- #1
mahler1
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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.
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.