Riemann's Condition Part 1

Riemann's Condition Part 1 - The Existence of Riemann-Stieltjes Integrals with Increasing Integrators

Recall that if $f$ is a function defined on $[a, b]$ and $\alpha$ is an increasing function on $[a, b]$ then for any partition $P = \{ a = x_0, x_1, ..., x_n = b \} \in \mathscr{P}[a, b]$ we have that:

(1)
\begin{align} \quad L(P, f, \alpha) \leq S(P, f, \alpha) \leq U(P, f, \alpha) \end{align}

As $P$ gets finer and $\| P \| \to 0$ we noted that $\displaystyle{L(P, f, \alpha) \to \underline{\int_a^b} f(x) \: d \alpha (x)}$ and $\displaystyle{U(P, f, \alpha) \to \overline{\int_a^b} f(x) \: d \alpha (x)}$. Since $L(P, f, \alpha) \leq S(P, f, \alpha) \leq U(P, f, \alpha)$ for all partitions $P$, we might expect a sort of "Squeeze theorem" for the existence of the Riemann-Stieltjes integral of $f$ with respect to $\alpha$ on $[a, b]$ provided that the difference $U(P, f, \alpha) - L(P, f, \alpha)$ can be made arbitrarily small. Fortunately, the extremely important Riemann's condition asserts just this.

The theorem below is rather long so we will prove that $a) \implies b)$ on this page and on the Riemann's Condition Part 2 - The Existence of Riemann-Stieltjes Integrals with Increasing Integrators page we will prove that $b) \implies c)$ and $c) \implies a)$.

 Theorem 1 (Riemann's Condition): Let $f$ be a function defined on $[a, b]$ and let $\alpha$ be an increasing function on $[a, b]$. Then the following statements are equivalent: a) $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$. b) For every $\epsilon > 0$ there exists a partition $P_{\epsilon}$ such that if $P$ is finer then $P_{\epsilon}$ ($P_{\epsilon} \subseteq P$) then $0 \leq U(P, f, \alpha) - L(P, f, \alpha) < \epsilon$ (Riemann's Condition). c) $\displaystyle{\overline{\int_a^b} f(x) \: d \alpha(x) = \underline{\int_a^b} f(x) \: d \alpha (x)}$.

It is important to remember that the Theorem above is guaranteed to hold when $\alpha$ is an increasing function!

Note that since $L(P, f, \alpha) \leq U(P, f, \alpha)$ that then $U(P, f, \alpha) - L(P, f, \alpha) > 0$ for all partitions $P \in \mathscr{P}[a, b]$.

Also note that Riemann's Condition provides a link between $f$ being Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$ and the upper and lower Riemann-Stieltjes integrals of $f$ with respect to $\alpha$ on $[a, b]$ equalling each other.

• Proof of $a) \implies b)$: Suppose that $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$. Then for some $A \in \mathbb{R}$ we have that:
(2)
\begin{align} \quad \int_a^b f(x) \: d \alpha (x) = A \end{align}
• Let $\epsilon > 0$ be given. Since $\alpha$ is an increasing function we have that since $a < b$ that then either $\alpha(a) = \alpha (b)$ or $\alpha(a) < \alpha(b)$.
• First consider the case when $\alpha (a) = \alpha (b)$. Since $\alpha$ is increasing on $[a, b]$ this implies that $\alpha$ is constant on $[a, b]$ and so for any partition $P = \{ a = x_0, x_1, ..., x_n = b \} \in \mathscr{P}[a, b]$ we have that $\Delta \alpha_k = \alpha(x_k) - \alpha(x_{k-1}) = 0$ for all $k \in \{1, 2, ..., n \}$. Moreover, for any partition $P$ we have that $U(P, f, \alpha) = 0$ and $L(P, f, \alpha) = 0$, so Riemann's condition is satisfied trivially since for any partition $P$, $0 = U(P, f, \alpha) - L(P, f, \alpha) < \epsilon$.
• For the second suppose, suppose tht $\alpha (a) < \alpha (b)$. Then since $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$ we have that for $\epsilon_1 = \frac{\epsilon}{3} > 0$ there exists a partition $P_{\epsilon_1} \in \mathscr{P}[a, b]$ such that if $P$ is finer than $P_{\epsilon_1}$ ($P_{\epsilon_1} \subseteq P$) and for any choices of $t_k, t_k' \in [x_{k-1}, x_k]$ we have that:
(3)
\begin{align} \quad \biggr \lvert \sum_{k=1}^{n} f(t_k) \Delta \alpha_k - A \biggr \rvert < \epsilon_1 = \frac{\epsilon}{3} \quad \mathrm{and} \quad \biggr \lvert \sum_{k=1}^{n} f(t_k') \Delta \alpha_k - A \biggr \rvert < \epsilon_1 = \frac{\epsilon}{3} \end{align}
• Using some properties of the absolute value we see that:
(4)
\begin{align} \quad \biggr \lvert \sum_{k=1}^{n} [f(t_k) - f(t_k')] \Delta \alpha_k \biggr \rvert = \biggr \lvert \sum_{k=1}^{n} f(t_k) - A - \left ( \sum_{k=1}^{n} f(t_k)\Delta \alpha_k - A \right ) \biggr \rvert \leq \biggr \lvert \sum_{k=1}^{n} f(t_k) \Delta \alpha_k - A \biggr \rvert + \biggr \lvert \sum_{k=1}^{n} f(t_k') \Delta \alpha_k - A \biggr \rvert & < \epsilon_1 + \epsilon_1 = \frac{\epsilon}{3} + \frac{\epsilon}{3} = \frac{2\epsilon}{3} \\ \end{align}
• Therefore $\biggr \lvert \sum_{k=1}^{n} [f(t_k) - f(t_k')] \Delta \alpha_k \biggr \rvert < \frac{2\epsilon}{3}$. $(*)$
• We want to somehow introduce the difference of upper and lower Riemann-Stieltjes sums into our proof. We first note that:
(5)
\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}
• So for all $x, x' \in [x_{k-1}, x_k]$ we have from the definition of the supremum that:
(6)
\begin{align} \quad M_k(f) - m_k(f) = \sup \{ f(x) - f(x') : x, x' \in [a, b] \} \geq f(x) - f(x') \end{align}
• So for all $h > 0$ there exists $t_k, t_k' \in [x_{k-1}, x_k]$ such that:
(7)
\begin{align} \quad M_k(f) - m_k(f) - h < f(x) - f(x') \\ \quad M_k(f) - m_k(f) < f(x) - f(x') + h \end{align}
• Choose $t_k, t_k' \in [x_{k-1}, x_k]$ such that for $h = \frac{\epsilon}{3[\alpha(b) - \alpha(a)]} > 0$, the inequality above holds. Let $P_{\epsilon} = P_{\epsilon_1}$. Then for $P$ finer than $P_{\epsilon}$ we have that $(*)$ holds and:
(8)
\begin{align} \quad U(P, f, \alpha) - L(P, f, \alpha) &= \sum_{k=1}^{n} M_k(f) \Delta \alpha_k - \sum_{k=1}^{n} m_k(f) \Delta \alpha_k \\ \quad &= \sum_{k=1}^{n} [M_k(f) - m_k(f)] \Delta \alpha_k \\ \quad & < \sum_{k=1}^{n} [f(t_k) - f(t_k') + h] \Delta \alpha_k \\ \quad & < \sum_{k=1}^{n} [f(t_k) - f(t_k')] \Delta \alpha_k + \sum_{k=1}^n h \Delta \alpha_k \\ \quad & < \frac{2\epsilon}{3} + h \sum_{k=1}^{n} \Delta \alpha_k \\ \quad & < \frac{2\epsilon}{3} + h [\alpha(b) - \alpha(a)] \\ \quad & < \frac{2\epsilon}{3} + \frac{\epsilon}{3[\alpha(b) - \alpha(a)]} \\ \quad & < \frac{2\epsilon}{3} + \frac{\epsilon}{3} \\ \quad & < \epsilon \end{align}