The Nonexistence of Certain R-S Integrals with Increasing Integrators

The Nonexistence of Certain Riemann-Stieltjes Integrals with Increasing Integrators

Recall from 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 that $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$ if Riemann's condition is satisifed, i.e., for all $\epsilon > 0$ there exists a partition $P_{\epsilon} \in \mathscr{P}[a, b]$ such that if $P$ is finer than $P_{\epsilon}$ ($P_{\epsilon} \subseteq P$ then:

(1)

\begin{align} \quad U(P, f, \alpha) - L(P, f, \alpha) < \epsilon \end{align}

Furthermore, we proved that if Riemann's condition is satisfied, then the upper Riemann-Stieltjes integral of $f$ with respect to $\alpha$ on $[a, b]$ equals the lower Riemann-Stieltjes integral of $f$ with respect to $\alpha$ on $[a, b]$:

(2)

\begin{align} \quad \overline{\int_a^b} f(x) \: d \alpha (x) = \underline{\int_a^b} f(x) \: d \alpha (x) \end{align}

We then showed that the equality of the upper and lower Riemann-Stieltjes integrals of $f$ with respect to $\alpha$ on $[a, b]$ implies that $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$ which completes our circle of equivalent statements.

In general, we can use Riemann's condition to show that certain Riemann-Stieltjes integrals exists, however, it is often times simpler to show that certain integrals do not exist.

For example, consider the function $f(x) = \left\{\begin{matrix} 0 & \mathrm{if} \: x \: \mathrm{is \: irrational}\\ 1 & \mathrm{if} \: x \: \mathrm{is \: rational} \end{matrix}\right.$ and the Riemann-Stieltjes integral $\int_a^b f(x) \: dx$ where $a, b \in \mathbb{R}$ and $a < b$. Note that $\alpha (x) = x$ which is an increasing function.

If $P = \{ a = x_0, x_1, ..., x_n = b \} \in \mathscr{P}[a, b]$ then we should note that for any $k \in \{1, 2, ..., n \}$ that since any subinterval $x_{k-1}, x_k]$ contains both rational and irrational numbers, and so:

(3)

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

(4)

\begin{align} \quad m_k (f) = \inf \{ f(x) : x \in [x_{k-1}, x_k] \} = 0 \end{align}

So for any partition $P \in \mathscr{P}[a, b]$ we have that:

(5)

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

So there exists $\epsilon_1 = \frac{b-a}{2} > 0$ such that for all partitions $P$ we have that:

(6)

\begin{align} \quad U(P, f, x) - L(P, f, x) = b - a > \frac{b-a}{2} = \epsilon_1 \end{align}

Hence Riemann's condition cannot be satisfied and so $f$ is not Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$.