Formal proof that cauchy integral formula for a function defined as follows 
Cauchy Integral formula
Let $\gamma:[0,1]\rightarrow \mathbb{C}$ be a rectifiable Jordan curve and $V$ be the interior of $\gamma$.
Let $f:\bar{V}\rightarrow \mathbb{C}$ be a continuous function which is holomorphic on $V$.
Then, $Wnd(\gamma,\alpha) f(\alpha)=\frac{1}{2\pi i} \int_\gamma \frac{f(z)}{z-\alpha} dz$, for each $\alpha\in V$.

This can be proved if we can construct a sequence of rectifiable curves inside $V$ which converges to $\gamma$ uniformly. If this sequence is constructed, we can apply the ordinary Cauchy integral formula. However, how do I construct a such sequence formally?
 A: This is just a long comment.
For any domain $D\subset \mathbb{R}^n$, an exhaustion of compact sets of $D$ with nice properties can be constructed as
$$K_n=\{x\in D:\|x\|\leq n,\rho(x,\partial D)\geq 1/n\}$$
such that $\cup_n K_n=D$. Since the function $\rho(x,\partial D)$ is Lipshitz, $\partial K_n$ is rectifiable(One may do even more: by covering $K_n$ with finitely many small balls contained in $D$, we may get an exhaustion of compact sets $K_n'$ with $\partial K_n'$ piecewise smooth).
Apply the procedure on $\mathrm{Int}\gamma$. Because $\gamma$ is bounded, for sufficiently large $n$ we see $K_n=\{x\in D:\rho(x,\partial D)\geq 1/n\}$. It's easy to check $\partial K_n\subseteq\{x\in D:\rho(x,\partial D)=1/n\}$. Hence $\partial K_n$ converges uniformly to $\partial D$.
But there is still one problem: $\partial K_n$ may consists of several curves. One may easily make $K_n$ connected(choosing any $z_0\in D$ and taking $K_n'$ to be the connected component containing $z_0$), but this seems not enough to make $\partial K_n$ a simple curve.
