Let $Z$ be a directed cash-flow (i.e. right-continuous, monotone increasing function $Z:[0,\infty) \rightarrow [0,\infty)$, furthermore let Z be absolutely continuous (i.e there exists a measurable function $g:[0,\infty) \rightarrow [0,\infty)$ such that $Z(t) = \int_{0}^{t} z(s) ds$ , $t \geq 0$.
Then, show that for every measurable function $g:[0,\infty) \rightarrow [0,\infty)$ holds $\int_{0}^{t} g(s) dZ(s)=\int_{0}^{t} g(s)z(s)ds$.
Can someone help me to solve this?
$\int_0^{t} I_{(a,b]} dZ(s)=Z(\min \{b,t\})-Z(a)=\int_0^{t}I_{(a,b]} z(s)ds$ so the equation holds when $g=I_{(a,b]}$. Hence it holds for $g=I_A$ where $A$ is a finite disjoint union of half closed intervals. Now $\int_0^{t} I_A dZ(s)$ and $\int_0^{t} I_A z(s)\, ds$ are both measures. Since they coincide on the algebra of finite disjoint union of half closed intervals they coincide on on Borel sets. Hence the equation golds whenever $g$ is a an indicator function. Take linear combinations to prove it for simple functions and then apply monotone convergence theorem to prove it for all measurable functions $g:[0,\infty) \to [0,\infty)$.