Proposition 11.6.2. Let $[a, b] \subset \real$, $E$ be a normed space, $G \in C^{1}([a, b]; E^{*})$, and $\mu_{G} \in M_{R}([a, b]; E^{*})$ be the associated Lebesgue-Stieltjes measure, then
\[\mu_{G}(dt) = DG(t)dt\]
Proof. By the Fundamental Theorem of Calculus, for any $[c, d] \subset [a, b]$,
\[\int_{a}^{b} \one_{[c, d]}dG = \int_{a}^{b} \one_{[c, d]}DG(t)dt\]
By Definition 11.6.1, applied to both $G$ and $\text{Id}$,
\[\int_{[a, b]}\one_{[c, d]}d\mu_{G} = \int_{a}^{b} \one_{[c, d]}dG = \int_{a}^{b} \one_{[c, d]}DG(t)dt = \int_{[a, b]}\one_{[c, d]}DG(t)dt\]
Thus by linearity, for any regulated step map $f: [a, b] \to E$,
\[\int_{[a, b]}f(t) \mu_{G}(dt) = \int_{[a, b]}\dpn{f(t), DG(t)}{E}dt\]
By Definition 11.5.3, the regulated step maps are dense in $\text{Reg}([a, b], G; E)$. Therefore by the Linear Extension Theorem, the above holds for all $f \in \text{Reg}([a, b], G; E)$. By uniqueness of the Lebesgue-Stieltjes measure, $\mu_{G}(dt) = DG(t)dt$.$\square$