Goursat's Lemma states that if $f$ is holomorphic inside and on a triangular path $\gamma$ then:
$\int_{\gamma}^{} f(z) dz = 0$
In the proof of Goursat's Lemma, it says:
'Now choose N large enough so that $\triangle_N \subseteq D(\omega;r)$'
Where $\triangle$ is a triangular path.
$\triangle_N$ is the Nth triangle that the path has been 'cut' into
$D(\omega;r)$ is the disk around $\omega$ with radius r
I know that the rough proof of Goursat's lemma is that we split the triangle into many smaller triangles. We then use the fact that it is holomorphic to show the integral around these small triangles is so small that even once we add up all the integrals around all the triangles all up its still very small.
However why does such an N exists that we can choose an N large enough so $\triangle_N \subseteq D(\omega;r)$?
Some points I know that could help are:
-I know the triangle with longest side length k is closed in a circle of radius k.
-$2k < l$ where %l% is the perimeter length
I know that if I use these in my reason I will also have to prove those but that isn't so much the issue.
At first you can look at $f_n(z) = n^{2} \int_0^{1/n} \int_0^{1/n} f(z+x+iy)dxdy$ to make the derivative (uniformly) continuous. Let $S_{b}$ be the boundary of the square $[0,b]\times [0,ib]$ then as $b\to 0$ $$\int_{a+S_{b}} f_n(z)dz =\int_{a+S_{b}}(f_n(a)+(z-a)f_n'(a)+o(z-a))dz = \int_{a+S_{b}}o(z-a)dz=o(b^2)$$ uniformly on $a$.
Thus taking $b_m \to 0$ and splitting $\text{interior}(\gamma)$ in $ L_m=O(b_m^{-2}\text{area}(\text{interior}(\gamma)) $ squares $a_{m,l}+S_{b_m}$ of size $b_m$ $$\int_\gamma f(z)dz =\lim_{n \to \infty}\int_\gamma f_n(z)dz = \lim_{n \to \infty}\lim_{m\to \infty} \sum_{l=1}^{L_m} \int_{a_{m,l}+S_{b_m}}f_n(z) dz = \lim_{n \to \infty}\lim_{m\to \infty} o_n(L_m b_m^2) =0$$