Riemann-Stieltjes Integrals with Single Disc. Step Func. as Integrators

# Riemann-Stieltjes Integrals with Single Discontinuity Step Functions as Integrators

Recall from the Step Functions page that a step function $\alpha$ on an interval $[a, b]$ is a function corresponding to a partition $P = \{ a = x_0, x_1, ..., x_n = b \} \in \mathscr{P}[a, b]$ such that $\alpha$ is constant on each open subinterval $(x_{k-1}, x_k)$ for each $k \in \{1, 2, ..., n \}$.

We will now look at a way to evaluate Riemann-Stieltjes integrals having a single discontinuity step functions as its integrator.

 Theorem 1: Let $f$ be a function on $[a, b]$ and let $\alpha$ be a step function having only one discontinuity at $c \in (a, b)$, i.e., $\alpha (x) = \left\{\begin{matrix} p & \mathrm{for \: all} \: x \in [a, c) \\ q & \mathrm{for} \: x = c\\ r & \mathrm{for \: all} \: x \in (c, b] \end{matrix}\right.$. Then $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$ if at least one of the following holds: 1) $f$ is continuous at $c$ on both sides. 2) $f$ is left continuous at $c$ and $\alpha$ is right continuous at $c$. 3) $f$ is right continuous at $c$ and $\alpha$ is left continuous at $c$. 4) $\alpha$ is continuous at $c$ on both sides. In either of the cases above, we'll also have that $\displaystyle{\int_a^b f(x) \: d \alpha(x) = f(c)[\alpha(c^+) - \alpha(c^-)]}$.
• Proof: Consider the partition $\{ a, c, b \} \in \mathscr{P}[a, b]$. Then if $P \in \mathscr{P}[a, b]$ is a refinement of $\{a, c , b \}$, that is, $\{a, c, b \} \subseteq \{ a = x_0, x_1, ..., x_n = b \} = P$ then $P$ is of the form:
(1)
\begin{align} \quad P = \{ a = x_0, x_1, ..., x_{j-1}, x_j = c, x_{j+1}, ..., x_n = b \} \end{align}
• Consider the corresponding Riemann-Stieltjes sum with respect to the partition $P$ and for $t_k \in (x_{k-1}, x_k)$ for each $k \in \{1, 2, ..., n \}$:
(2)
\begin{align} \quad S(P, f, \alpha) &= \sum_{k=1}^{n} f(t_k) \Delta \alpha_k \\ \quad S(P, f, \alpha) &= \sum_{k=1}^{n} f(t_k) (\alpha(x_k) - \alpha(x_{k-1}) \\ \quad S(P, f, \alpha) &= f(t_1)[\alpha(x_1) - \alpha(x_0)] + f(t_2)[\alpha(x_2) - \alpha(x_1)] + ... + f(t_j)[\alpha(x_j) - \alpha(x_{j-1})] + f(t_{j+1})[\alpha(x_{j+1}) - \alpha(x_j)] + ... + f(t_n)[\alpha(x_n) - \alpha(x_{n-1})] \\ \quad S(P, f, \alpha) &= f(t_1)[\alpha(x_1) - \alpha(x_0)] + f(t_2)[\alpha(x_2) - \alpha(x_1)] + ... + f(t_j)[\alpha(c) - \alpha(x_{j-1})] + f(t_{j+1})[\alpha(x_{j+1}) - \alpha(c)] + ... + f(t_n)[\alpha(x_n) - \alpha(x_{n-1})] \end{align}
• In the above sum we see that many of the terms are equal to $0$ since $\alpha(x) = p$ for all $x \in [a, c)$ and $\alpha(x) = r$ for all $x \in (c, b]$, so:
(3)
\begin{align} \quad S(P, f, \alpha) &= \underbrace{f(t_1)[\alpha(x_1) - \alpha(x_0)] + f(t_2)[\alpha(x_2) - \alpha(x_1)] + ...}_{=0 \: \mathrm{since} \: x_1, x_2, ..., x_{j-1} \in [a, c) \: \mathrm{and} \: \alpha(x) = p \: \forall x \in [a, c)} + f(t_j)[\alpha(x_j) - \alpha(x_{j-1})] + f(t_{j+1})[\alpha(x_{j+1}) - \alpha(x_j)] \underbrace{+ ... + f(t_n)[\alpha(x_n) - \alpha(x_{n-1})]}_{=0 \: \mathrm{since} \: x_{j+1}, ..., x_n \in (c, b] \: \mathrm{and} \: \alpha(x) = r \: \forall x \in (c, b]} \\ \quad S(P, f, \alpha) &= f(t_j)[\alpha(c) - \alpha(x_{j-1})] + f(t_{j+1})[\alpha(x_{j+1}) - \alpha(c)] \\ \end{align}
• Since $x_{j-1} \in [a, b)$ we have that $\alpha(x_{j-1}) = p$ and since $x_{j+1} \in (c, b]$ we have that $\alpha(x_{j+1}) = r$. Furthermore, $\alpha(c) = q$ and so:
(4)
\begin{align} \quad S(P, f, \alpha) &= f(t_j)[q - p] + f(t_{j+1})[r - q] \end{align}
• We will now consider the limit of these Riemann-Stieltjes sums as $P$ gets finer and finer (i.e., $\| P \| \to 0$) for each of the four cases.
• Case 1: Suppose that $f$ is continuous on both sides. Then take the limit at $\| P \| \to 0$:
(5)
\begin{align} \quad \lim_{\| P \| \to 0} S(P, f, \alpha) = f(c)[q - p] + f(c)[r - q] \\ \quad &= f(c)[(q - p) + (r - q)] \\ \quad &= f(c) [r - p] \\ \quad &= f(c)[\alpha(c^+) - \alpha(c^-)] \end{align}
• Case 2: Suppose that $f$ is left continuous at $c$ and $\alpha$ is right continuous at $c$. Since $\alpha$ is right continuous at $c$ we have that $q = r$. Therefore:
(6)
\begin{align} \quad S(P, f, \alpha) &= f(t_j)[q - p] \end{align}
• Now take the limit as $\| P \| \to 0$:
(7)
\begin{align} \quad \lim_{\| P \| \to 0} S(P, f, \alpha) = f(c)[\alpha (c^+) - \alpha(c^-)] \end{align}
• Case 3: Suppose that $f$ is right continuous at $c$ and $\alpha$ is left continuous at $c$. Since $\alpha$ is left continuous at $c$ we have that $p = q$. Therefore:
(8)
\begin{align} \quad S(P, f, \alpha) &= f(t_{j+1})[r - q] \end{align}
• Now take the limit as $\| P \| \to 0$:
(9)
\begin{align} \quad \lim_{\| P \| \to 0} S(P, f, \alpha) = f(c)[r - q] \\ \quad \lim_{\| P \| \to 0} S(P, f, \alpha) = f(c)[\alpha(c^+) - \alpha(c^-)] \end{align}
• Case 4: Last suppose that $\alpha$ is continuous at $c$ on both sides. Then $p = q = r$. So $\alpha$ is actually constant on all of $[a, b]$ and so:
(10)
\begin{align} \quad S(P, f, \alpha) = 0 = f(c)\underbrace{[\alpha(c^+) - \alpha(c^-)]}_{=0} \end{align}
• In all cases we have that:
(11)
\begin{align} \quad \int_a^b f(x) \: d \alpha (x) = f(c)[\alpha(c^+) - \alpha(c^-)] \quad \blacksquare \end{align}