Naturalness of definition of line integral

Question

Let $I = [a,b]$ denote some interval, and $f : I \rightarrow \mathbb{C}$ a continuous function of bounded variation, in other words, $f$ is a parameterisation of $\gamma = f(I)$, a rectifiable curve. Let $\mathcal{l}(\gamma) = L$ be its length, and write $\hat{f} : [0,L] \rightarrow \mathbb{C}$ for the normal parameterisation (that is, $\hat{f}$ is obtainable from $f$ via a continuous, non-decreasing change of parameter, and the length function of $\hat{f}$ is given by $\sigma(s) = s$ for $0 \leq s \leq L$).

Suppose $\rho : \mathbb{C} \rightarrow \mathbb{C}$ is Borel measurable and define: $$ \int_{\gamma} \rho |dz| = \begin{cases} \displaystyle\int_{[0,L]} \rho \circ \hat{f} dm = \int_{\gamma} \rho \hat{f}_{\ast}(dm)&\text{if }L>0\\ \\ 0 &\text{otherwise} \end{cases} $$ where $dm$ is the one-dimensional Lebesgue measure.

This is independent of the parameterisation $f$, in the sense that if $g = f \circ \phi$, where $\phi : [c,d] \rightarrow I$ is non-decreasing and continuous then $\hat{g} = \hat{f}$.

$\textbf{Question}$: Is it necessarily the case that $\hat{f}_{\ast}(dm) = \mathcal{H}^{1} \restriction \gamma$? Here $\mathcal{H}^1$ is the one-dimensional Hausdorff measure in $\mathbb{C} \equiv \mathbb{R}^2$ and $\hat{f}_{\ast}(dm)$ is the pushforward of the one-dimensional Lebesgue measure on $[0,L]$ onto $\gamma$. If $f$ is not injective this can't be true, but is there still a way to relate everything to $\mathcal{H}^1$ with multiplicity?

Answer 1

This is not an entirely complete answer. In particular I'll assume that $\rho$ is $L^1$ and that $\hat{f}$ is Lipschitz.

Consider the following version of the Area formula (where $\mathcal{L}_n$ denotes Lebesgue measure on $\mathbb{R}^n$):

Theorem (Area formula). Let $f:\mathbb{R}^n\to\mathbb{R}^m$ be Lipschitz, where $n\leq m$, and let $g\in L^1(\mathbb{R}^n)$. Then $$\int_{\mathbb{R}^n}gJf\,\mathrm{d}\mathcal{L}_n=\int_{\mathbb{R}^n}\sum_{x\in f^{-1}(\{y\})}g(x)\,\mathrm{d}\mathcal{H}^n(y).$$

Let us suppose that $\rho$ is $L^1$ and that $\hat{f}$ is Lipschitz and such that $\lVert \hat{f'}(x)\rVert=1$ for a.e. $x\in[0,L]$. Then $J\hat{f}(x)=1$ for a.e. $x\in[0,L]$, and so we can use the Area formula to obtain that

$$\int_\gamma\rho(z)\,\lvert\mathrm{d}z\rvert=\int_{[0,L]}(\rho\circ\hat{f})J\hat{f}\,\mathrm{d}\mathcal{L}_1=\int_{\gamma}\sum_{x\in \hat{f}^{-1}(\{y\})}\rho(\hat{f}(x))\,\mathrm{d}\mathcal{H}^1(y).$$

Define now a multiplicity function

$$N_\gamma:\gamma\to\mathbb{N},\quad y\mapsto \lvert \hat{f}^{-1}(\{y\})\rvert,$$

which is well defined a.e. as the above quantity is invariant of our choice of $\hat{f}$. It is then clear that

$$\sum_{x\in\hat{f}^{-1}(\{y\})}\rho(\hat{f}(x))=\rho(y)N_\gamma(y),$$

and so in particular this gives us that

$$\int_\gamma\rho(z)\,\lvert\mathrm{d}z\rvert=\int_\gamma\rho N_\gamma\,\mathrm{d}\mathcal{H}^1.$$

In particular you get that

$$\frac{\mathrm{d}(\hat{f}_*\mathcal{L}_1)}{\mathrm{d}(\mathcal{H}^1\rvert\gamma)}=N_\gamma\qquad \text{(a.e.)}$$

In particular, for the case when we can a.e. injectively parametrize $\gamma$ we get that $\hat{f}_*\mathcal{L}_1=\mathcal{H}^1\rvert\gamma$, as you would expect.