Riemann-Stieltjes Integrals with Integrators of Bounded Variation Review

# Riemann-Stieltjes Integrals with Integrators of Bounded Variation Review

We will now review some of the recent material regarding Riemann-Stieltjes integrals with integrators of bounded variation.

• Recall that a function $f$ is said to be of Bounded Variation on $[a, b]$ if there exists an $M \in \mathbb{R}$, $M > 0$ such that f or every partition $P = \{ a = x_0, x_1, ..., x_n = b \} \in \mathscr{P}[a, b]$ we have that the variation of $f$ associated with the partition $P$ is less than or equal to $M$, that is:
(1)
\begin{align} \quad V_f(P) = \sum_{k=1}^{n} \mid f(x_k) - f(x_{k-1}) \mid \leq M \end{align}
• If a function was of bounded variation on $[a, b]$ then we said that the Total Variation of $f$ on $[a, b]$ is the supremum (least upper bound) of the variations of $f$ associated with $P \in \mathscr{P}[a, b]$, i.e.,:
(2)
\begin{align} \quad V_f(a, b) = \sup \{ V_f(P) : P \in \mathscr{P}[a, b] \} \end{align}
• We also recalled that if $f$ is of bounded variation on $[a, b]$ then $f$ can be expressed as the difference of two increasing functions. If $V(x) = V_f(a, x)$ is the Total Variation Function of $f$ on $[a, b]$ then we saw that $V$ and $V - f$ are both increasing on $[a, b]$ and:
(3)
\begin{align} \quad f = V - (V - f) \end{align}
• We noted that this representation is not unique though since if $g$ is any increasing function on $[a, b]$ then $V + g$ and $V - f + g$ are also increasing functions and $f = (V + g) - (V - f + g)$.
• On the Riemann Stieltjes-Integrals with Integrators of Bounded Variation we saw that if $f$ is a bounded function on $[a, b]$, $\alpha$ is a function of bounded variation on $[a, b]$, and $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$ the $f$ is also Riemann-Stieltjes integrable with respect to $V$ and $V - f$ on $[a, b]$.
• We then looked at an important theorem on The First Mean-Value Theorem for Riemann-Stieltjes Integrals page which said that if $f$ is bounded on $[a, b]$, $\alpha$ is increasing on $[a, b]$, $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$ and $M = \sup \{ f(x) : x \in [a, b] \}$, $m = \inf \{ f(x) : x \in [a, b] \}$ then there exists a $c \in [m, M]$ such that:
(4)
\begin{align} \quad \int_a^b f(x) \: d \alpha (x) = c[\alpha(b) - \alpha(a)] \end{align}
• If $f$ is furthermore continuous then there exists an $x_0 \in [a, b]$ such that $f(x_0) = c$ and the formula above can be rewritten as:
(5)
\begin{align} \quad \int_a^b f(x) \: d \alpha (x) = f(x_0) [\alpha(b) - \alpha(a)] \end{align}
(6)
\begin{align} \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) \end{align}