R-S Integrability of Functions on Subintervals with Integrators of B.V.

# Riemann-Stieltjes Integrability of Functions on Subintervals with Integrators of Bounded Variation

We will now look at a very important theorem which will tell us that if $f$ is a function defined on $[a, b]$, $\alpha$ is of bounded variation on $[a, b]$, and $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$ then $f$ is also Riemann-Stieltjes integrable with respect to $\alpha$ on any subinterval $[c, d] \subseteq [a, b]$.

Theorem 1: Let $f$ be a function defined on $[a, b]$ and let $\alpha$ be a function of bounded variation on $[a, b]$. If $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$ then $f$ is also Riemann-Stieltjes integrable with respect to $\alpha$ on every subinterval $[c, d] \subseteq [a, b]$. |

**Proof:**If $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$, then as we mentioned on the Riemann-Stieltjes Integrals with Integrators of Bounded Variation page, if $V(x) = V_{\alpha}(a, x)$ is the total variation function of $\alpha$ on $[a, b]$ then $V$ and $V - f$ are increasing functions and $f$ is Riemann-Stieltjes integrable with respect to $V$ and $V - f$ on $[a, b]$. If we can show that Theorem 1 is true for increasing functions, then by the Linearity of the Integrator of Riemann-Stieltjes Integrals we have that:

\begin{align} \quad \int_c^d f(x) \: dV(x) - \int_c^d f(x) \: d [V(x) - \alpha(x)] = \int_c^d f(x) \: d \alpha (x) \end{align}

- Assume that $\alpha$ is an increasing function on $[a, b]$. We first show that $\int_a^c f(x) \: d \alpha (x)$ exists. Let $c \in (a, b)$.

- Now since $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$ then Riemann's condition is satisfied, and for every $\epsilon > 0$ there exists a partition $P_{\epsilon} \in \mathscr{P}[a, b]$ such that if $P$ is finer than $P_{\epsilon}$ then $U(P, f, \alpha) - L(P, f, \alpha) < \epsilon$. If $c \in P_{\epsilon}$, leave $P_{\epsilon}$ how it is. If $c \not \in P_{\epsilon}$ then add $c$ to $P_{\epsilon}$ (which is alright since $P_{\epsilon}$ will just be finer and still suffice to satisfy Riemann's condition).

- So take $P$ finer than $P_{\epsilon}$. Then $P = P' \cup P''$ where $P' \in \mathscr{P}[a, c]$ and $P'' \in \mathscr{P}[c, b]$. Let $P^*$ be finer than $P'$. Then $P^* \cup P'' is finer than [[$ P$ and:

\begin{align} \quad L(P', f, \alpha) \leq L(P^*, f, \alpha) \leq U(P^*, f, \alpha) \leq U(P', f, \alpha) \end{align}

- Hence we see that $U(P^*, f, \alpha) - L(P^*, f, \alpha) \leq U(P', f, \alpha) - L(P', f, \alpha)$ and so:

\begin{align} \quad \epsilon &> \underbrace{U(P, f, \alpha) - L(P, f, \alpha)}_{P \in \mathscr{P}[a, b]} \\ &= [\underbrace{U(P', f, \alpha)}_{P' \in \mathscr{P}[a, c]} + \underbrace{U(P'', f, \alpha)}_{P'' \in \mathscr{P}[c, b]}] - [\underbrace{L(P', f, \alpha)}_{P' \in \mathscr{P}[a, c]} + \underbrace{L(P'', f, \alpha)}_{P'' \in \mathscr{P}[c, b]}] \\ &= [\underbrace{U(P', f, \alpha) - L(P', f, \alpha)}_{P' \in \mathscr{P}[a, c]}] + [\underbrace{U(P'', f, \alpha) - L(P'', f, \alpha)}_{P'' \in \mathscr{P}[c, b]}] \\ &\geq \underbrace{U(P', f, \alpha) - L(P', f, \alpha)}_{P' \in \mathscr{P}[a, c]} \\ &\geq \underbrace{U(P^*, f, \alpha) - L(P^*, f, \alpha)}_{P^* \in \mathscr{P}[a, c]} \end{align}

- So for each $\epsilon > 0$ choose any partition $P^* \in \mathscr{P}[a, c]$ finer than $P' \in \mathscr{P}[a, c]$. Then Riemann's condition is satisfied and $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, c]$.

- Furthermore, if $d \in (a, b)$ then we also have that $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, d]$. Therefore if $c < d$ then:

\begin{align} \quad \int_a^d f(x) \: d \alpha(x) - \int_a^c f(x) \: d \alpha (x) = \int_c^d f(x) \: d \alpha (x) \end{align}

- Both of the Riemann-Stieltjes integrals exist on the lefthand side of the equation above, and so $\int_c^d f(x) \: d \alpha (x)$ exists. So $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[c, d] \subseteq [a, b]$. $\blacksquare$