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-Proposition: Let $I \subset \mathbb{R} $ be an interval and $Z: I \to \mathbb{R}$ monotonically increasing and rightside continuous. If

  1. $I=(a,b)$ or $I=(a,b]$ then there exist a unique measure $m_Z$ on $\mathcal{B}(I)$ (Borel-$\sigma$-Algebra on $I$) such that $$m_Z((s,t])=Z(t)-Z(s)\;, s\leq t.$$
  2. $I=[a,b)$ 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 $m_Z([\tilde{a},a))>0$.

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    $\begingroup$ You've dropped something in your definition of the Lebesgue-Stieltjes integral. $m_Z$ is a set function, but in your definition, you are using it as a function on the elements of $I$. That makes no sense. $m_Z$ is a measure. You should know how Lebesgue measure with respect to an arbitrary measure is expressed. This is not it. $\endgroup$
    – Paul Sinclair
    Nov 7, 2021 at 2:15
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    $\begingroup$ Second answer: $m_{Z|_J}([a,t])\ne Z|_J(t)$. Instead $m_{Z|_J}([a,t])= Z|_J(t) - Z|_J(a)$ $\endgroup$
    – Paul Sinclair
    Nov 7, 2021 at 2:19
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    $\begingroup$ Both are bad notation, particularly the first, which, again, makes no sense at all. Concerning your other question. I admit I was mistaken. I had overlooked that $\sigma$-algebras are closed under complementation. The $\sigma$-algebra generated by $\{(s,t]\mid s,t \in I\}$ includes the sets $[x, b] = \bigcap_n (x-\frac 1n, b]$, and therefore $[a, x)$ for all $x\in (a,b)$. From there is easy to see that it contains all subintervals of $I$ and so must be $\mathcal B(I)$. Similar remarks occur for $\{(s,t)\mid s,t \in I\}$. $\endgroup$
    – Paul Sinclair
    Nov 7, 2021 at 13:57
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    $\begingroup$ I never said anything about $a - \frac 1n$. Note in fact that I only claimed to show $[x,b]$ is in the algebra for $x \in (a,b)$. $x = a$ is not included. However, notice that I started by pointing out my earlier mistake: $\sigma$-algebras are closed under complementation. Now. what is $I \setminus [x, b]$ when $I = [a,b]$? If your book uses $d\lambda(x)$, that is a bit bad, but also quite common. It misleads about what is going on with $\lambda$ and is only used because of its resemblance to Leibniz notation. But $m_Z(s)$ is much worse. It outright lies about the nature of this integration. $\endgroup$
    – Paul Sinclair
    Nov 7, 2021 at 15:07
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    $\begingroup$ There often is a function that could be used where you used $m_Z(s)$. That is the Rado-Nikodym derivative $\frac{dm_Z}{d\lambda}$. But that is not $m_Z$, and only exists for some $Z$ (I believe when $Z$ is continuous, but I may be forgetting something again). But the L.S. integral, which is actually defined by $\int_I f(x)\,dm_Z$ (and there even I use the $dm_Z$ notation despite its short-comings) exists for all $Z$ as described, not just the continuous ones. $\endgroup$
    – Paul Sinclair
    Nov 7, 2021 at 15:13

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