Proof of the Contour Integral Formula as a Riemann Sum A couple days ago I asked a question concerning complex integrals
The person that answered explained to me that the integral of the contour $C$ defined by $γ(t),a≤t≤b$ $ $ can be written as follows
$\int_Cf(z)dz$ is equal to the limit as $n$ tends to infinity of
$f(z_1)[z_1-z_0] + f(z_2)[z_2-z_1] + ... + f(z_n)[z_n-z_{n-1}]$ 
where $z_n = $$C$$(t_n)$, $a = t_0<t_1<t_2<\cdots<t_n=b$

The issue with the latter is that it implies that as long as all the $z_j-z_{j-1}$ tend to $0$, we will get the same result. I suppose this assumption is correct, but I haven´t been able to find any proof on the internet.
And, I´m in need of understanding this basic assumption in order to then prove the equality:
$\int_Cf(z)dz$ = $∫f(z(t))z′(t)dt$ (from $a$ to $b$)
So if someone could direct me to or show me the most clear proof you know without using the latter equality as an axiom I would truly appreciate it!
 A: 
The issue with the latter is that it implies that as long as all the $z_j-z_{j-1}$ tend to $0$, we will get the same result.

It's the mesh of the partition that must go to zero. The meaning of the limit is the following:

To every $\epsilon > 0$ there corresponds a $\delta > 0$ such that if $P:a = t_0 < t_1 < \cdots < t_n = b$ is a partition of $[a,b]$ with mesh less than $\delta$, then $\lvert\sum_{i = 1}^n f(z_i) \Delta z_i- \int_C f(z)\, dz\rvert < \epsilon.$ 

To prove $\int_C f(z)\, dz = \int_a^b f(z(t))z'(t)\, dt$, I'll use $\gamma(t)$ in place of $z(t)$. Set $M = \max\limits_{a\le t \le b}\lvert \gamma'(t)\rvert$ and let $\epsilon > 0$. There exists $\delta > 0$ such that if $P = \{s_0,\ldots, s_m\}$ is a partition of $[a,b]$ with $\|P\| < \delta$, then $\left\lvert \int_C f(z)\, dz - \sum_{i = 1}^m f(\gamma(s_i))\Delta z_i\right\rvert < \epsilon$. Since $f\circ \gamma$ is continuous on $[a,b]$, it is uniformly continuous on $[a,b]$. Hence, there is a $\delta' > 0$ such that for all $t, t'\in [a,b]$, $\lvert t - t'\rvert < \delta'$ implies $\lvert f(\gamma(t)) - f(\gamma(t'))\rvert < \frac{\epsilon}{b-a}$. Let $n\in \Bbb N$ such that $\frac{1}{n} < \min\{\delta,\delta'\}$, and set $t_i = a + (b-a)i/n$ for $i = 0,1,\ldots, n$. Then $Q = \{t_0,t_1,\ldots, t_n\}$ is a partition of $[a,b]$ with $\|Q\| = \frac{1}{n} < \delta$. Therefore $$\left\lvert \int_C f(z)\, dz - \sum_{i = 1}^n f(\gamma(t_i))\Delta z_i \right\rvert < \epsilon\tag{1}$$  As $\Delta z_i = \gamma(t_i) - \gamma(t_{i-1}) = \int_{t_{i-1}}^{t_i} \gamma'(t)\, dt$ for each $i$,
\begin{align}\sum_{i = 1}^n f(\gamma(t_i)) \Delta z_i - \int_a^b f(\gamma(t)) \gamma'(t)\, dt &= \sum_{i = 1}^n \int_{t_{i-1}}^{t_i} f(\gamma(t_i))\gamma'(t)\, dt - \sum_{i = 1}^n \int_{t_{i-1}}^{t_i} f(\gamma(t))\gamma'(t)\, dt\\
&=\sum_{i = 1}^n \int_{t_{i-1}}^{t_i} [f(\gamma(t_i)) - f(\gamma(t))]\gamma'(t)\, dt\tag{2}
\end{align}
Since $\|Q\| < \delta'$, the length of each interval $[t_{i-1},t_i]$ is less than $\delta'$. So for fixed $i$, $\lvert f(\gamma(t_i)) - f(\gamma(t))\rvert < \frac{\epsilon}{b-a}$ for all $t\in [t_{i-1},t_i]$. Consequently, 
$$\left\lvert \int_{t_{i-1}}^{t_i} [f(\gamma(t_i)) - f(\gamma(t))]\gamma'(t)\, dt\right\rvert < \frac{\epsilon M(t_i-t_{i-1})}{b-a}\quad (i = 1,\ldots, n)$$ It then follows from $(2)$ and the triangle inequality that 
$$\left\lvert \sum_{i = 1}^n f(\gamma(t_i)) \Delta z_i - \int_a^b f(\gamma(t))\gamma'(t)\, dt\right\rvert < \epsilon M\tag{3}$$
Combining $(1)$, $(3)$, and the triangle inequality, 
$$\left\lvert \int_C f(z)\, dz - \int_a^b f(\gamma(t))\gamma'
(t)\, dt\right\rvert < \epsilon(1+M)$$ Since $\epsilon$ was arbitrary, 
$$\int_C f(z)\, dz = \int_a^b f(\gamma(t))\gamma'(t)\, dt$$
A: To define an integral over a curve requires a parametric representation of the curve. Nothing else really makes sense.
The standard way to define $\int_{\gamma}f(z)dz$ is through the use of a continuous parametrized curve $\gamma : [a,b]\rightarrow \mathbb{C}$. Starting with an augmented partition $\mathscr{P}$ consisting of division points
$$
             a = t_0 < t_1 < t_2 < \cdots < t_N = b
$$
and augmentation points $t_j^* \in [t_{j-1},t_{j}]$, one forms the Riemann-Stieltjes sum
$$
    \sum_{\mathscr{P}} f(\gamma(t_j^*))\{ \gamma(t_j)-\gamma(t_{j-1})\}
$$
The integral is then defined as
$$
      \int_{a}^{b}f(\gamma(t))d\gamma(t) = \lim_{\|\mathscr{P}\|\rightarrow 0}\sum_{\mathscr{P}}f(\gamma(t_j^*))\{\gamma(t_{j})-\gamma(t_{j-1})\},
$$
provided such a limit exists. The limit will exist, for example, if $f(\gamma(t))$ is continuous and $\gamma(t)$ is continuous and of bounded variation. And you can show that the limit is independent of continuous parametrization of the curve $\gamma$, so long as it remains of bounded variation. More specificially, if $\rho$ is a continuous non-decreasing function from $[a,b]$ onto itself, then
$$
        \int_{a}^{b}f(\gamma(t))d\gamma(t)) = \int_{a}^{b} f(\gamma(\rho(t)))d\gamma(\rho(t))
$$
If $\gamma$ is continuously differentiable on $[a,b]$, then
$$
         \int_{a}^{b}f(\gamma(t))d\gamma(t)) = \int_{a}^{b}f(\gamma(t))\gamma'(t)dt.
$$
More precisely, if $f$ is a bounded function, then the Riemann-Stieltjes integral on the left exists iff the Riemann integral on the right exists and, in that case, the two are equal. These results taken together give precise meaning to $\int_{\gamma} fdz$.
