The Second Mean-Value Theorem for Riemann-Stieltjes Integrals

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

Recall from The First Mean-Value Theorem for Riemann-Stieltjes Integrals page that if $f$ is a bounded function on $[a, b]$, $\alpha$ is an increasing function on $[a, b]$, and furthermore let $M = \sup \{ f(x) : x \in [a, b] \}$ and $m = \inf \{ f(x) : x \in [a, b] \}$.

If $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$ then there exists a $c \in [m, M]$ such that:

(1)
\begin{align} \quad \int_a^b f(x) \: d \alpha (x) = c[\alpha(b) - \alpha(a)] \end{align}

Furthermore, if $f$ is continuous on $[a, b]$ then there exists an $x_0 \in [a, b]$ such that $f(x_0) = c$ and so:

(2)
\begin{align} \quad \int_a^b f(x) \: d \alpha (x) = f(x_0)[\alpha(b) - \alpha(a)] \end{align}

We will now look at another theorem known as the second mean-value theorem for Riemann-Stieltjes integrals.

 Theorem 1 (The Second Mean-Value Theorem for Riemann-Stieltjes Integrals): Let $f$ be an increasing function on $[a, b]$ and let $\alpha$ be continuous on $[a, b]$. Then there exists a point $x_0 \in [a, b]$ such that $\displaystyle{\int_a^b f(x) \: d \alpha (x) = f(a) \int_a^{x_0} d \alpha(x) + f(b) \int_{x_0}^b \: d \alpha (x)}$.
• Proof: If $\alpha$ is a continuous function on $[a, b]$ and $f$ is an increasing function on $[a, b]$ then $f$ is of bounded variation on $[a, b]$ and so $\alpha$ is Riemann-Stieltjes integrable with respect to $f$ which implies that $f$ is Riemann-Stieltjes integrable with respect to $\alpha$. Using the The Formula for Integration by Parts of Riemann-Stieltjes Integrals and we have that:
(3)
\begin{align} \quad \int_a^b f(x) \: d \alpha(x) + \int_a^b \alpha(x) \: d f(x) &= f(b)\alpha(b) - f(a)\alpha(a) \\ \quad \int_a^b f(x) \: d \alpha(x) &= f(b)\alpha(b) - f(a)\alpha(a) - \int_a^b \alpha(x) \: d f(x) \end{align}
• Since $\alpha$ is a continuous function on the closed and bounded interval $[a, b]$ we must have that $\alpha$ is bounded on $[a, b]$. So by using the corollary to the First Mean-Value Theorem for Riemann-Stieltjes integrals we have that there exists an $x_0 \in [a, b]$ such that:
(4)
\begin{align} \quad \int_a^b \alpha(x) \: d f(x) = \alpha(x_0)[f(b) - f(a)] \end{align}
• Substituting this into the formula for the integration by parts and we get:
(5)
\begin{align} \quad \int_a^b f(x) \: d \alpha(x) &= f(b)\alpha(b) - f(a)\alpha(a) - \alpha(x_0)[f(b) - f(a)] \\ \quad \int_a^b f(x) \: d \alpha(x) &= f(b)\alpha(b) - f(a)\alpha(a) - f(b)\alpha(x_0) + f(a)\alpha(x_0) \\ \quad \int_a^b f(x) \: d \alpha (x) &= f(a)[\alpha(x_0) - \alpha(a)] + f(b)[\alpha(b) - \alpha(x_0)] \\ \quad \int_a^b f(x) \: d \alpha (x) &= f(a) \int_a^{x_0} \: d \alpha(x) + f(b) \int_{x_0}^b \: d \alpha (x) \quad \blacksquare \end{align}