Two relations between bounded variation and Riemann-Stieltjes integral

[mahler1]

I am reading Apostol's section on Riemann-Stieltjes integral and I have doubts on one statement:

Let $α$ be a function of bounded variation on $[a,b]$ and suppose $f \in R(α)$ on $[a,b]$. We define $F$ as $F(x)=\int_a^x f(x)dα$ if $x \in [a,b]$, then $F$ is a function of bounded variation on $[a,b]$.

He proves this statement but in his proof he says: 'it's sufficient if we assume $α$ is monotone increasing on $[a,b]$.

The proof (assuming $α$ is monotone increasing) is short and simple:
First he uses a theorem he proves in a previous page which says: Let$α$ is monotone increasing and $f \in R(α)$ on $[a,b]$. If $m$ and $M$ are the sup and inf of the set $\{f(x): x \in [a,b]\}$, then there exists $c$ with $m\leq c \leq M$ such that:

$\int_a^b f(x)dα(x)=c\int_a^b dα(x)=c[α(b)-α(a)]$.

Using this, we have $F(y)-F(x)=\int_x^y fdα=c[α(y)-α(x)]$ and from here one can deduce that $F$ is of bounded variation.

I understood this proof, but I don't understand why it is sufficient to prove that the statement holds for the case $α$ monotone increasing. I don't see why it would have to hold for an arbitrary function $α$ of bounded variation such that $f \in R(α)$ if we've already proved it for $α$ monotone increasing.

 

[R136a1]

What is the relation between montone increasing functions and functions of bounded variation?

 

[mahler1]

What is the relation between montone increasing functions and functions of bounded variation?

Well, he says it is sufficient to prove the statement holds in the case $α$ is a monotone increasing function. I don't see why if the statement holds for this case, then one can deduce the statement holds for an arbitrary $α$ of bounded variation (not necessarily monotone increasing).

Oh, and in my title says 'two relations...' but it's actually one statement/relation

 

[R136a1]

Well, he says it is sufficient to prove the statement holds in the case $α$ is a monotone increasing function. I don't see why if the statement holds for this case, then one can deduce the statement holds for an arbitrary $α$ of bounded variation (not necessarily monotone increasing)

Are all functions of bounded variation also monotone increasing? If the answer is no, can you represent function of bounded variation in some way using monotone increasing functions?

 

[mahler1]

Are all functions of bounded variation also monotone increasing? If the answer is no, can you represent function of bounded variation in some way using monotone increasing functions?

Oh, I could express $α=V_a^x-(V_a^x-α)$, which are two monotone increasing functions and the case reduces again to monotone increasing functions.

 

[R136a1]

Oh, I could express $α=V_a^x-(V_a^x-α)$, which are two monotone increasing functions and the case reduces again to monotone increasing functions.

Yes. This is called the Jordan decomposition and is very important in the study of bounded variation. Using the Jordan decomposition, do you see why Apostols statement is true?

 

[mahler1]

Yes. This is called the Jordan decomposition and is very important in the study of bounded variation. Using the Jordan decomposition, do you see why Apostols statement is true?

Just in case I want to check:

I suppose that $\int_a^x f(x)dα=\int_a^x f(x)dV_a^x-\int_a^xf(x)d(V_a^x-α)$ and since both $V$ and $V-α$ are monotone increasing on $[a,b]$, then both are of bounded variation. If I call $g(x)=\int_a^x f(x)dα=\int_a^x f(x)dV_a^x$ and $h(x)=\int_a^xf(x)d(V_a^x-α)$, then these functions are of bounded variation, and as $F(x)=g(x)-h(x)$, then $F$ is also of bounded variation.
I said I suppose, but I don't have totally clear if $\int_a^x f(x)dα=\int_a^x f(x)dV_a^x-\int_a^xf(x)d(V_a^x-α)$

 

[R136a1]

That's correct. For the thing you're unsure of, you should have seen somewhere in Apostol that

\int h d(\alpha + \beta) = \int hd\alpha + \int h d\beta

 

[mahler1]

That's correct. For the thing you're unsure of, you should have seen somewhere in Apostol that

\int h d(\alpha + \beta) = \int hd\alpha + \int h d\beta

You've said so, I should. Now I'll look up in Apostol. Thank you!