Riemann-Stieltjes Integral Defined Functions

Riemann-Stieltjes Integral Defined Functions

Recall from the Riemann-Stieltjes Integrability of Functions on Subintervals with Integrators of Bounded Variation 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 $[c, d] \subseteq [a, b]$.

So $\int_a^{\tilde{x}} f(x) \: d \alpha (x)$ is defined for all $\tilde{x} \in [a, b]$. Hence we can define an integral-defined function $F : [a, b] \to \mathbb{R}$ given for all $\tilde{x} \in [a, b]$ by:

\begin{align} \quad F(\tilde{x}) = \int_a^{\tilde{x}} f(x) \: d \alpha (x) \end{align}

This function has some nice properties which we prove in the following theorem.

Theorem 1: Let $f$ be a function defined on $[a, b]$ and let $\alpha$ be a function of bounded variation on $[a, b]$. Furthermore, let $f$ be Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$ so that the function $F : [a, b] \to \mathbb{R}$ defined for all $\tilde{x} \in [a, b]$ by $\displaystyle{F(\tilde{x}) = \int_a^{\tilde{x}} f(x) \: d \alpha (x)}$ is well defined. Then:
a) $F$ is of bounded variation on $[a, b]$.
b) If $\alpha$ is continuous at $x_0 \in [a, b]$ then $F$ is continuous at $x_0$.
c) If $f$ is continuous at $x_0 \in [a, b]$, $\alpha$ is increasing on $[a,b]$, and $\alpha'(x_0)$ exists then $F'(x_0)$ exists.
  • Proof of a): Since $\alpha$ is of bounded variation on $[a, b]$ it is sufficient to show that (a) holds with an increasing integrator. Assume that $\alpha$ is increasing on $[a, b]$.
  • Let $P = \{a = x_0, x_1, ..., x_n = b \} \in \mathscr{P}[a, b]$ and consider the variation of $F$ associated with $P$:
\begin{align} \quad V_F(P) = \sum_{k=1}^{n} \mid F(x_k) - F(x_{k-1}) \mid \end{align}
  • Note that $x_{k-1} < x_k$ for $k \in \{1, 2, ..., n \}$ and so $F(x_{k-1}) = \int_a^{x_{k-1}} f(x) \: d \alpha (x)$ and $F(x_k) = \int_a^{x_k} f(x) \: d \alpha (x)$ and furthermore, for each $k \in \{ 1, 2, ..., n \}$ we have that:
\begin{align} \quad \mid F(x_k) - F(x_{k-1}) \mid = \biggr \lvert \int_a^{x_k} f(x) \: d \alpha (x) - \int_a^{x_{k-1}} f(x) \: d \alpha (x) \biggr \rvert = \biggr \lvert \int_{x_{k-1}}^{x_k} f(x) \: d \alpha (x) \biggr \rvert \end{align}
\begin{align} \quad \int_{x_{k-1}}^{x_k} f(x) \: d \alpha (x) = c_k[\alpha(x_k) - \alpha(x_{k-1})] \end{align}
  • Therefore:
\begin{align} \quad \mid F(x_k) - F(x_{k-1}) \mid = \mid c_k[\alpha(x_k) - \alpha(x_{k-1})\mid = \mid c_k \mid \mid \alpha (x_k) - \alpha(x_{k-1}) \mid \end{align}
  • Let $M^* = \sup \{ f(x) : x \in [a, b] \}$. Then noting that $\alpha(x_k) - \alpha(x_{k-1}) \geq 0$ for all $k \in \{1, 2, ..., n \}$ from the assumption that $\alpha$ is increasing and we get that::
\begin{align} \quad V_F(P) = \sum_{k=1}^{n} \mid c_n \mid \mid \alpha(x_k) - \alpha(x_{k-1}) \mid \\ \quad V_F(P) \leq \sum_{k=1}^{n} M^* (\Delta \alpha_k) \\ \quad V_F(P) \leq M^* \sum_{k=1}^{n} \Delta \alpha_k \\ \quad V_F(P) \leq M^* [\alpha (b) - \alpha(a)] \end{align}
  • So if we let $M = M^*[\alpha (b) - \alpha(a)]$ then for all partitions $P \in \mathscr{P}[a, b]$ we have that $V_F(P) \leq M$, so $F$ is of bounded variation on $[a, b]$. $\blacksquare$
  • Proof of b): Suppose further that $\alpha$ is continuous at $x_0 \in [a, b]$. Let $M = \sup \{ f(x) : x \in [a, b] \}$. By applying the first Mean-Value theorem for Riemann-Stieltjes integrals and we have that there exists a $c_{\tilde{x}} \in [x_0, x]$ such that:
\begin{align} \quad \mid F(x) - F(x_0) \mid = \biggr \lvert \int_{x_0}^{\tilde{x}} f(x) \: d \alpha (x) \biggr \rvert = \mid c_{\tilde{x}} \mid \mid \alpha(x) - \alpha(x_0) \mid \leq \mid M \mid \alpha (x) - \alpha(x_0) \mid \end{align}
  • Therefore we have that:
\begin{align} \quad 0 \leq \lim_{x \to x_0} \mid F(x) - F(x_0) \mid \leq \lim_{x \to x_0} \mid M \mid \mid \alpha (x) - \alpha(x_0) \mid \end{align}
  • Since $\alpha$ is continuous at $x_0$ we have that $\lim_{x \to x_0} \mid \alpha (x) - \alpha(x_0) \mid = 0$ and therefore:
\begin{align} \quad 0 \leq \lim_{x \to x_0} \mid F(x) - F(x_0) \mid \leq 0 \end{align}
  • By the Squeeze theorem we see that therefore $\lim_{x \to x_0} F(x) = F(x_0)$ so $F$ is continuous at $x_0$. $\blacksquare$
  • Proof of c): Let $f$ be continuous at $x_0$, $\alpha$ be an increasing function, and let $\alpha'(x_0)$ exist. Consider the following quotient:
\begin{align} \quad \frac{F(x) - F(x_0)}{x - x_0} = \frac{\int_{x_0}^{x} f(x) \: d \alpha (x)}{x - x_0} \end{align}
  • Applying the first Mean-Value theorem again we have that that for $c_{\tilde{x}} \in [\inf \{ f(x) : \tilde{x} \in [x_0, x] \}, \sup \{ f(x) : \tilde{x} \in [x_0, x] \}]$ we have that:
\begin{align} \quad \frac{F(x) - F(x_0)}{x - x_0} = \frac{c_{\tilde{x}} [\alpha (x) - \alpha(x_0)]}{x - x_0} = c_{\tilde{x}} \frac{\alpha (x) - \alpha(x_0)}{x - x_0} \end{align}
  • Therefore, since $\lim_{\tilde{x} \to x_0} \left ( \inf \{ f(x) : \tilde{x} \in [x, x_0] \}\right ) = f(x_0)$ and $\lim_{\tilde{x} \to x_0} \left ( \sup \{ f(x) : \tilde{x} \in [x, x_0] \} \right ) = f(x_0)$ we have by the Squeeze theorem that:
\begin{align} \quad \lim_{\tilde{x} \to x_0} \left ( \inf \{ f(x) : \tilde{x} \in [x, x_0] \} \right ) \leq \lim_{\tilde{x} \to x_0} c_{\tilde{x}} \leq \lim_{\tilde{x} \to x_0} \left ( \sup \{ f(x) : \tilde{x} \in [x, x_0] \} \right ) \\ \quad f(x_0) \leq \lim_{\tilde{x} \to x_0} c_{\tilde{x}} \leq f(x_0) \end{align}
  • So $\lim_{\tilde{x} \to x_0} c_{\tilde{x}} = f(x_0)$ and therefore:
\begin{align} \quad \lim_{\tilde{x} \to x_0} \frac{F(x) - F(x_0)}{x - x_0} &= \lim_{\tilde{x} \to x_0} c_{\tilde{x}} \frac{\alpha (x) - \alpha(x_0)}{x - x_0} \\ \quad F'(x_0) &= f(x_0) \alpha'(x_0) \end{align}
  • Therefore $F'(x_0)$ exists and $F'(x_0) = f(x_0) \alpha'(x_0)$. $\blacksquare$
Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License