0

BV space is defined as a function whose weak derivative Borel measure. I don't understand how the function can be a measure. Measure is usually defined on the subset of real line, and the function is defined on each point of the real line. Can someone please help?

jk001
  • 781
  • Why a downvote? – jk001 Mar 23 '22 at 05:20
  • As in, I expect that a "function" $f$ being a "measure" would translate to the measure $\mu$ defined by $\mu((a,b)) = f(b)-f(a)$, with $\mu$ then extended to the Borel sigma-algebra. – Sarvesh Ravichandran Iyer Mar 23 '22 at 05:27
  • Thanks for your answer. Sorry, I don't seem to understand.. does the function have to be increasing? What do you mean "extended to the Borel sigma-algebra"? – jk001 Mar 23 '22 at 05:30
  • Do you have the "exact" mathematical definition of the BV space available to you? I can't really go further without knowing your notation and your details : I don't want to confuse you,essentially. – Sarvesh Ravichandran Iyer Mar 23 '22 at 05:35
  • It is possible for functions in $L^1(\mathbb R^d)$ to have a weak derivative that is not a function, but a measure. This comes from the definition of a weak derivative being a "distribution", and it's possible to interpret a measure as a distribution. Functions in $BV$ are those whose weak derivatives are measures which are finite and Radon. – Sarvesh Ravichandran Iyer Mar 23 '22 at 05:38
  • 1
    Do you know what is a weak derivative? – Arctic Char Mar 23 '22 at 05:42
  • Yes, I know the definition. What I don't understand is how a function can be a measure. – jk001 Mar 23 '22 at 05:46
  • I can see how a function can be a measure in a distribution sense.. sorry for nonsense question – jk001 Mar 23 '22 at 05:52
  • No problem, we all ask them I suppose! – Sarvesh Ravichandran Iyer Mar 23 '22 at 05:54

1 Answers1

2

To say a measure is a Borel measure just means that it assigns a measure to every open set, and to every set which can be obtained by complements, countable unions and countable intersections of open sets. There are some sets which you can't assign a Lebesgue measure to, and calling a measure a Borel measure just means that it assigns a measure to the same sets that Lebesgue measure is defined for. A Borel measure can be completely defined by its value on open sets: there is only one way to assign a measure to any non-open sets which is consistent with those measures on the open sets.

The variation of a function $f:\mathbb{R} \to \mathbb{R}$ can be defined as a Borel measure: for open sets $A$ $Var(f)(A):=\sup\{\int_A f \, div(g)\,dx:g \in C^1(\mathbb{R}),g(\mathbb{R} \backslash A)=\{0\}\}$. I think this is what you're calling the "weak derivative", although what I would call the weak derivative is something different, a signed measure. If $f \in C^1(\mathbb{R})$, then the variation equals $\int_A |\nabla f| \,dx$. If $f$ has a non-removable discontinuity, then $Var(f)$ assigns a point mass to that point. The variation is a measure in the sense that it assigns a value in $[0,\infty]$ to each Borel subset of $\mathbb{R}$, i.e. it assigns a value to each subset of $\mathbb{R}$ which can be obtained by complements and countable unions or intersections of open sets.

In fact, nonnegative $L^1$ functions are already defined as measures. An "$L^1$ function" is actually an equivalence class of functions, which are equal to each other except on sets of measure zero, which are defined precisely by the fact that they all give rise to the same measure $A \mapsto \int_A f\,dx$. If $f$ is nonnegative, then this is a measure. (If $f$ can be negative, then it's a signed measure).

1Rock
  • 2,046