-Proposition: Let $I \subset \mathbb{R} $ be an interval and $Z: I \to \mathbb{R}$ monotonically increasing and rightside continuous. If
- $I=(a,b)$ or $I=(a,b]$ then there exist a unique measure $m_Z$ on $\mathcal{B}(I)$ (Borel-$\sigma$-Algebra on $I$) such that $$m_Z((s,t])=Z(t)-Z(s)\;, s\leq t.$$
- $I=[a,b)$ or $I=[a,b]$ and $Z(a)\geq0$ then there exist a unique measure $m_Z$ on $\mathcal{B}(I)$ such that $$m_Z([a,t])=Z(t)\;, t\geq a$$
-First Question: Is it correct that $\{(s,t]|s\leq t, s,t \in I\}$ induces the $\sigma$-algebra $\mathcal{B}(I)$ in the case that $I=(a,b)$ or $I=(a,b]$ but not if $I=[a,b)$ or $I=[a,b]$?
-Lebesgue-Stieltjes integral: For a $\mathcal{B}(I)$-measurable function $f:I \to \mathbb{R}$ we can define the Lebesgue-stieltjes integral as $$\int_I f(s) dZ(s):=\int_I f(s) m_Z(s) ds$$ if the Lebesgue integral on the right hand side is defined.
-2.Question: Sometimes we want to integrate over a subinterval $J\subset I$. From the defintion of the Lebesgue integral this is defined as $\int_J f(s) dZ(s):=\int_J f|_J(s) \;m_Z|_{\mathcal{B}(J)}(s)$ (or equivalently: $\int_J f(s) dZ(s):=\int_I f(s) \mathbb{1}_J\;m_Z(s)$). But do we obtain the same result if we instead consider the measure $m_{Z|_J}$ on $\mathcal{B}(J)$ that is induced by the restriction $Z|_J: J \to \mathbb{R}$, i.e $$\int_J f|_J(s) dZ|_J(s)\;\;\quad?$$ Equivalently I can formulate the question as follows: Is $m_{Z|_J}=m_Z|_{\mathcal{B}(J)}$ on $\mathcal{B}(J)$?
-Considerations for the 2. Question: This is true if $J=(a,b)$ or $J=(a,b]$ no matter what form $I$ has: $$m_{Z|_J}((s,t])= Z|_J(t)-Z|J(s)=Z(t)-Z(s)=m_Z((s,t])=m_Z|_{\mathcal{B}(J)}((s,t]).$$ But if $J=[a,b]$ or $J=[a,b)$ this isn't generally true: Let for example $I=[\tilde{a},\tilde{b}]$ then $$m_{Z|_J}([a,t])= Z|_J(t)=Z(t)=m_Z([\tilde{a},t])=m_Z([\tilde{a},a))+m_Z([a,t])=m_Z([\tilde{a},a))+m_Z|_{\mathcal{B}(J)}([a,t]) \neq m_Z|_{\mathcal{B}(J)}([a,t])$$ if $m_Z([\tilde{a},a))>0$.