📂Analysis

Continuous Functions are Riemann-Stieltjes Integrable

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This article is based on the Riemann-Stieltjes integral. If set as $\alpha=\alpha (x)=x$, it is the same as the Riemann integral.

Theorem

If function $f$ is continuous on $[a,b]$, then it is Riemann(-Stieltjes) integrable on $[a,b]$.

Proof

Suppose $\epsilon >0$ is given. And let’s say we chose $\eta>0$ that satisfies $\left[ \alpha (b) - \alpha (a) \right] \eta < \epsilon$. Since $[a,b]$ is compact as it is closed and bounded, and continuous functions on a compact set are uniformly continuous, $f$ is uniformly continuous. Therefore, by the definition of uniform continuity, there exists $\delta >0$ for which the following equation holds.

$$|x-t|<\delta \implies |f(x)-f(t)|<\eta\quad \forall x, t \in [a,b]$$

By the definition of uniform continuity, any positive number can satisfy the place of $\eta$, so the $\eta$ we chose above satisfies it naturally.

Let’s say the partition $P$ of $[a,b]$ was given to satisfy $\Delta x_{i} <\delta (i=1,\cdots,n)$. Also, let’s consider the following.

$$M_{i}=\sup\limits_{[x_{i-1},x_{i}]}f(x) \quad \text{and} \quad m_{i}=\inf\limits_{[x_{i-1},x_{i}]}f(x)$$

Then by the condition that $f$ is uniformly continuous, the following is true.

$$M_{i}-m_{i} \le \eta \quad (i=1,\cdots,n)$$

Then we obtain the following equation.

$$\begin{align*} U(P,f,\alpha) - L(P,f,\alpha) &= \sum \limits_{i=1}^n (M_{i}-m_{i})\Delta \alpha_{i} \\ & \le \sum \limits _{i=1} ^n \eta \Delta \alpha_{i} \\ &= \eta \sum \limits_{i=1}^n \Delta \alpha_{i} \\ &= \eta \left[ \big( \alpha ( x_{2}) -\alpha (a) \big) + \cdots + \big( \alpha ( b) -\alpha (x_{n-1}) \big)\right] \\ &=\eta \left[ \alpha ( b) - \alpha (a) \right] \\ &< \epsilon \end{align*}$$

As for the beginning part of the proof, we chose $\eta$ such that it satisfies the last equation, so it is natural that the last line holds. This equation is the equivalence condition for being integrable, therefore $f$ is integrable.

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