1
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ The answer is affirmative, as already shown, and such measure can be written in terms of the Dirac generalized function. For example, the arc length measure on the unit circle in $\mathbb R^2$ is given by $$\delta\left(1- \sqrt{x^2+y^2}\right)dxdy.$$See math.stackexchange.com/a/56998/8157 $\endgroup$ Oct 27, 2020 at 13:47

1 Answer 1

3
$\begingroup$

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).

$\endgroup$
4
  • $\begingroup$ Yes, more generally, $$ \mu(A) = \int_{\Omega} \chi_A(x)\,\mu(\mathrm d x) $$ allows to do the correspondence between measures and linear forms acting on continuous functions. $\endgroup$
    – LL 3.14
    Oct 26, 2020 at 17:38
  • $\begingroup$ @supinf Could you give me a hint how I can see this? At first I was like 'Easy, just define some measure with a specific density function.', but now when I try to validate the required relation I do have some troubles. $\endgroup$
    – bongobums
    Oct 27, 2020 at 13:20
  • $\begingroup$ @bongobums I added something. Maybe there is a simpler way, but this was the first thing that comes to mind. $\endgroup$
    – supinf
    Oct 27, 2020 at 13:44
  • $\begingroup$ @supinf Thank you very much! $\endgroup$
    – bongobums
    Oct 27, 2020 at 14:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .