TITLE: Cauchy's theorem: what about non-smooth homotopies?
QUESTION [19 upvotes]: This morning I realized I have never understood a technical issue about Cauchy's theorem (homotopy form) of complex analysis. To illustrate, let me first give a definition.
(In what follows $\Omega$ will always denote an open subset of the complex plane.)
Definition Let $\gamma, \eta\colon [0, 1]_t \to \Omega$ be two piecewise smooth closed curves (or circuits). We say that $\gamma, \eta$ are $\Omega$-homotopic if there exists a continuous mapping $H \colon [0, 1]_\lambda \times [0, 1]_t \to \Omega$ s.t. 

$H(0, t)=\gamma(t)$ and $H(1, t)=\eta(t), \quad \forall t \in [0, 1]$;
$H(\lambda, 0)=H(\lambda, 1), \quad \forall \lambda \in [0, 1]$.

Theorem (Cauchy) Let $f\colon \Omega \to \mathbb{C}$ be holomorphic. If $\gamma, \eta$ are $\Omega$-homotopic circuits, then 
$$\int_{\gamma} f(z)\, dz= \int_{\eta}f(z)\, dz.$$

Problem The function $H$ above is only continuous and need not be smooth. So for $0< \lambda < 1$, the closed curve $H(\lambda, \cdot)$ may be pretty much everything (a Peano curve, for example). Does this void the validity of theorem as it is stated above? How can integration be defined over such a pathological object? 

The proof of Cauchy's theorem that I have in mind goes as follows. To begin, one observes that for a sufficiently small value of $\lambda_1$, the circuits $\gamma=H(0, \cdot)$ and $H(\lambda_1, \cdot)$ are close toghether; that is, they can be covered by a finite sequence of disks not leaving $\Omega$ like in the following figure:

Since $f$ is locally exact, its integrals over every single disk depend only on the local primitive. Playing a bit with this, one arrives at 
$$\int_\gamma f(z)\, dz= \int_{H(\lambda_1, \cdot)} f(z)\, dz.$$
Then one repeats this process, yielding a $\lambda_2$ greater than $\lambda_1$ and such that
$$\int_{H(\lambda_1,\cdot)} f(z)\, dz= \int_{H(\lambda_2, \cdot)} f(z)\, dz.$$
And so on. A compactness argument finally shows that this algorithm ends in a finite number of steps. 
Problem is: this proof assumes implicitly that $H(\lambda_1, \cdot), H(\lambda_2, \cdot) \ldots$ are piecewise smooth, to make sense of integrals $$\int_{H(\lambda_j, \cdot)}f(z)\, dz.$$ 
This, however, does not follow from the definition if $H$ is only assumed to be continuous. Therefore this proof works only for smooth $H$. 
Is this regularity condition necessary?

REPLY [10 votes]: As has already been pointed out by Akhil in the comments, any two smooth curves $\gamma_0, \; \gamma_1$ which are continuously homotopic are also smoothly homotopic. The point here is to approximate a continuous homotopy, rather than the curves themselves.
More concretely, given homotopic smooth curves $\gamma_0, \gamma_1: I \to \Omega$,    let $H(s,t): I\times [0,1] \to \Omega$ be a continuous homotopy between $\gamma_0$ and $\gamma_1$. 
By the approximation theorem of your choice (mine is due to Whitney), we can find a smooth map $G: I\times[0,1] \to \Omega$, which coincides with $H$ on $I \times \{0\} \cup I \times \{1\}$ (since $H$ is smooth there). 
This means that we find a smooth homotopy $G$ between $\gamma_0$ and $\gamma_1$, so we are done in this special case.
Now, if we start out with rectifiable curves rather than smooth ones, we can find two smooth curves $\tilde \gamma_0$ and $\tilde \gamma_1$, which approximate our initial $\gamma_0$ and $\gamma_1$, respectively, and such that the linear homotopy
$$H_\lambda(s,t) := t\gamma_\lambda(s) + (1-t)\tilde \gamma_\lambda(s) \qquad \lambda = 0, 1$$
maps into $\Omega$ (choose a $\epsilon$-neighborhood of the image $\gamma_\lambda(I)$ of $I$ which is contained in $\Omega$ and take $\tilde \gamma_\lambda$ to be contained within this neighborhood).
It is not difficult to see that $H_\lambda$ will then be a "rectifiable" homotopy. 
But with $\tilde \gamma_0$, $\tilde \gamma_1$ we are again in the first situation, so there is a smooth homotopy between them.
Now we can build a rectifiable homotopy between $\gamma_0$ and $\gamma_1$ in three steps

Homotop $\gamma_0$ to $\tilde \gamma_0$ by the linear homotopy.
Use a smooth homotopy between $\tilde \gamma_0$ and $\tilde \gamma_1$
Go from $\tilde \gamma_1$ to $\gamma_1$ by the straight line homotopy.

Thus proving that all notions of "homotopic" agree.