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:
\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:
\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:
\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:
\begin{align} \quad L(P, f, \alpha) \leq S(P, f, \alpha) \leq U(P, f, \alpha) \end{align}
- On the Properties of Upper and Lower Riemann-Stieltjes Sums page we looked at two very important properties. We noted that if $P'$ is finer than $P$ then:
\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:
\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.
- With this in mind, on the Upper and Lower Riemann-Stieltjes Integrals page we defined the Upper Riemann-Stieltjes Integral of $f$ with respect to $\alpha$ on $[a, b]$ as:
\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:
\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}
- On the Splitting Upper and Lower Riemann-Stieltjes Integrals page, using some basic properties of the supremum and infimum of a bounded set, we proved that if $c \in (a, b)$ then:
\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}
- On the Upper and Lower Riemann-Stieltjes Integrals of Sums of Integrands we also proved that:
\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}
- On the Riemann's Condition Part 1 - The Existence of Riemann-Stieltjes Integrals with Increasing Integrators and Riemann's Condition Part 2 - The Existence of Riemann-Stieltjes Integrals with Increasing Integrators pages we proved a very important theorem that has an extremely useful condition known as Riemann's Condition which provides a necessary condition for a function $f$ with an increasing integrator $\alpha$ to be Riemann-Stieltjes integrable on $[a, b]$. We proved that if $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$ then Riemann's condition is satisfied, that is, for all $\epsilon > 0$ there exists a $P_{\epsilon} \in \mathscr{P}[a, b]$ such that for any $P$ finer than $P_{\epsilon}$ we have that:
\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.
- On The Comparison Theorem for Riemann-Stieltjes Integrals with Increasing Integrators page we saw that if further $f(x) \leq g(x)$ for all $x \in [a, b]$, then we can compare the Riemann-Stieltjes integrals of $f$ and $g$, and:
\begin{align} \quad \int_a^b f(x) \: d \alpha(x) \leq \int_a^b g(x) \: d \alpha (x) \end{align}
- Using Riemann's condition, on The Absolute Value of Riemann-Stieltjes Integrals with Increasing Integrators page we saw that further $\mid f \mid$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$ and that:
\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}
- Also using Riemann's condition, on The Squares of Riemann-Stieltjes Integrable Functions with Increasing Integrators page we also saw that $f^2$ is then Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$
- 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.