R-S Integrability of Cts. Functions with Integrators of Bounded Variation

# Riemann-Stieltjes Integrability of Continuous Functions with Integrators of Bounded Variation

Recall from the Riemann-Stieltjes Integrals with Integrators of Bounded Variation page that if $f$ is a bounded function defined on $[a, b]$, $\alpha$ is a function of bounded variation on $[a, b]$ and $V(x) = V_{\alpha}(a, x)$ is the total variation function on $\alpha$ then if $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ then $f$ is Riemann-Stieltjes integrable with respect to $V$ and $V - f$.

We will now look at a whole class of functions that are Riemann-Stieltjes integrable - namely continuous integrands accompanied by integrators of bounded variation.

Theorem 1: Let $f$ be a continuous function on $[a, b]$ and let $\alpha$ be a function of bounded variation on $[a, b]$. Then $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$. |

**Proof:**Since $f$ is continuous on the closed bounded interval $[a, b]$ we have that $f$ is also bounded on $[a, b]$ by the Boundedness Theorem.

- Since $\alpha$ is of bounded variation on $[a, b]$ we have that $\alpha = V - (V - \alpha)$ where $V$ and $V - \alpha$ are increasing functions with respect to $V$ be the total variation function of $\alpha$. If we show that the theorem holds for increasing functions, then we have shown that it holds for $V$ and $V - \alpha$ and hence holds for any function of bounded variation.

- Assume that $\alpha$ is an increasing function. If $\alpha(a) = \alpha(b)$ then $\Delta \alpha_k = 0$ and the theorem is trivially true since $\alpha$ will be a constant integrator and we've already dealt with this case on the Riemann-Stieltjes Integrals with Constant Integrators page.

- Instead assume that $\alpha(a) < \alpha(b)$. Then $\alpha(b) - \alpha(a) > 0$. We will show that Riemann's condition holds under these hypotheses.

- Since $f$ is continuous on $[a, b]$ we have that $f$ is also uniformly continuous on $[a, b]$, i.e., for all $\epsilon > 0$ there exists a $\delta > 0$ such that if $\mid x - y \mid < \delta$ then $\mid f(x) - f(y) \mid < \epsilon$.

- So for $\epsilon_1 = \frac{\epsilon}{\alpha(b) - \alpha(a)} > 0$ there exists a $\delta_1 > 0$ such that if $\mid x - y \mid < \delta_1$ then:

\begin{align} \quad \mid f(x) - f(y) \mid < \epsilon_1 = \frac{\epsilon}{\alpha(b) - \alpha(a)} \quad \end{align}

- Now let $P \in \mathscr{P}[a, b]$ be any partition and consider the difference of the upper and lower Riemann-Stieltjes sums:

\begin{align} \quad U(P, f, \alpha) - L(P, f, \alpha) = \sum_{k=1}^{n} [M_k(f) - m_k(f)] \Delta \alpha_k \end{align}

- Since $f$ is continuous on $[a, b]$ it is also continuous on each subinterval $[x_{k-1}, x_k]$ of $[a, b]$. Furthermore, each of these subintervals $[x_{k-1}, x_k]$ are bounded and so there exists $t_k, t_k' \in [x_{k-1}, x_k]$ such that:

\begin{align} \quad M_k = \sup \{ f(x) : x \in [x_{k-1}, x_k] \} = f(t_k) \quad \mathrm{and} \quad m_k(f) = \inf \{ f(x) : x \in [x_{k-1}, x_k] \} = f(t_k') \end{align}

- Hence the difference of the upper and lower Riemann-Stieltjes sums can be written as:

\begin{align} \quad U(P, f, \alpha) - L(P, f, \alpha) = \sum_{k=1}^{n} [f(t_k) - f(t_k')] \Delta \alpha_k \end{align}

- Now make $P$ finer and finer such that $\| P \| < \delta$. Then the length of each subinterval will be less than $\delta$, i.e., $\mid x_k - x_{k-1} \mid < \delta$, and since $t_k, t_k' \in [x_{k-1}, x_k]$ we see that $\mid t_k - t_k' \mid < \delta$. Hence:

\begin{align} \quad \mid f(t_k) - f(t_k') \mid < \epsilon_1 = \frac{\epsilon}{\alpha(b) - \alpha(a)} \quad (*) \end{align}

- So choose $P_{\epsilon}$ such that $\| P_{\epsilon} \| < \delta$. Then for all $P$ finer than $P_{\epsilon}$ we have that $(*)$ holds and so:

\begin{align} \quad U(P, f, \alpha) - L(P, f, \alpha) &= \sum_{k=1}^{n} [M_k(f) - m_k(f)] \Delta \alpha_k \\ \quad &= \sum_{k=1}^{n} [f(t_k) - f(t_k')] \Delta \alpha_k \\ \quad &< \sum_{k=1}^{n} \epsilon_1 \Delta \alpha_k \\ \quad &< \sum_{k=1}^{n} \frac{\epsilon}{\alpha(b) - \alpha(a)} \Delta \alpha_k \\ \quad &< \frac{\epsilon}{\alpha(b) - \alpha(a)} \sum_{k=1}^{n} \Delta \alpha_k \\ \quad &< \frac{\epsilon}{\alpha(b) - \alpha(a)} [\alpha(b) - \alpha(a)] \\ \quad &< \epsilon \end{align}

- So for every $\epsilon > 0$ there exists a partition $P_{\epsilon} \in \mathscr{P}[a, b]$ such that if $P$ is finer than $P_{\epsilon}$ then $U(P, f, \alpha) - L(P, f, \alpha) < \epsilon$, so Riemann's condition is satisfied and $f$ is Riemann-Stieltjes integrable with respect to $\alpha$. $\blacksquare$