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In the text Complex Analysis by Stein and Shakarchi: Add to Goursat's Theorem is stated as follows: If $\Omega$ is an open set in $C$ and $T \subset \Omega$ a triangle whose interior is also contained in $\Omega$ then the following occurs:

$$ 1.) \, \, \, \, \, \, \, \, \, \, \int_{T}f(z)dz = 0$$

The initial proof of 1.) topologically speaking is done by considering the original triangle, and by giving a bisection of the triangle into it's respective pieces i.e( new tringle are created from the bisection $T_{1}^{(1)}$,$T_{2}^{(2)}$,$T_{3}^{(3)}$,$T_{4}^{(4)}$). It is also important to note the visually the new tringles remain consistent of that of the original triangle. After this I observed that the author takes the integral of each of the triangles and combines them as follows in 2.)

$$2.) \, \, \, \int_{T_{0}^{(0)}}f(z)dz = \int_{T_{1}^{(1)}}f(z)dz \, + \int_{T_{2}^{(2)}}f(z)dz \, + \int_{T_{3}^{(3)}}f(z)dz + \int_{T_{4}^{(4)}}f(z)dz$$

The following procedure in 2.) as well as the topological basis can be visualized below in the image.What I failed to observe within the proof is why the following is stated within 4.)

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$\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ Bisection of a Tringle T

$\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ For some j we must have the following:

$$ 4.) \, \, \, \, |\int_{T_{0}^{(0)}}f(z)dz | \leq 4 \, |\int_{T_{j}^{(1)}}f(z)dz |$$

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    I believe Stein/Shakarchi used notation $T_1^{(1)}, T_2^{(1)}, \ldots$, so that the superscript (in parentheses) was the depth of triangles we had created. – 2017-02-05
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    Interesting I didn't catch that at first, I initially thought that it was used to donate each of the triangles and their respective areas. – 2017-02-05
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    The RH diagram represents 4 integrals $a,b,c,d$ adding to the integral $e$ represented by the LH diagram. Let $f=\max (|a|,|b|,|c|,|d|).$ ..... Then $|e|=|a+b+c+d|\leq |a|+|b|+|c|+|d|\leq 4f.$.... And $f $ is one of $|a|,|b|,|c|,|d|$...Which is all that the last line says. – 2017-02-05

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Okay, I think you're asking why we can justify the equality in (4), so I'll take a stab.

Since $|\int_{T_1^{(1)}} f(z)\ dz + \int_{T_2^{(1)}} f(z)\ dz + \int_{T_3^{(1)}} f(z)\ dz + \int_{T_4^{(1)}} f(z)\ dz| = |\int_{T_0} f(z)\ dz|$ the triangle inequality gives us

$$|\int_{T_1^{(1)}} f(z)\ dz| + |\int_{T_2^{(1)}} f(z)\ dz| + |\int_{T_3^{(1)}} f(z)\ dz| + |\int_{T_4^{(1)}} f(z)\ dz| \geq |\int_{T_0} f(z)\ dz|$$ so that $\frac{1}{4}|\int_{T_0} f(z)\ dz| \leq |\int_{T_1^{(j)}} f(z)\ dz|$ for some $j$. Multiply by 4 on both sides and you're there!

Obviously, we can repeat this for the nested triangles of $T_j^{(1)}$ and obtain the generalized statement that

$$|\int_{T_0} f(z)\ dz| \leq 4^n|\int_{T_j^{(n)}} f(z)\ dz| $$

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    Thank you I didn't see that step that could be used to justify the proof, is their another approach to proving equality 4.) besides using tools form the theory of inequalities. – 2017-02-06
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    I don't know of one off the top of my head but I'm sure there is some alternate formulation. However, this is the standard analytic way, and being a text on Complex Analysis... :D – 2017-02-10