If the variation of a measure $μ$ is smaller than the variation of a measure $ν$ on any compact interval, then $μ$ is absolutely continuous wrt $ν$

Question

Let

• $a,b\in\mathbb R$ with $a<b$
• $I:=(a,b]$
• $\operatorname{Var}_{[s,\:t]}g$ denote the variation of a function $g:\overline I\to\mathbb R$ on $[s,t]\subseteq\overline I$
• $g,h:\overline I\to\mathbb R$ be right-continuous and of bounded variation with $$\operatorname{Var}_{[s,\:t]}g\le\operatorname{Var}_{[s,\:t]}h\;\;\;\text{for all }[s,t]\subseteq\overline I\tag1$$
• ${\rm d}g$ and ${\rm d}h$ denote the unique measures on $\mathcal B(I)$ with $${\rm d}g((s,t])=g(t)-g(s)\tag2$$ and $${\rm d}h((s,t])=h(t)-h(s)\tag3$$ for all $(s,t]\subseteq I$
• $|{\rm d}g|:\mathcal B(I)\to\mathbb R$ and $|{\rm d}h|:\mathcal B(I)\to\mathbb R$ denote the variation meaures of ${\rm d}g$ and ${\rm d}h$

How can we prove that any ${\rm d}h$-null set $N$ is a ${\rm d}g$ null set?

Answer 1

Recall that a $dh$-null set is a set $N$ such that $|dh|(N)=0$ (rather than $dh(N)=0$). Indeed, $|dh|(N) = 0$ is equivalent to $(dh)^+(N)=(dh)^-(N)=0$, and the latter is equivalent to $dh(B)=0$ for all $B\subset N$, due to $dh=(dh)^+-(dh)^-$.

The assumption (1) says that $|dg|(I)\le |dh|(I)$ for every interval $I$. In other words, the measure $|dh|-|dg|$ is nonnegative on every interval. Since intervals generate the Borel $\sigma$-algebra, the measure $|dh|-|dg|$ is nonnegative for all Borel sets $E$. Consequently $|dg|(E)\le |dh|(E)$, which implies that $|dg| $ vanishes whenever $|dh|$ does.