Reducing Riemann-Stieltjes Integrals to Riemann Integrals

# Reducing Riemann-Stieltjes Integrals to Riemann Integrals

So far we have not really been able to evaluate Riemann-Stieltjes integral directly - however, we should be rather familiar with evaluating standard Riemann integrals from calculus. In certain circumstances, we can reduce Riemann-Stieltjes integrals to regular Riemann integrals. The following theorem provides us with one such method provided that $f$ is bounded and $\alpha'$ exists and is continuous.

 Theorem 1: Let $f$ be a Riemann-Stieltjes integrable function with respect to $\alpha$ on the interval $[a, b]$. Furthermore, let $f$ be bounded on $[a, b]$, let $\alpha'$ exist and be continuous on $[a, b]$. Then $\displaystyle{\int_a^b f(x) \: d \alpha (x) = \int_a^b f(x) \alpha'(x) \: dx}$.
• Proof: Let $f$ be a Riemann-Stieltjes integrable function with respect to $\alpha$ on the interval $[a, b]$. Then for some $A \in \mathbb{R}$ we have that $\int_a^b f(x) \: d \alpha (x) = A$. We want to show that then for all $\epsilon > 0$ there exists a partition $P_{\epsilon} \in \mathscr{P}[a, b]$ such that for all partitions $P \in \mathscr{P}[a, b]$ finer than $P_{\epsilon}$ ($P_{\epsilon} \subseteq P)$ we have that:
(1)
\begin{align} \quad \mid S(P, f\alpha', x) - A \mid < \epsilon \end{align}
• This would then imply that $\int_a^b f(x) \alpha'(x) \: dx = A$.
• Let $\epsilon > 0$ be given. Since $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$ then for $\epsilon_1 = \frac{\epsilon}{2} > 0$ there exists a partition $P_{\epsilon_1} \in \mathscr{P}[a, b]$ such that for all partitions $P = \{a = x_0, x_1, ..., x_n = b \} \in \mathscr{P}[a, b]$ finer than $P_{\epsilon_1}$ ($P_{\epsilon_1} \subseteq P$) we have that:
(2)
\begin{align} \quad \mid S(P, f, \alpha) - A \mid < \epsilon_1 = \frac{\epsilon}{2} \quad (*) \end{align}
• By applying the triangle inequality we get:
(3)
\begin{align} \quad \mid S(P, f\alpha', x) - A \mid & = \mid S(P, f\alpha', x) - S(P, f, \alpha) + S(P, f, \alpha) - A \mid \\ \quad \mid S(P, f\alpha', x) - A \mid & \leq \mid S(P, f\alpha', x) - S(P, f, \alpha) \mid + \mid S(P, f, \alpha) - A \mid \\ \quad \mid S(P, f\alpha', x) - A \mid & \leq \biggr \lvert \sum_{k=1}^{n} f(t_k) \alpha'(t_k) \Delta x_k - \sum_{k=1}^{n} f(t_k) \Delta \alpha_k \biggr \rvert + \mid S(P, f, \alpha) - A \mid \end{align}
• We want to simplify the sum above. Since $\alpha'$ exists and is continuous on $(a, b)$ then by the Mean Value Theorem there exists $\xi_k \in (x_{k-1}, x_k)$ for each $k \in \{1, 2, ..., n \}$ we have that:
(4)
\begin{align} \quad \Delta \alpha_k = \alpha(x_k) - \alpha(x_{k-1}) = \alpha'(\xi_k)[x_k - x_{k-1}] = \alpha'(\xi_k) \Delta x_k \end{align}
• Therefore we have that $\sum_{k=1}^{n} f(t_k) \Delta \alpha_k = \sum_{k=1}^{n} f(t_k) \alpha' (\xi_k) \Delta x_k$ and putting this in for the sum yields:
(5)
\begin{align} \quad \biggr \lvert \sum_{k=1}^{n} f(t_k) \alpha'(t_k) \Delta x_k - \sum_{k=1}^{n} f(t_k) \Delta \alpha_k \biggr \rvert & = \biggr \lvert \sum_{k=1}^{n} f(t_k) \alpha'(t_k) \Delta x_k - \sum_{k=1}^{n} f(t_k)\alpha'(\xi_k)\Delta x_k \biggr \rvert \\ & = \biggr \lvert \sum_{k=1}^{n} f(t_k)(\alpha'(t_k) - \alpha'(\xi_k))\Delta x_k \biggr \rvert \\ \end{align}
• Since $f$ is bounded on $[a, b]$ we have that there exists an $M > 0$ such that $\mid f(x) \mid \leq M$ for all $x \in [a, b]$ so $\mid f(t_k) \mid \leq M$ for all $k \in \{1, 2, ..., n \}$ since each $t_k \in [a, b]$. So:
(6)
\begin{align} \quad \biggr \lvert \sum_{k=1}^{n} f(t_k)(\alpha'(t_k) - \alpha'(\xi_k))\Delta x_k \biggr \rvert \leq M \biggr \lvert \sum_{k=1}^{n} (\alpha'(t_k) - \alpha'(\xi_k))\Delta x_k \biggr \rvert \end{align}
• Now since $\alpha'$ exists and is continuous on $[a, b]$ we must have that $\alpha'$ is uniformly continuous on $[a, b]$ and so we have that for $\epsilon_2 = \frac{\epsilon}{2M(b - a)} > 0$ there exists a $\delta > 0$ such that if $\mid x - y \mid < \delta$ then $\mid \alpha'(x) - \alpha'(y) \mid < \frac{\epsilon}{2M(b - a)}$.
• Let $P_{\epsilon_2}$ be sufficiently fine such that the mesh of $\| P_{\epsilon_2} \| < \delta$. So for all partitons $P \in \mathscr{P}[a, b]$ finer than $P_{\epsilon_2}$ ($P_{\epsilon_2} \subseteq P$) we have that $\| P \| \leq \| P_{\epsilon_1} \| < \delta$. So $\mid t_k - \xi_k \mid < \delta$ for all $k \in \{1, 2, ..., n \}$ so $\mid \alpha'(t_k) - \alpha'(\xi_k) \mid < \epsilon_2 = \frac{\epsilon}{2M(b - a)}$ i.e.,
(7)
\begin{align} \quad M \sum_{k=1}^{n} \mid \alpha'(t_k) - \alpha'(\xi_k) \mid \Delta x_k \leq M \sum_{k=1}^{n} \frac{\epsilon}{2M(b-a)} \Delta x_k = M\frac{\epsilon}{2M(b-a)} \sum_{k=1}^{n} \Delta x_k = \frac{\epsilon}{2(b - a)} (b - a) = \frac{\epsilon}{2} \quad (**) \end{align}
• Let $P_{\epsilon} = P_{\epsilon_1} \cup P_{\epsilon_2}$. Hence we have that for all partitions $P \in \mathscr{P}[a, b]$ finer than $P_{\epsilon}$ ($P_{\epsilon} \subseteq P$) we have that $(*)$ and $(**)$ hold and so:
(8)
\begin{align} \quad \mid S(P, f\alpha', x) - A \mid & \leq \biggr \lvert \sum_{k=1}^{n} f(t_k) \alpha'(t_k) \Delta x_k - \sum_{k=1}^{n} f(t_k) \Delta \alpha_k \biggr \rvert + \mid S(P, f, \alpha) - A \mid \\ \quad & < \epsilon_2 + \epsilon_1 = \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align}
• But $\epsilon > 0$ is arbitrary, and hence $S(P, f\alpha', x) - A = 0$, so:
(9)
\begin{align} \quad S(P, f\alpha', x) = A \\ \quad \int_a^b f(x)\alpha'(x) \: dx = \int_a^b f(x) \: d \alpha (x) \quad \blacksquare \end{align} Corollary 1: If $f$ and $\alpha$ are functions defined on the interval $[a, b]$ where $f(x) = 1$ and let $\alpha'$ exist and be continuous on $[a, b]$. Then $\int_a^b \: d \alpha (x) = \int_a^b \alpha'(x) \: dx = \alpha(b) - \alpha (a)$.

Notice that this correlation is half of the Fundamental Theorem of Calculus.

• Proof: Applying Theorem 1 immediately yields the result above. $\blacksquare$