Riemann-Stieltjes Integrability on Subintervals

# Riemann-Stieltjes Integrability on Subintervals

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 $\int_a^b f(x) \: d\alpha(x) = A$.

We will now look at the relation for $c \in (a, b)$ of the existence of the integrals $\displaystyle{\int_a^b f(x) \: d \alpha (x)}$, $\displaystyle{\int_a^c f(x) \: d \alpha(x)}$ and $\displaystyle{\int_c^b f(x) \: d \alpha (x)}$.

 Theorem 1: Let $f$ and $\alpha$ be functions defined on the interval $[a, b]$ and let $c \in (a,b)$. Then: a) $\displaystyle{\int_a^b f(x) \: d \alpha (x)}$ exists if both $\displaystyle{\int_a^c f(x) \: d \alpha (x)}$ and $\displaystyle{\int_c^b f(x) \: d \alpha (x)}$ exists. b) $\displaystyle{\int_a^c f(x) \: d \alpha (x)}$ exists if both $\displaystyle{\int_a^b f(x) \: d \alpha (x)}$ and $\displaystyle{\int_c^b f(x) \: d \alpha (x)}$ exists. c) $\displaystyle{\int_c^b f(x) \: d \alpha (x)}$ exists if both $\displaystyle{\int_a^c f(x) \: d \alpha (x)}$ and $\displaystyle{\int_a^b f(x) \: d \alpha (x)}$ exists.

We will only prove a) since b) and c) can be proved analogously.

• Proof of a) Suppose that $\int_a^c f(x) \: d \alpha (x) = A$ and $\int_c^b f(x) \: d \alpha (x) = B$ for some $A, B \in \mathbb{R}$. Let $\epsilon > 0$ be given.
• Since $\int_a^c f(x) \: d \alpha (x) = A$ we have that for $\epsilon_1 = \frac{\epsilon}{2} > 0$ there exists a partition $P_{\epsilon_1} \in \mathscr{P}[a, c]$ such that for all partitions $P' \in \mathscr{P}[a, c]$ finer than $P_{\epsilon_1}$, ($P_{\epsilon_1} \subseteq P'$) and for any choice of $t_k$'s in each $k^{\mathrm{th}}$ subinterval we have that:
(2)
\begin{align} \quad \mid S(P', f, \alpha) - A \mid < \epsilon_1 = \frac{\epsilon}{2} \quad (*) \end{align}
• Similarly, since $\int_c^b f(x) \: d \alpha (x) = B$ we have that for $\epsilon_2 = \frac{\epsilon}{2} > 0$ there exists a partition $P_{\epsilon_2} \in \mathscr{P}[c, b]$ such that for all partitions $P'' \in \mathscr{P}[c, b]$ finer than $P_{\epsilon_2}$, $(P_{\epsilon_2} \subseteq P''$) and for any choice of $u_k$'s in each $k^{\mathrm{th}}$ subinterval we have that:
(3)
\begin{align} \quad \mid S(P'', f, \alpha) - B \mid < \epsilon_2 = \frac{\epsilon}{2} \quad (**) \end{align}
• Let $P_{\epsilon} = P_{\epsilon_1} \cup P_{\epsilon_2}$. Then $P_{\epsilon}$ is a partition of $[a, b]$ and for all partitions $P \in \mathscr{P}[a, b]$ finer than $P_{\epsilon}$, ($P_{\epsilon} \subseteq P$) we must have that $(*)$ and $(**)$ hold. Then for any choice of $v_k$'s in each $k^{\mathrm{th}}$ subinterval we have that:
(4)
\begin{align} \quad \mid S(P, f, \alpha) - (A + B) \mid = \mid S(P', f, \alpha) + S(P'', f, \alpha) - (A + B) \mid \leq \mid S(P', f, \alpha - A \mid + \mid S(P'', f, \alpha) - B \mid < \epsilon_1 + \epsilon_2 = \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align}
• Hence $\int_a^b f(x) \: d \alpha (x)$ exists and:
(5)
\begin{align} \quad \int_a^b f(x) \: d \alpha (x) = \int_a^c f(x) \: d \alpha (x) + \int_c^b f(x) \: d \alpha (x) \quad \blacksquare \end{align}