Linearity of the Integrator of Riemann-Stieltjes Integrals

# Linearity of the Integrator of Riemann-Stieltjes Integrals

Recall from the Riemann-Stieltjes Integrals page that if $[a, b]$ is a closed interval and $f, \alpha$ are functions on $[a, b]$, then $f$ is said to be Riemann-Stieltjes integral with respect to $\alpha$ on $[a, b]$ if there exists an $A \in \mathbb{R}$ such that for all $\epsilon > 0$ and such that for all partitions $P \in \mathscr{P}[a, b]$ finer than $P_{\epsilon} \in \mathscr{P}[a, b]$ ($P_{\epsilon} \subseteq P$) we have that:

(1)
\begin{align} \quad \mid S(P, f, \alpha) - A \mid < \epsilon \end{align}

Where $\displaystyle{S(P, f, \alpha) = \sum_{k=1}^{n} f(t_k) \Delta \alpha_k}$ is a Riemann-Stieltjes sum for the partition $P = \{ a = x_0, x_1, ..., x_n = b \} \in \mathscr{P}[a, b]$, $t_k \in [x_{k-1}, x_k]$ for all $k \in \{1, 2, ..., n \}$, and $\Delta \alpha_k = \alpha(x_k) - \alpha(x_{k-1})$.

If such an $A \in \mathbb{R}$ exists, we write $\int_a^b f(x) \: d\alpha(x) = A$.

On the Linearity of the Integrand of Riemann-Stieltjes Integrals we proved that if $f$ and $g$ were Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$ where $\int_a^b f(x) \: d \alpha(x) = A$ and $\int_a^b g(x) \: d \alpha (x) = B$ and if $c \in \mathbb{R}$ then $f + g$ and $cf$ was Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$ and $\int_a^b [f(x) + g(x)] \: d \alpha (x) = A + B$ and $\int_a^b cf(x) \: d \alpha (x) = cA$.

We will now look at some nice linearity properties of the integrator of Riemann-Stieltjes integrals.

 Theorem 1: Let $f$ be Riemann-Stieltjes integrable with respect to both $\alpha$ and $\beta$ on the interval $[a, b]$ where $\int_a^b f(x) \: d \alpha(x) = A$ and $\int_a^b f(x) \: d \beta(x) = B$. Then $f$ is Riemann-Stieltjes integrable with respect to $\alpha + \beta$ on $[a, b]$ and $\int_a^b f(x) \: d (\alpha + \beta)(x) = A + B$.
• Proof: Let $\epsilon > 0$ be given and let $f$ be Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$. Then there exists an $A \in \mathbb{R}$ such that for all $\epsilon_1 = \frac{\epsilon}{2} > 0$ and such that for all partitions $P \in \mathscr{P} [a, b]$ finer than $P_{\epsilon_1} \in \mathscr{P}[a, b]$ ($P_{\epsilon_1} \subseteq P$) we have that:
(2)
\begin{align} \quad \mid S(P, f, \alpha) - A \mid < \epsilon_1 = \frac{\epsilon}{2} \end{align}
• Let $f$ also be Riemann-Stieltjes integrable with respect to $\beta$ on $[a, b]$. Then there exists a $B \in \mathbb{R}$ such that for all $\epsilon_2 = \frac{\epsilon}{2} > 0$ and such that for all partitions $P \in \mathscr{P}[a, b]$ finer than $P_{\epsilon_2} \in \mathscr{P}[a, b]$ ($P_{\epsilon_2} \subseteq P$) we have that:
(3)
\begin{align} \quad \mid S(P, f, \beta) - B \mid < \epsilon_2 = \frac{\epsilon}{2} \end{align}
• Then for all partitions $P = \{ a = x_0, x_1, ..., x_n = b \} \in \mathscr{P}[a, b]$ finer than $P_{\epsilon} = P_{\epsilon_1} \cup P_{\epsilon_2}$ and for $t_k \in [x_{k-1}, x_k]$ for each $k \in \{1, 2, ..., n \}$ we have that:
(4)
\begin{align} \quad \mid S(P, f, \alpha + \beta) - (A + B) \mid &= \biggr \lvert \sum_{k=1}^{n} f(t_k) \Delta (\alpha + \beta)_k - (A + B)\biggr \rvert \\ \quad &= \biggr \lvert \sum_{k=1}^{n} f(t_k) [\alpha(x_k) + \beta(x_k) - (\alpha(x_{k-1}) + \beta(x_{k-1}))] - (A + B) \biggr \rvert \\ \quad &= \biggr \lvert \sum_{k=1}^{n} f(t_k) [(\alpha(x_k) - \alpha(x_{k-1})) + (\beta(x_k) - \beta(x_{k-1}))] - (A + B) \biggr \rvert \\ \quad &= \biggr \lvert \sum_{k=1}^{n} f(t_k) [\Delta \alpha_k + \Delta \beta_k] - (A + B) \biggr \rvert \\ \quad &= \biggr \lvert \sum_{k=1}^{n} f(t_k) \Delta \alpha_k + \sum_{k=1}^{n} f(t_k) \Delta \beta_k - (A + B) \biggr \rvert \\ \quad & < \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 \beta_k - B \biggr \rvert \\ \quad & < \epsilon_1 + \epsilon_2 = \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align}
• Therefore $f$ is Riemann-Stieltjes integrable with respect to $\alpha + \beta$ on $[a, b]$ and:
(5)
\begin{align} \quad \int_a^b f(x) \: d (\alpha + \beta)(x) = A + B \end{align}
 Theorem 2: Let $f$ be Riemann-Stieltjes integrable with respect to $\alpha$ on the interval $[a, b]$ where $\int_a^b f(x) \: d \alpha(x) = A$ and let $c \in \mathbb{R}$. Then $f$ is Riemann-Stieltjes integrable with respect to $c\alpha$ on $[a, b]$ and $\int_a^b f(x) \: d (c \alpha)(x) = c A$.
• Proof: Let $\epsilon > 0$ be given and let $f$ be Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$ such that $\int_a^b f(x) \: d \alpha (x) = A$. First assume that $c \in \mathbb{R}$ and $c \neq 0$. Then there exists an $A \in \mathbb{R}$ such that for $\epsilon_1 = \frac{\epsilon}{\mid c \mid} > 0$ and such that for all partitions $P \in \mathscr{P}[a, b]$ finer than $P_{\epsilon_1} \in \mathscr{P}[a, b]$ ($P_{\epsilon} \subseteq P$) we have that:
(6)
\begin{align} \quad \mid S(P, f, \alpha) - A \mid < \epsilon_1 = \frac{\epsilon}{2} \end{align}
• Then for all partitions $P = \{ a = x_0, x_1, ..., x_n = b \} \in \mathscr{P}[a, b]$ finer than $P_{\epsilon} = P_{\epsilon_1}$ and for $t_k \in [x_{k-1}, x_k]$ for each $k \in \{1, 2, ..., n \}$ we have that:
(7)
\begin{align} \quad \mid S(P, f, c\alpha) - cA \mid &= \biggr \lvert \sum_{k=1}^{n} f(t_k) \Delta (c\alpha)_k - cA \biggr \rvert \\ \quad &= \biggr \lvert \sum_{k=1}^{n} f(t_k)[c\alpha(x_k) - c\alpha(x_{k-1}] - cA \biggr \rvert \\ \quad &= \biggr \lvert c \sum_{k=1}^{n} f(t_k) \Delta \alpha_k - cA \biggr \rvert \\ \quad &= \mid c \mid \biggr \lvert \sum_{k=1}^{n} f(t_k) \Delta \alpha_k - A \biggr \rvert \\ \quad &< \mid c \mid \epsilon_1 = \mid c \mid \cdot \frac{\epsilon}{\mid c \mid} = \epsilon \end{align}
• Therefore $f$ is Riemann-Stieltjes integrable with respect to $c\alpha$ on $[a, b]$ and:
(8)
\begin{align} \quad \int_a^b f(x) \: d (c \alpha)(x) = cA \end{align}
• Now if $c = 0$, then:
(9)
\begin{align} \quad \mid S(P, f, 0) - 0 \mid = \biggr \lvert \sum_{k=1}^{n} f(t_k) \Delta (0)_k \biggr \rvert = 0 < \epsilon \end{align}
• So $f$ is Riemann-Stieltjes integrable with respect to $0\alpha$ on $[a, b]$ and once again, $\int_a^b f(x) \: d (0\alpha(x)) = 0A = 0$. $\blacksquare$