Borel Measures and Bounded Variation
Solution 1
Let $f\in BV[a,b]$ Let $f=\phi_1  \phi_2$ be the Jordan decomposition. We can now define two premeasures $\mu_0'$ 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 signchanging 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 nonnegative such function.
We just established a bijection between the two sets. q.e.d.
Solution 2
Given a rightcontinuous function of bounded variation on a closed interval, you can define a unique borel measure that gives rise to the LebesgueStieltjes integral. Look at the link for more details.
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superAnnoyingUser
Updated on February 11, 2020Comments

superAnnoyingUser almost 3 years
What is the connection between finite Borel signchanging measures and the functions with bounded variation on the same interval? Proof would be appreciated.

Ilya almost 10 yearsA proof of what?
