A function $g: [a,b] \rightarrow \mathbb{R}$ a said to be of bounded variation on the interval $[a,b]$ if
$$ \sup_{P: a=x_0 < x_1 \ldots < x_i < \ldots < x_{n_P}=b} \sum_{i=1}^{n_P} |g(x_{i}) -g(x_{i-1}))| < \infty, $$
where the supremum if taken over all partitions $P$ of the interval $[a,b]$.
One can show that a (right-continous) function $g: [a,b] \rightarrow \mathbb{R}$ is of bounded variation if and only if there exist two monotone non-decreasing (right-continuous) functions $g^+$ and $g^-$ such that $g= g^+ -g^-$. (However, this decomposition is not unique, as one can for example add any monotone non-decreasing (right-continuous) function to both $g^+$ and $g^-$). This is called a Jordan decomposition of $g$.
Moreover, given any $g: [a,b] \rightarrow \mathbb{R}$ which is monotone, non-decreasing and right-continuous function, there exists a unique measure $dg$ on $[a,b]$ such that
$$ dg((c,d])=g(d)-g(c) $$
for all $(c,d] \in [a,b]$ and $dg(\{a\})=0$. This measure is called the Lebesgue-Stieltjes measure of $g$.
Given a bounded function $f: [a,b] \rightarrow \mathbb{R}$ and a right-continuous function of bounded variation $g : [a,b] \rightarrow \mathbb{R}$, one can define
$$ \int_{[a,b]} f\,dg := \int_{[a,b]} f\,dg^+ – \int_{[a,b]} f\,dg^-, $$
where $g= g^+ -g^-$ is a Jordan decomposition as above. This is called the Lebesgue-Stieltjes integral of $f$ w.r.t. $g$.
My question is, why is $\int_{[a,b]} f\,dg$ well-defined, i.e. independent of the chosen Jordan decomposition?
Best Answer
Let $g:[a,b]\rightarrow\mathbb{R}$ be right-continuous and of bonded variation. Suppose that we have decompositions $g=g_{1}-g_{2}=g_{3}-g_{4}$ for some right-continuous, increasing (non-strict sense) functions $g_{1},g_{2},g_{3},g_{4}$ defined on $[a,b]$. For each $i=1,2,3,4$, let $\mu_{i}:\mathcal{B}((a,b])\rightarrow\mathbb{R}$ be a Borel measure induced by $g_{i}$ such that for any $a\leq c<d\leq b$, $\mu_{i}\left((c,d]\right)=g_{i}(d)-g_{i}(c)$. We go to show that $\int fd\mu_{1}-\int fd\mu_{2}=\int fd\mu_{3}-\int fd\mu_{4}$ for any bounded Borel function $f:(a,b]\rightarrow\mathbb{R}$. Define Borel measures $\nu=\mu_{1}+\mu_{4}$ and $\nu'=\mu_{2}+\mu_{3}$. Note that $\nu$ and $\nu'$ are finite measures. Let $\mathcal{P}=\{(c,d]\mid a\leq c<d\leq b\}\cup\{\emptyset\}$ and $\mathcal{L}=\{A\in\mathcal{B}\left((a,b]\right)\mid\nu(A)=\nu'(A)\}$. Clearly $\mathcal{P}$ is a $\pi$-class (in the sense that $A\cap B\in\mathcal{P}$ whenever $A,B\in\mathcal{P}$) and $\mathcal{L}$ is a $\lambda$-class (in the sense that: $\emptyset\in\mathcal{L}$; $A^{c}\in\mathcal{L}$ whenever $A\in\mathcal{L}$; $\cup_{n=1}^{\infty}A_{n}\in\mathcal{L}$ whenever $A_{1},A_{2}\ldots\in\mathcal{L}$ are pairwisely disjoint.). It is routine to show that $\mathcal{P}\subseteq\mathcal{L}$. For, let $(c,d]\in\mathcal{P}$, then \begin{eqnarray*} \mu_{1}(c,d]-\mu_{2}(c,d] & = & \left[g_{1}(d)-g_{1}(c)\right]-\left[g_{2}(d)-g_{2}(c)\right]\\ & = & \left[g_{1}(d)-g_{2}(d)\right]-\left[g_{1}(c)-g_{2}(c)\right]\\ & = & g(d)-g(c)\\ & = & \left[g_{3}(d)-g_{4}(d)\right]-\left[g_{3}(c)-g_{4}(c)\right]\\ & = & \left[g_{3}(d)-g_{3}(c)\right]-\left[g_{4}(d)-g_{4}(c)\right]\\ & = & \mu_{3}(c,d]-\mu_{4}(c,d]. \end{eqnarray*} Re-arranging terms, we have $\nu(c,d]=\nu'(c,d]$. This shows that $\mathcal{P}\subseteq\mathcal{L}$. By Dynkin's $\pi-\lambda$ theorem, we have $\sigma(\mathcal{P})\subseteq\mathcal{L}$. However, it is well known that $\sigma(\mathcal{P})=\mathcal{B}((a,b])$, so we have $\sigma(\mathcal{P})=\mathcal{L=\mathcal{B}}(a,b])$.
For a Borel function of the form $f=1_{A}$, where $A\in\mathcal{B}((a,b])$, we have \begin{eqnarray*} \int fd\mu_{1}+\int fd\mu_{4} & = & \mu_{1}(A)+\mu_{4}(A)\\ & = & \nu(A)\\ & = & \nu'(A)\\ & = & \int fd\mu_{2}+\int fd\mu_{3}. \end{eqnarray*} By linearlity, the above holds for all simple functions $f$. If $f$ is a non-negative, bounded Borel function, we may choose a sequence of simple functions $(f_{n})$ such that $0\leq f_{1}\leq f_{2}\leq\ldots\leq f$ and $f_{n}\rightarrow f$ pointwisely. By monotone convergence theorem, the above inequlity holds for all non-negative, bounded Borel functions. Finally, if $f$ is a bounded Borel function, consider $f=f^{+}-f^{-}$ and we can prove that the above also holds for bounded Borel function $f$. Finally, re-arrange terms, we have $\int fd\mu_{1}-\int fd\mu_{2}=\int fd\mu_{3}-\int fd\mu_{4}$ or in the Stieltjes notation: $\int fdg_{1}-\int fdg_{2}=\int fdg_{3}-\int fdg_{4}$.