Riemann Stieltjes-Integrals with Integrators of Bounded Variation

Riemann Stieltjes-Integrals with Integrators of Bounded Variation

Recall from the Functions of Bounded Variation page that if $f$ is a function defined on the closed interval $[a, b]$ and if $P = \{ a = x_0, x_1, ..., x_n = b \} \in \mathscr{P}[a, b]$, then the variation of $f$ associated with $P$ is:

(1)
\begin{align} \quad V_f(P) = \sum_{k=1}^{n} \mid f(x_k) - f(x_{k-1}) \mid \end{align}

If there exists an $M \in \mathbb{R}$, $M > 0$ such that $V_f(P) \leq M$ for all $P \in \mathscr{P}[a, b]$ then we said that $f$ is of bounded variation on $[a, b]$.

On the Total Variation of a Function page we said that if $f$ was of bounded variation then the least upper bound of the variation of $f$ ranging through all partitions $P \in \mathscr{P}[a, b]$ is the total variation of $f$ and we denote it:

(2)
\begin{align} \quad V_f(a, b) = \sup \{ V_f(P) : P \in \mathscr{P}[a, b] \} \end{align}

On the Decomposition of Functions of Bounded Variation as the Difference of Two Increasing Functions page we saw that if $f$ is of bounded variation on $[a, b]$ and we define the function for all $x \in [a, b]$ by $V(x) = V_f(a, x)$ then $f$ can be expressed as the difference of two increasing functions, namely:

(3)
\begin{align} \quad f = \underbrace{V}_{\mathrm{increasing}} - (\underbrace{V - f}_{\mathrm{increasing}}) \end{align}

This decomposition is not unique though. For any increasing function $g$ we have that $f = (V + g) - (V - f + g)$.

Now let $\alpha$ be any function of bounded variation. Then $\alpha = \alpha_1 - \alpha_2$ where $\alpha_1$ and $\alpha_2$ are increasing functions. If $f$ is Riemann-Stieltjes integrable with respect to $\alpha_1$ and $\alpha_2$ then by Linearity of the Integrator of Riemann-Stieltjes Integrals page we must have that $f$ is Riemann-Stieltjes integrable with respect to $\alpha$. We can of course use any of the important theorems regarding the existence of Riemann-Stieltjes integrals of functions of increasing integrators.

However, if $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ then $f$ need not be Riemann-Stieltjes integrable with respect to both $\alpha_1$ and $\alpha_2$.

The following theorem tells us that for the special decomposition $V$ and $V - \alpha$ that if $f$ is bounded on $[a, b]$, and Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$ then it is also Riemann-Stieltjes integrable with respect to $V$ (and $V - \alpha$).

Theorem 1: Let $f$ be a bounded function defined on $[a, b]$ and let $\alpha$ be a function of bounded variation on $[a, b]$. Furthermore let $V(x) = V_{\alpha}(a, x)$ be the total variation function on $\alpha$. Then if $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$ then $f$ is Riemann-Stieltjes integrable with respect to $V$ and $V - \alpha$ on $[a, b]$.

We will not prove Theorem 1 as it is rather lengthy to show.

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