How to show that the integral $\frac{1}{2\pi i} \int_{\gamma} \frac{dz}{z - a}$ is integer-valued when the curve $\gamma$ is not piecewise smooth?

Question

In Conway's Functions of One Complex Variable, there is a proposition which is as follows:

5.1 Proposition. If $\gamma\colon [0,1] \to \mathbb{C}$ is a closed rectifiable curve and $a \notin \{\gamma\}$ then $$\frac{1}{2\pi i} \int_{\gamma}\frac{dz}{z - a}$$ is an integer.

Proof. This is only proved under the hypothesis that $\gamma$ is differentiable.

For those unfamiliar with Conway's book, $\{\gamma\}$ is his notation for the image set of the curve $\gamma$ and differentiable (when referring to curves) means $\gamma'$ exists and is continuous. I know the latter as smooth. The proposition is meant to motivate his definition of the winding number.

From Conway's proof, it's trivial to generalize to closed piecewise smooth curves. This leads me to my first question: how can one prove the proposition in its full generality for closed rectifiable curves?

In general, the integral is a Riemann-Stieltjes integral $$\frac{1}{2\pi i} \int_{0}^{1}\frac{1}{\gamma(t) - a}\,d\gamma(t)$$ whose existence follows from the fact that $1/(\gamma(t) - a)$ is continuous and $\gamma(t)$ is of bounded variation.

Seeing as how the winding number about $a$ can be defined using a continuous choice of argument for all closed curves not passing through $a$, my second question is: if the hypothesis is relaxed to include all closed curves, does the integral necessarily exist and is it still integer-valued?

Answer 1

Following the way suggested in Conway's book, let me try to write down a full proof here. The idea is simple: as is indicated in the book(if I recall correctly), the generalization from piecewise differentiable curves to rectifiable curves usually involves approximation by polygons.

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The required result for piecewise differentiable curves follows almost the same line of the original proof, so we assume it here. To generalize to a rectifiable curve $\gamma$, we first make some simple observations:

Observation 1. $\gamma$ is a continuous map defined on a compact set, hence is uniformly-continuous. In particular, for any $\delta>0$, there is a partition $0=t_1<t_2<\ldots<t_n=1$ such that $|\gamma(s)-\gamma(t)|<\delta$ whenever $t_k\le s, t\le t_{k+1}$.

Obsevation 2. For the same reason in Observation 1, $\gamma$ has compact images. Thus $d:=\operatorname{dist}(a,\{\gamma\})>0$.

Combining these two observations, we may find a refinement of the partition in Observation 1 with $|\gamma(s)-\gamma(t)|<\frac12\min(\delta, d)$. Note that $f(z):=1/(z-a)$ is uniformly continuous in $N(\{\gamma\},\frac12d)$, the $\frac12d$-neighborhood of $\{\gamma\}$, by a standard argument. Thus we may choose $\delta$ sufficiently small such that $|f(\xi)-f(\eta)|<\epsilon$ whenever $\xi,\eta\in N(\{\gamma\},\frac12d)$ and $|\xi-\eta|<\delta$.

We now have all the ingredients for an approximation. Let $\Gamma:[0,1]\to\mathbb C$ be the polygon with vertices $\Gamma(t_k)=\gamma(t_k)$. That is, $\Gamma$ on $[t_k,t_{k+1}]$ is the segment connecting $\gamma(t_k),\gamma(t_{k+1})$. This implies $|\Gamma(t)-\gamma(t_k)|<\min(\delta, d)$. We first approximate $\int_\gamma f$ by its value at the end point. This follows from the uniform continuity of $f$: \begin{align} &\left|\int_{t_k}^{t_{k+1}}f(\gamma(t))d\gamma(t)-f(\gamma(t_k))(\gamma(t_{k+1})-\gamma(t_k))\right|\\ =&\left|\int_{t_k}^{t_{k+1}}(f(\gamma(t))-f(\gamma(t_k)))d\gamma(t)\right|\\ \le&\epsilon V_{t_k}^{t_{k+1}}(\gamma) \end{align} where $V$ is the total variation of $\gamma$. Consequently $$\left|\int_\gamma f-\sum_{k=0}^{n-1}f(\gamma(t_k))(\gamma(t_{k+1})-\gamma(t_k))\right|\le \epsilon V(\gamma)$$ We also need an explicit formula for $\int_\Gamma f$ to proceed. This can be computed directly $$\int_{t_k}^{t_{k+1}}f(\Gamma(t))d\Gamma(t)=\frac{\gamma(t_{k+1})-\gamma(t_k)}{t_{k+1}-t_k}\int_{t_k}^{t_{k+1}}f(\Gamma(t))dt$$ Putting all these together we have \begin{align} \left|\int_\Gamma f-\int_\gamma f\right|&\le\epsilon V(\gamma)+\sum_{k=0}^{n-1}\left|\int_{t_k}^{t_{k+1}}f(\Gamma(t))d\Gamma(t)-f(\gamma(t_k))(\gamma(t_{k+1})-\gamma(t_k))\right|\\ &=\epsilon V(\gamma)+\sum_{k=0}^{n-1}\frac{|\gamma(t_{k+1})-\gamma(t_k)|}{t_{k+1}-t_k}\left|\int_{t_k}^{t_{k+1}}(f(\Gamma(t))-f(\gamma(t_k)))dt\right|\\ &\le 2\epsilon V(\gamma) \end{align} Since $\int_\Gamma f$ is an integer, this yields $\int_\gamma f$ is an integer.