Can the line integral be induced by a measure

Question

For $\Omega\subset\mathbb{R}^d$, $x_0\in\mathbb{R}^d$ and $v\in C(\Omega)$, the evaluation of $v$ at the point $x_0$ can be expressed by
$$v(x_0)=\int_\Omega v(x)\,d\delta_{x_0}(x),$$
where $\delta_{x_0}:\mathcal{B}(\Omega)\to\{0,1\}$ is the Dirac measure
$$\delta(A):=\begin{cases}1,&x_0\in A,\\0,&x_0\notin A.\end{cases}$$
If one imagines the curve integral along the curve $\Gamma:[0,1]\to\Omega$,
$$ \int_\Gamma v\,ds := \int_0^1v(\Gamma(x))|\Gamma'(x)|\,dx ,$$
to be a generalization of the above point evaluation, can one find a measure $\mu:\mathcal{B}(\Omega)\to[0,\infty]$ such that the relation
$$ \int_\Gamma v\,ds = \int_\Omega v(x)\,d\mu(x)$$
holds?

Answer 1

Yes.

The measure $\mu$ can be defined via $$ \mu(A) = \int_0^1 \chi_A(\Gamma(x)) |\Gamma'(x)| \mathrm dx, $$ where $A\in\mathcal B(\Omega)$, and $\chi_A:\Omega \to \{0,1\}$ is the function that is $1$ on $A$ and $0$ on $\Omega\setminus A$.

Then it can be shown that $$ \int_\Gamma v\,ds := \int_0^1v(\Gamma(x))|\Gamma'(x)|\,dx =\int_\Omega v(x)\,d\mu(x)$$ holds not only for continuous functions, but measurable functions (which are integrable with respect to $\mu$). One possible way to see this is using a standard procedure from measure theory: first show it for characteristic functions, then for linear combinations of these (sometimes called simple functions), then for nonnegative functions by using supremum of a monotone sequence of simple functions, then for arbitrary functions (using positive and negative parts).