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:

\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.,:

\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:

\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 $[a, b]$.

- On the
**Riemann-Stieltjes Integrability of Continuous Functions with Integrators of Bounded Variation**page we saw that if $f$ is continuous on $[a, b]$ and $\alpha$ is of bounded variation on $[a, b]$ then $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$. This is an extremely useful theorem that has the normal Riemann integrals that we're most familiar with as a subfamily satisfying these conditions.

- On the
**Riemann Integrability of Continuous Functions and Functions of Bounded Variation**page we applied the theorem above to Riemann integrals. We saw that if $f$ is continuous on $[a, b]$ then $\int_a^b f(x) \: dx$ exists. We also saw that if $\alpha$ is of bounded variation on $[a, b]$ then $\int_a^b f(x) \: dx$ exists too.

- On the
**Riemann-Stieltjes Integrability of Functions on Subintervals with Integrators of Bounded Variation**page we saw that if $f$ is any function defined on $[a, b]$, $\alpha$ is of bounded variation on $[a, b]$, and $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$ then $f$ is also Riemann-Stieltjes integrable with respect to $\alpha$ on any subinterval $[c, d] \subseteq [a, b]$.

- 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:

\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:

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

- On
**The Second Mean-Value Theorem for Riemann-Stieltjes Integrals**page we looked at another Mean-Value theorem which says that if $f$ is an increasing function on $[a, b]$ and $\alpha$ is continuous on $[a, b]$ then there exists an $x_0 \in [a, b]$ such that:

\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}