1

Let $\phi: [0,1] \to R$ be a function of bounded variation that jumps at one point $x=1/2.$ I know that by the structure theorem
$D\phi= \nabla \phi L^1 + D^c \phi+( \phi(x^+) - \phi(x^-))\delta_x.$

Now I want to integrate a BV function $\psi $ that jumps also at $x=1/2$ with respect to the measure $D\phi.$ What confuses me is the jump part of the measure. If $\psi$ was continuous I would get $( \phi(x^+) - \phi(x^-))\psi(1/2) $ but $\psi$ here is a BV function so it is defined as a class of functions!!!!

Adrian Keister
  • 10,099
  • 13
  • 30
  • 43
user527849
  • 21

1 Answers1

0

The question is based on a misunderstanding: even if $\psi\in\mathrm{BV}([{0,1}])$ it is nevertheless a function, not a class. It can be chosen as the representative of a given class in $\mathrm{BV}([{0,1}])$, and as such its pointwise structure is irrelevant (as long as it is relevant only on a set of measure zero), from the point of view of the calculation of the associated Radon measure, but not when you use it for doing other operations. The exact knowledge of $\psi$ allows in principle to integrate it respect to the Radon measure generated by another $\phi\in\mathrm{BV}([{0,1}])$ (which can be determined by using any function in the same equivalence class in $\mathrm{BV}$) by using the standard formula $$ \int\limits_{[0,1]}\psi \mathrm{d}(D\phi)=\int\limits_{[0,1]}\psi\nabla\phi\mathrm{d}\mu_L+\int\limits_{[0,1]}\psi \mathrm{d}(D^c\phi)+\left(\phi(x^+) - \phi(x^-)\right)\psi(x)\quad x=\frac{1}{2} \tag{1}\label{1} $$ where $\mu_L$ is the standard Lebesgue measure.

It is nevertheless true that, if you only know that $\psi$ is a function of bounded variation you cannot evaluate \eqref{1}, but this happens in every case when the knowledge of a property of a function is not equivalent to the knowledge of the property needed to perform a particular operation. In sum, to integrate a function respect to a given general Radon measure you should know its values on the singular set.

Daniele Tampieri
  • 9,727