First we need to define some concepts to explan my question:
- Curve $C\subset\mathbb{C}$ have length
- $P=\{P_1,...,P_{n+1}\}$ satisfy $P_i\in C$ and lines at the direction of integration, saperate $L$ into $n$ parts
- $\Delta s_i$ is the length of $i$'th partition by $P$
- $\lambda=\max(\Delta s_i)$
- $\xi_i$ is a random points on partition $P_iP_{i+1}$ of $C$
- $f(z)$ is a limited function on $C$
- $\Delta z_i=z_{i+1}-z_i$ is the minus of complex number on points $P_{i+1}$ and $P_i$
So if $\int_Cf(z)dz$ exists, it stats that $\forall\epsilon>0,\exists\delta>0$ and $\forall P,\{\xi_i\}$ (that is, dispite of way of partition $P$ and pick of points $\xi_i$) as long as $\lambda<\delta$ we have $\left|\sum_{i=1}^{n-1}f(\zeta_i)\Delta z_i-\int_C f(z)dz\right|<\epsilon$
Based on above condition, I would like to prove that will $\exists A$ that satisfy $\forall\epsilon>0,\exists\delta>0$ and $\forall P,\{\xi_i\}$ as long as $\lambda<\delta$ we have $\left|\sum_{i=1}^{n-1}|f(\zeta_i)||\Delta z_i|-A\right|<\epsilon$
I think the claim is wrong for curves $C$ which are not rectifiable, i.e., have infinite length. Consider $$C:\quad t\mapsto z(t):=t\left(1+ i\,\sin{1\over t}\right)\qquad(0< t\leq1),\qquad z(0):=0$$ and $f(t):\equiv1$. Then $$\int_C f(z)\,dz=\int_C dz=z(1)-z(0)=1+i\,\sin 1\ ,$$ and all your Riemann sums have this value. But the integral $$\int_C|f(z)|\,|dz|=\int_C|dz|=\infty\ .$$ For a nice "positive result" one would have to know more about $f$ and $C$, and why the integral $\int_C f(z)dz$ exists.