Borel Measures and Bounded Variation

1,336

Solution 1

Let $f\in BV[a,b]$ Let $f=\phi_1 - \phi_2$ be the Jordan decomposition. We can now define two pre-measures $\mu_0'$ and $\mu_0''$. From which we construct the outer measures $\mu'^\star$ and $\mu''^\star$ and by Carathéodory's extension theorem we get $\mu'$ and $\mu''$. Now define $\mu=\mu'+\mu''$ by Jordan’s theorem for compositions of measures.

Our construction is obvoiusly injective. We will show that it is surjective as well. Let $\mu$ be a finite Borel sign-changing measure on $[a,b]$. Let $\mu=\mu^+-\mu^+$ be the Jordan decomposition. Set $\phi_1(x)=\mu^+([a,x))$ and $\phi_2(x)=\mu^-([a,x))$. Now $f=\phi_1 - \phi_2$ is a function of bounded variation as the difference of two monotonic non-negative such function.

We just established a bijection between the two sets. q.e.d.

Solution 2

Given a right-continuous function of bounded variation on a closed interval, you can define a unique borel measure that gives rise to the Lebesgue-Stieltjes integral. Look at the link for more details.

1,336

superAnnoyingUser

Updated on February 11, 2020

Comments

superAnnoyingUser almost 3 years

What is the connection between finite Borel sign-changing measures and the functions with bounded variation on the same interval? Proof would be appreciated.
- Ilya almost 10 years
  
  A proof of what?