The First Mean-Value Theorem for Riemann-Stieltjes Integrals

# The First Mean-Value Theorem for Riemann-Stieltjes Integrals

We will now look at a very useful theorem known as the First Mean-Value Theorem for Riemann-Stieltjes integrals. The proof is relatively simple too!

 Theorem 1 (The First Mean-Value Theorem for Rieman-Stieltjes Integrals): Let $f$ be a bounded function on $[a, b]$ and let $\alpha$ be an increasing function on $[a, b]$. Furthermore, let $f$ be Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$. If $M = \sup \{ f(x) : x \in [a, b] \}$ and $m = \inf \{ f(x) : x \in [a, b] \}$ then there exists a $c \in [m, M]$ such that $\displaystyle{\int_a^b f(x) \: d \alpha(x) = c[\alpha(b) - \alpha(a)]}$.
• Proof: Suppose that $\alpha(a) = \alpha(b)$. Then since $\alpha$ is increasing on $[a, b]$ we must have that $\alpha$ is constant on $[a, b]$ and so $\int_a^b f(x) \: d \alpha (x) = 0$. Furthermore, $\alpha(b) - \alpha(a) = 0$, so any $c \in [m, M]$ satisfies the equality.
• Instead suppose that $\alpha(a) < \alpha(b)$ and let $P = \{ a = x_0, x_1, ..., x_n = b \} \in \mathscr{P}[a, b]$ and consider the following Riemann-Stieltjes sum:
(1)
\begin{align} \quad S(P, f, \alpha) = \sum_{k=1}^{n} f(t_k) \Delta \alpha_k \end{align}
• Since $m = \inf \{ f(x) : x \in [a, b] \}$ and $M = \sup \{ f(x) : x \in [a, b] \}$ we see that $m \leq f(t_k) \leq M$ for all $t_k \in [x_{k-1}, x_k] \subseteq [a, b]$, and so:
(2)
\begin{align} \quad \sum_{k=1}^{n} m \Delta \alpha_k \leq \sum_{k=1}^{n} f(t_k) \Delta \alpha_k \leq \sum_{k=1}^{n} M \Delta \alpha_k \\ \quad m \sum_{k=1}^{n} \Delta \alpha_k \leq \sum_{k=1}^{n} f(t_k) \Delta \alpha_k \leq M \sum_{k=1}^{n} \Delta \alpha_k \end{align}
• But $\sum_{k=1}^{n} \Delta \alpha_k = \sum_{k=1}^{n} [\alpha(x_k) - \alpha(x_{k-1}) = \alpha(b) - \alpha(a)$ and so:
(3)
\begin{align} \quad m[\alpha(b) - \alpha(a)] \leq \sum_{k=1}^{n} f(t_k) \Delta \alpha_k \leq M[\alpha(b) - \alpha(a)] \end{align}
• This inequality holds for all partitions $P \in \mathscr{P}[a, b]$, and since $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$ we have that:
(4)
\begin{align} \quad m[\alpha(b) - \alpha(a)] \leq \int_a^b f(x) \: d \alpha (x) \leq M[\alpha(b) - \alpha(a)] \\ \quad m \leq \frac{1}{\alpha(b) - \alpha(a)} \int_a^b f(x) \: d \alpha(x) \leq M \end{align}
• Let $c = \frac{1}{\alpha(b) - \alpha(a)} \int_a^b f(x) \: d \alpha(x)$. Then $c \in [m, M]$ by the inequality above and it satisfies the following equation:
(5)
\begin{align} \quad \int_a^b f(x) \: d \alpha(x) = c[\alpha(b) - \alpha(a)] \quad \blacksquare \end{align}
 Corollary 1: Let $f$ be a continuous and bounded function on $[a, b]$ and let $\alpha$ be an increasing function on $[a, b]$. Furthermore, let $f$ be Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$. If $M = \sup \{ f(x) : x \in [a, b] \}$ and $m = \inf \{ f(x) : x \in [a, b] \}$ then there exists an $x_0 \in [a, b]$ such that $\displaystyle{\int_a^b f(x) \: d \alpha(x) = f(x_0)[\alpha(b) - \alpha(a)]}$.
• From the first Mean-Value theorem for Riemann-Stieltjes integrals we have that there exists a $c \in [m, M]$ such that:
(6)
\begin{align} \quad \int_a^b f(x) \: d \alpha (x) = c[\alpha(b) - \alpha(a) ]]. \end{align} • Sincef$is continuous on the closed bounded interval$[a, b]$we have that$M = \sup \{ f(x) : x \in [a, b] \} = f(x')$for some$x' \in [a, b]$and similarly$m = \inf \{ f(x) : x \in [a, b] \} = f(x'')$for some$x'' \in [a, b]$, so$c \in [f(x''), f(x')]$, i.e.,$f(x'') \leq c \leq f(x')$. • By the intermediate value theorem, since$f$is continuous there exists an$x_0 \in [a, b]$such that$f(x_0) = c\$ and so:
(7)
\begin{align} \quad \int_a^b f(x) \: d \alpha (x) = f(x_0) [\alpha(b) - \alpha(a)] \quad \blacksquare \end{align}
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