Riemann-Stieltjes Integral by Norm of Subdivision
Theorem
Let $f, \alpha$ be real functions that are bounded on $\closedint a b$.
Suppose there exists some $A \in \R$ where, for every $\epsilon > 0$, there exists some $\delta_\epsilon > 0$ such that:
- For every finite subdivision $P$ of $\closedint a b$, if the norm $\norm P < \delta_\epsilon$, then:
- For every Riemann-Stieltjes sum $\map S {P, f, \alpha}$:
- $\size {\map S {P, f, \alpha} - A} < \epsilon$
- For every Riemann-Stieltjes sum $\map S {P, f, \alpha}$:
Then, $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $\closedint a b$ and:
- $\ds \int_a^b f \rd \alpha = A$
Corollary
Let $f$ be a real function that is bounded on $\closedint a b$.
Suppose $f$ is Riemann integrable on $\closedint a b$.
Let $\iota$ be the identity mapping on $\closedint a b$.
Then, $f$ is Riemann-Stieltjes integrable with respect to $\iota$ on $\closedint a b$ and:
- $\ds \int_a^b f \rd \iota = \int_a^b \map f x \rd x$
where the integral on the right is understood to denote the Riemann integral.
Proof
Let $\epsilon > 0$ be arbitrary.
By hypothesis, let $\delta_\epsilon > 0$ be such that, for every subdivision $P$ with $\norm P < \delta_\epsilon$:
- $\size {\map S {P, f, \alpha} - A} < \epsilon$
By Existence of Subdivision with Small Norm, let $P_\epsilon$ be a subdivision of $\closedint a b$ such that:
- $\norm {P_\epsilon} < \delta_\epsilon$
Let $P$ be a subdivision that is finer than $P_\epsilon$.
Then:
\(\ds \norm P\) | \(\le\) | \(\ds \norm {P_\epsilon}\) | Norm of Refinement is no Greater than Norm of Subdivision | |||||||||||
\(\ds \) | \(<\) | \(\ds \delta_\epsilon\) | Definition of $P_\epsilon$ |
Thus, by hypothesis:
- $\size {\map S {P, f, \alpha} - A} < \epsilon$
But, since $P$ finer than $P_\epsilon$ and $\epsilon > 0$ were arbitrary:
- $\ds \int_a^b f \rd \alpha = A$
by definition of the Riemann-Stieltjes integral.
$\blacksquare$
Sources
- 1974: Tom M. Apostol: Mathematical Analysis (2nd ed.): Chapter $7$ The Riemann-Stieltjes Integral: Exercise $7.3$