text: # Can the line integral be induced by a measure?
  
  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?
  
  • 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 Oct 27, 2020 at 13:47
  
  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).
  • 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. Oct 26, 2020 at 17:38
explanation: The document contains a detailed mathematical discussion that starts with a question about whether a line integral can be induced by a measure. It then provides an answer, explaining how such a measure can be constructed and showing the equivalence of the line integral and the integral with respect to this measure. The document also includes additional comments that further elaborate on the topic.
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distilabel_metadata.content: # Can the line integral be induced by a measure?
  
  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?
  
  • 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 Oct 27, 2020 at 13:47
  
  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).
  • 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. Oct 26, 2020 at 17:38
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a detailed mathematical discussion that starts with a question about whether a line integral can be induced by a measure. It then provides an answer, explaining how such a measure can be constructed and showing the equivalence of the line integral and the integral with respect to this measure. The document also includes additional comments that further elaborate on the topic.
  
  Label: question-and-answer