The Absolute Value of R-S Integrals with Increasing Integrators

The Absolute Value of Riemann-Stieltjes Integrals with Increasing Integrators

Theorem 1: Let $f$ be a function defined on $[a, b]$ and let $\alpha$ be an increasing function on $[a, b]$. Then if $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$ then $\mid f \mid$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$ and $\displaystyle{\biggr \lvert \int_a^b f(x) \: d \alpha (x) \biggr \rvert \leq \int_a^b \mid f(x) \mid \: d \alpha (x)}$.

• Proof: Since $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$ and $\alpha$ is an increasing function on $[a, b]$, we have that by Riemann's Condition that for all $\epsilon > 0$ that there exists a partition $P_{\epsilon}^* \in \mathscr{P}[a, b]$ such that for all partitions $P = \{ a = x_0, x_1, ..., x_n = b \} \in \mathscr{P}[a, b]$ finer than $P_{\epsilon}^*$ ($P_{\epsilon}^* \subseteq P$ we have that then:

(1)

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

• We note that:

(2)

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

• Recall that for any real numbers $a, b \in \mathbb{R}$ that $\mid \mid a \mid - \mid b \mid \mid \leq \mid a - b \mid$. Therefore:

(3)

\begin{align} \quad \mid \mid f(x) \mid - \mid f(x') \mid \mid \leq \mid f(x) - f(x') \mid \end{align}

• We note that $\sup \{ f(x) - f(x') : x, x' \in [x_{k-1}, x_k] \} > 0$, and so using the inequality above we see that $M_k(\mid f \mid) - m_k(\mid f \mid) \leq M_k (f) - m_k (f)$ $(*)$. Let $P_{\epsilon} = P_{\epsilon}^*$. Then for all partitions $P$ finer than $P_{\epsilon}$ we have that $(*)$ holds and so:

(4)

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

• So for all $\epsilon > 0$ there exists a partition $P_{\epsilon} \in \mathscr{P}[a, b]$ such that for all partitions $P$ finer than $P_{\epsilon}$ we have that $U(P, \mid f \mid, \alpha) - L(P, \mid f \mid, \alpha) < \epsilon$, so Riemann's condition is satisfied as $\mid f \mid$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$.

• Now note that for all $x \in [a, b]$ that $- \mid f(x) \mid \leq f(x) \leq \mid f(x) \mid$ (a proof can be found on The Triangle Inequality page). Therefore, by The Comparison Theorem for Riemann Stieltjes Integrals with Increasing Integrators, since $f$ and $\mid f \mid$ are Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$ we see that:

(5)

\begin{align} \quad \int_a^b - \mid f(x) \mid \: d \alpha (x) \leq \int_a^b f(x) \: d \alpha (x) \leq \int_a^b \mid f(x) \mid \: d \alpha (x) \\ \quad - \int_a^b \mid f(x) \mid \: d \alpha (x) \leq \int_a^b f(x) \: d \alpha (x) \leq \int_a^b \mid f(x) \mid \: d \alpha (x) \\ \end{align}

• The inequalities above can be combined to conclude that:

(6)

\begin{align} \quad \biggr \lvert \int_a^b f(x) \: d \alpha(x) \biggr \rvert \leq \int_a^b \mid f(x) \mid \: d \alpha (x) \quad \blacksquare \end{align}