oriented in the usual way and that everything is taking place in the plane.
First recall the Jordan curve theorem.
Theorem 12.4.1 Let J be a Jordan curve in the plane. That is J is γ
The following lemma will be of importance in what follows. To say two sets are homeomorphic is to say there is a one to one continuous onto mapping from one to the other which also has continuous inverse. Clearly the statement that two sets are homeomorphic defines an equivalence relation. Then from Lemma 2.9.12, a Jordan curve is just a curve which is homeomorphic to a circle.
From now on, U_{o} is the outside component of a simple closed curve Γ which is unbounded and U_{i} is the inside component. Also I will write γ^{∗} or Γ to emphasize that the set of points in the curve is being considered rather than some parametrization. The following lemma is about cutting in two pieces a simple closed curve with its inside.
Lemma 12.4.2 In the situation of Theorem 12.4.1, let Γ be a simple closed curve and let γ^{∗} be a straight line segment such that the open segment, γ_{o}^{∗},γ^{∗} without its endpoints, is contained in U_{i} and the intersection of γ^{∗} with Γ equals

Furthermore, U_{1i} ∩ U_{2i} = ∅.
Proof: To aid in the discussion, for α a real number, denote by
Since Γ is a simple closed curve, there is γ : S^{1} → Γ which is one to one, onto and continuous. Say γ

where p =
Thus there are two simple curves

Since the line segment is in U_{i}, this specifies a simple closed curve γ_{1}
Refer to these simple closed curves as Γ_{1},Γ_{2}. They intersect in the line segment γ^{∗}. Thus, by the Jordan curve theorem, these simple closed curves have insides U_{1i} and U_{2i} respectively. Are these insides contained in U_{i}, the inside of the original Jordan curve?
Claim 1: U_{1i},U_{2i} ⊆ U_{i}, U_{1i},U_{2i} contain no points of Γ_{2} ∪ Γ_{1}.
Proof of claim 1: Since U_{1i} is connected, it cannot have points which are in U_{i} and points which are in U_{o}, the insides and outsides of the original Jordan curve Γ. However, it does contain points of U_{i} because γ^{∗} ⊆ U_{i} and every point of γ^{∗} is a limit of points of U_{1i}. Therefore, U_{1i} ⊆ U_{i}. Similarly U_{2i} ⊆ U_{i}. Thus no point of U_{1i} has any point of
Now consider a point x ∈ U_{i}. It could be a point of γ_{o}^{∗} so suppose it is not. x is also not a point of
To see that U_{1i} ∩ U_{2i} = ∅, note that these are open conected sets. Also, it was argued above that U_{2i} is contained in U_{i} so it has no points of Γ. Neither does it have points of γ^{∗}. Therefore, U_{2i} =
The following lemma has to do with decomposing the inside and boundary of a simple closed rectifiable curve into small pieces. In doing this I will refer to a region as the union of a connected open set with its boundary. Also, two regions will be said to be non overlapping if they either have empty intersection or the intersection is contained in the intersection of their boundaries.The height of a set A equals

The width of A will be defined similarly.
We have a rough idea of what it means for things like circles and squares and triangles to be oriented in counter clockwise direction. You move from left to right on the bottom and from right to left on top. This is motivation for the lemma which follows in which a line segment will split the simple closed curve into two pieces, one having the top point and the other having the bottom point. Add in the line segment to obtain two simple closed curves. Then along the line, it will be oriented to move from left to right as part of the simple closed curve which has the point at the top and from right to left as part of the simple closed curve containing the point at the bottom. This produces an orientation on the original simple closed curve as well as an orientation on the resulting smaller simple closed curves.
This lemma is about splitting things into simple regions as just described and in estimating the number of regions which intersect Γ. The main idea consists of showing that you can divide a simple closed curve into two simple closed curves, one of which has the top point and the other having the bottom point and then doing the same thing relative to the left and right points of the simple closed curve.
The following argument will be used without explicit reference in the argument. Here γ is a parametrization and γ_{2} its second component. Let γ_{2}

This equals
In the following lemma, the term “region” will be used to denote a simple closed curve along with its inside.
Lemma 12.4.3 Let Γ be a simple closed rectifiable curve. Also let δ > 0 be given such that 2δ is smaller than both the height and width of Γ. Then there exist finitely many non overlapping regions

border regions. The construction also yields an orientation for Γ and for all these regions, the orientations for any segment shared by two regions are opposite.
Proof: Let x ∈ U_{i} and let l be the horizontal line through x where x =
The problem is, we might have picked the wrong segment. This is illustrated in the picture. The segment shown does separate Γ into two simple curves but one of them contains both the top and bottom points. The other is very short. It is desired to get two of these such that one of them contains a top point and excludes at least one bottom point. This prevents one of the new curves being very short.
Let γ be an orientation of Γ such that γ :

If c is not less than d, you could simply change the orientation or modify the argument. This orientation is used only to describe the special line segment. The final orientation will be obtained later after the line segment has been found.
The connected components of l ∩ U_{i} are the segments
Now here is the illustration of the correct segment. γ^{∗} is the segment shown. It divides Γ into two simple curves, one containing
By Lemma 12.4.2, this divides Γ into two simple closed curves having γ^{∗} as their intersection. Do the same process to each of these two simple closed curves. Each time you do this process, you get two simple closed curves in place of one. If the height of the curve is h > δ, then both new simple closed curves have length at least
Using Proposition 4.4.12, use the orientation on Γ obtained earlier to orient the resulting horizontal line segments. Thus, from this proposition, each segment has opposite orientation as part of the two simple closed curves resulting from its inclusion.
Now follow the same process just described on each of the non overlapping “short” regions just obtained using vertical rather than horizontal lines, letting the orientation of the vertical edges be determined from the orientation already obtained, but this time feature width instead of height and let the lines be vertical of the form x = k
How many border regions are there? Denote by V

The resulting arcs are each contained in a box having sides of length no more than δ in length. Each of these N arcs can’t intersect any more than four of the rectangles of the above construction because of their short length. Therefore, at most 4N boxes of the construction can intersect Γ. Thus there are no more than

border regions. This proves the lemma. ■
Note that this shows that the part of U_{i} which is included by the boundary boxes has area at most equal to
 (12.6) 
which converges to 0 as δ → 0. Also, this has shown that Γ is contained in the union of boxes having sides of length δ whose total Lebesgue measure is no more than 12.6. Since δ is arbitrary, this shows that the m_{2}
Proposition 12.4.4 The two dimensional Lebesgue measure of a simple closed curve having finite total variation is 0.
The following is a proof of Green’s theorem based on the above. Included in this formula is a way to analytically define the orientation of a simple closed curve in the plane. Recall that there were two orientations of a simple closed curve depending on the fact that there are two orientations for a circle in the plane. Green’s theorem can distinguish between these two orientations.
Lemma 12.4.5 Let R =
Then letting γ denote the oriented boundary of R as shown,

where

In this context the line integral is usually written using the notation

If the bounding curve were oriented in the opposite direction, then the area integral would be

Proof: This follows from direct computation. A parametrization for the bottom line of R is

A parametrization for the top line of R with the given orientation is

A parametrization for the line on the right side is

and a parametrization for the line on the left side is

Now it is time to do the computations using Theorem 4.2.5.

Changing the variables and combining the integrals, this equals

With this lemma, it is possible to prove Green’s theorem and also give an analytic criterion which will distinguish between different orientations of a simple closed rectifiable curve. First here is a discussion which amounts to a computation.
Let Γ be a rectifiable simple closed curve with inside U_{i} and outside U_{o}. Let
Let ℬ_{δ} be the set of border regions and let ℐ_{δ} be the rectangles contained in U_{i}. Thus in taking the sum of the line integrals over the boundaries of the interior rectangles, the integrals over the “interior edges” cancel out and you are left with a line integral over the exterior edges of a polygon which is composed of the union of the squares in ℐ_{δ}.
Now let f

and that for

the following pointwise convergence holds for δ denoting a sequence converging to 0.

By the dominated convergence theorem,

where m_{2} denotes two dimensional Lebesgue measure discussed earlier. Let ∂R denote the boundary of R for R one of these regions of Lemma 12.4.3 oriented as described. Let w_{δ}
 (12.7) 
whenever δ is small enough. Always let δ be this small.
Also since the line integrals cancel on shared edges
 (12.8) 
Consider the second sum on the left. From 12.7,

Denote by Γ_{R} the part of Γ which is contained in R ∈ℬ_{δ} and V

where

of these border regions. Furthermore, the sum of the lengths of all four edges of one of these is no more than 4δ and so

Thus

Let ε_{n} → 0 and let δ_{n} be the corresponding sequence of δ such that δ_{n} → 0 also. Hence
