proof of Goursat's theorem

We argue by contradiction. Set $$\eta = \oint_{\partial R} f(z)\, dz,$$ and suppose that $\eta\neq 0$ . Divide $R$ into four congruent rectangles $R_1, R_2, R_3, R_4$ (see Figure 1), and set $$\eta_i = \oint_{\partial R_i} f(z)\, dz.$$

Now subdivide each of the four sub-rectangles, to get 16 congruent sub-sub-rectangles $R_{i_1i_2},\; i_1,i_2=1\ldots 4$ , and then continue ad infinitum to obtain a sequence of nested families of rectangles $R_{i_1\ldots i_k}$ , with $\eta_{i_1\ldots i_k}$ the values of $f(z)$ integrated along the corresponding contour.

Orienting the boundary of $R$ and all the sub-rectangles in the usual counter-clockwise fashion we have $$\eta = \eta_1+\eta_2+\eta_3+\eta_4,$$ and more generally $$\eta_{i_1\ldots i_k} = \eta_{i_1\ldots i_k1}+\eta_{i_1\ldots i_k2}+\eta_{i_1\ldots i_k3}+\eta_{i_1\ldots i_k4}.$$ In as much as the integrals along oppositely oriented line segments cancel, the contributions from the interior segments cancel, and that is why the right-hand side reduces to the integrals along the segments at the boundary of the composite rectangle.

Let $j_1\in\{1,2,3,4\}$ be such that $\vert\eta_{j_1}\vert$ is the maximum of $\vert \eta_i \vert,\, i=1,\ldots,4$ . By the triangle inequality we have $$\vert\eta_1\vert+\vert\eta_2\vert+\vert\eta_3\vert+\vert\eta_4\vert \geq \vert\eta\vert,$$ and hence $$\vert\eta_{j_1}\vert\geq 1/4 \vert\eta\vert.$$ Continuing inductively, let $j_{k+1}$ be such that $\vert\eta_{j_1\ldots j_k j_{k+1}}\vert$ is the maximum of $\vert\eta_{j_1 \ldots j_k i}\vert,\, i=1,\ldots,4$ . We then have

Now the sequence of nested rectangles $R_{j_1 \ldots j_k}$ converges to some point $z_0\in R$ ; more formally $$\{z_0\} = \bigcap_{k=1}^\infty R_{j_1\ldots j_k}.$$ The derivative $f'(z_0)$ is assumed to exist, and hence for every $\epsilon>0$ there exists a $k$ sufficiently large, so that for all $z\in R_{j_1\ldots j_k}$ we have $$\vert f(z)-f'(z_0)(z-z_0) \vert \leq \epsilon \vert z-z_0\vert.$$ Now we make use of the following. The first of these assertions follows by the Fundamental Theorem of Calculus; after all the function $az+b$ has an anti-derivative. The second assertion follows from the fact that the absolute value of an integral is smaller than the integral of the absolute value of the integrand -- a standard result in integration theory.

Using the Lemma and the fact that the perimeter of a rectangle is greater than its diameter we infer that for every $\epsilon>0$ there exists a $k$ sufficiently large that $$\eta_{j_1\ldots j_k} = \left\vert \oint_{\partial R_{j_1\ldots j_k}}\hskip -2em f(z)\, dz \right\vert \leq \epsilon \vert\partial R_{j_1\ldots j_k}\vert^2 = 4^{-k}\, \vert \partial R\vert^2 \epsilon.$$ where $\vert\partial R\vert$ denotes the length of perimeter of the rectangle $R$ . This contradicts the earlier estimate (1). Therefore $\eta=0$ .