Dtyjlui

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?

Let $g:[a,b]rightarrowmathbb{R}$ 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 $mu_{i}:mathcal{B}((a,b])rightarrowmathbb{R}$ be a Borel
measure induced by $g_{i}$ such that for any $aleq c<dleq b$,
$mu_{i}left((c,d]right)=g_{i}(d)-g_{i}(c)$. We go to show that
$int fdmu_{1}-int fdmu_{2}=int fdmu_{3}-int fdmu_{4}$ for
any bounded Borel function $f:(a,b]rightarrowmathbb{R}$. Define
Borel measures $nu=mu_{1}+mu_{4}$ and $nu'=mu_{2}+mu_{3}$.
Note that $nu$ and $nu'$ are finite measures. Let $mathcal{P}={(c,d]mid aleq c<dleq b}cup{emptyset}$
and $mathcal{L}={Ainmathcal{B}left((a,b]right)midnu(A)=nu'(A)}$.
Clearly $mathcal{P}$ is a $pi$-class (in the sense that $Acap Binmathcal{P}$
whenever $A,Binmathcal{P}$) and $mathcal{L}$ is a $lambda$-class
(in the sense that: $emptysetinmathcal{L}$; $A^{c}inmathcal{L}$
whenever $Ainmathcal{L}$; $cup_{n=1}^{infty}A_{n}inmathcal{L}$
whenever $A_{1},A_{2}ldotsinmathcal{L}$ are pairwisely disjoint.).
It is routine to show that $mathcal{P}subseteqmathcal{L}$. For,
let $(c,d]inmathcal{P}$, 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
$mathcal{P}subseteqmathcal{L}$. By Dynkin's $pi-lambda$ theorem,
we have $sigma(mathcal{P})subseteqmathcal{L}$. 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 $Ainmathcal{B}((a,b])$,
we have
begin{eqnarray*}
int fdmu_{1}+int fdmu_{4} & = & mu_{1}(A)+mu_{4}(A)\
& = & nu(A)\
& = & nu'(A)\
& = & int fdmu_{2}+int fdmu_{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 $0leq f_{1}leq f_{2}leqldotsleq f$
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 $int fdmu_{1}-int fdmu_{2}=int fdmu_{3}-int fdmu_{4}$
or in the Stieltjes notation: $int fdg_{1}-int fdg_{2}=int fdg_{3}-int fdg_{4}$.