Riemann-Stieltjes Integrals with Increasing Integrators Review

# Riemann-Stieltjes Integrals with Increasing Integrators Review

We will now review some of the recent content regarding Riemann-Stieltjes integrals with increasing integrators.

Let $f$ and $g$ be functions defined on $[a, b]$ and let $\alpha$ be an increasing function on $[a, b]$.

• On the Upper and Lower Riemann-Stieltjes Sums page we considered a special type of Riemann-Stieltjes sum. For every partition $P = \{ a = x_0, x_1, ..., x_n = b \} \in \mathscr{P}[a, b]$, for each $k \in \{1, 2, ..., n \}$ we defined the numbers $M_k(f)$ and $m_k(f)$ as:
(1)
\begin{align} \quad M_k (f) = \sup \{ f(x) : x \in [x_{k-1}, x-k] \} \quad \mathrm{and} \quad m_k(f) = \inf \{ f(x) : x \in [x_{k-1}, x_k] \} \end{align}
• Then we said that the Upper Riemann-Stieltjes Sum corresponding to the partition $P$ is:
(2)
\begin{align} \quad U(P, f, \alpha) = \sum_{k=1}^{n} M_k(f) \Delta \alpha_k \end{align}
• Similarly, we said that the Lower Riemann-Stieltjes Sum corresponding to the partition $P$ is:
(3)
\begin{align} \quad L(P, f, \alpha) = \sum_{k=1}^{n} m_k(f) \Delta \alpha_k \end{align}
• We noted that for all partitions $P$, that since $\alpha$ is an increasing function that then $\Delta \alpha_k \geq 0$, and for all $t_k \in [x_{k-1}, x_k]$, since $m_k(f) \leq f(t_k) \leq M_k(f)$, we see that:
(4)
\begin{align} \quad L(P, f, \alpha) \leq S(P, f, \alpha) \leq U(P, f, \alpha) \end{align}
(5)
\begin{align} \quad U(P', f, \alpha) \leq U(P, f, \alpha) \quad \mathrm{and} \quad \mathrm{and} L(P, f, \alpha) \leq L(P', f, \alpha) \end{align}
• We also noted that if $P_1, P_2 \in \mathscr{P}[a, b]$ are ANY partitions on $[a, b]$ then we always have that:
(6)
\begin{align} \quad L(P_1, f, \alpha) \leq U(P_2, f, \alpha) \end{align}
• We concluded that the set of upper Riemann-Stieltjes sums is bounded below by any lower Riemann-Stieltjes sum, and similarly, the set of all lower Riemann-Stieltjes sums is bounded above by any upper Riemann-Stieltjes sum.
(7)
\begin{align} \quad \overline{\int_a^b} f(x) \: d \alpha(x) = \inf \{ U(P, f, \alpha) : P \in \mathscr{P}[a, b] \} \end{align}
• Similarly, we defined the Lower Riemann-Stieltjes Integral of $f$ with respect to $\alpha$ on $[a, b]$ as:
(8)
\begin{align} \quad \underline{\int_a^b} f(x) \: d \alpha(x) = \sup \{ L(P, f, \alpha) : P \in \mathscr{P}[a, b] \} \end{align}
(9)
\begin{align} \quad \overline{\int_a^b} f(x) \: d \alpha (x) = \overline{\int_a^c} f(x) \: d \alpha (x) + \overline{\int_c^b} f(x) \: d \alpha (x) \end{align}
(10)
\begin{align} \quad \underline{\int_a^b} f(x) \: d \alpha (x) = \underline{\int_a^c} f(x) \: d \alpha (x) + \underline{\int_c^b} f(x) \: d \alpha (x) \end{align}
(11)
\begin{align} \quad \overline{\int_a^b} [f(x) + g(x)] \: d \alpha (x) \leq \overline{\int_a^b} f(x) \: d \alpha(x) + \overline{\int_a^b} f(x) \: d \alpha (x) \end{align}
(12)
\begin{align} \quad \underline{\int_a^b} [f(x) + g(x)] \: d \alpha (x) \geq \underline{\int_a^b} f(x) \: d \alpha (x) + \underline{\int_a^b} f(x) \: d \alpha (x) \end{align}
(13)
\begin{align} \quad U(P, f, \alpha) - L(P, f, \alpha) < \epsilon \end{align}
• We then proved that if Riemann's condition holds, then $\overline{\int_a^b} f(x) \: d \alpha(x) = \underline{\int_a^b} f(x) \: d \alpha (x)$, i.e., the upper and lower Riemann-Stieltjes integrals of $f$ with respect to $\alpha$ on $[a, b]$ are equal to each other. We proved that the equality of the upper and lower Riemann-Stieltjes integrals then implies the existence of the regular Riemann-Stieltjes integral of $f$ with respect to $\alpha$ on $[a, b]$ which completes the equivalence of these three statements.
(14)
\begin{align} \quad \int_a^b f(x) \: d \alpha(x) \leq \int_a^b g(x) \: d \alpha (x) \end{align}
(15)
\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) \end{align}
• With the theorem directly above, on The Product of Riemann-Stieltjes Integrable Functions with Increasing Integrators page we were able to show that the product $fg$ is also Riemann-Stieltjes integrable by noting that $f(x)g(x) = \frac{1}{2}[f(x)]^2 + \frac{1}{2}[g(x)]^2 - \frac{1}{2}[f(x) + g(x)]^2$ and noting that Riemann-Stieltjes integrals of all terms on the righthand side of this equation exist by previous theorems we've looked at.