why is the Lebesgue-Stieltjes integral well-defined? 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?
 A: 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: $\int fdg_{1}-\int fdg_{2}=\int fdg_{3}-\int fdg_{4}$.
