Riemann-Stieltjes Integral of Constant Integrand

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Theorem

Let $\alpha$ be a real function that is bounded on $\closedint a b$.

Let $f : \closedint a b \to \R$ be defined as:

$\map f x = c$

for some $c \in \R$.


Then, $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $\closedint a b$, and:

$\ds \int_a^b f \rd \alpha = c \map \alpha b - c \map \alpha a$


Proof

Let $\epsilon > 0$

Define $P_\epsilon = \set {a, b}$.

Let $P = \set {x_0, \dotsc, x_n}$ be a subdivision of $\closedint a b$ that is finer than $P_\epsilon$.

Then:

\(\ds \map S {P, f, \alpha}\) \(=\) \(\ds \sum_{k \mathop = 1}^n \map f {t_k} \paren {\map \alpha {x_k} - \map \alpha {x_{k - 1} } }\) Definition of Riemann-Stieltjes Sum
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^n \paren {c \map \alpha {x_k} - c \map \alpha {x_{k - 1} } }\) Definition of $f$
\(\ds \) \(=\) \(\ds c \map \alpha {x_n} - c \map \alpha {x_0}\) Telescoping Series
\(\ds \) \(=\) \(\ds c \map \alpha b - c \map \alpha a\) Definition of Finite Subdivision

Therefore:

\(\ds \size {\map S {P, f, \alpha} - \paren {c \map \alpha b - c \map \alpha a} }\) \(=\) \(\ds \size {\paren {c \map \alpha b - c \map \alpha a} - \paren {c \map \alpha b - c \map \alpha a} }\) Above
\(\ds \) \(=\) \(\ds 0\)
\(\ds \) \(<\) \(\ds \epsilon\) Definition of $\epsilon$

Therefore, by definition of the Riemann-Stieltjes integral:

$\ds \int_a^b f \rd \alpha = c \map \alpha b - c \map \alpha a$

$\blacksquare$


Sources

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