📂Analysis

Monotone Functions are Riemann-Stieltjes Integrable

About Riemann Integration

Let’s suppose that the function $f$ is monotonic over $[a,b]$. Then $f$ is Riemann integrable.

Proof

Assume that $f$ is a monotonically increasing function¹. Let $\epsilon >0$ be given. Consider a partition $P= \left\{ x_{i} : a=x_{0} < x_{1} < x_{2} < \cdots < x_{n}=b \right\}$ of the interval $[a,b]$ that satifies the following for any natural number $n$:

$$\Delta x_{i} = x_{i}-x_{i-1} = \dfrac{b-a}{n},\quad (i=1,2,\dots,n)$$

In other words, $P$ is a partition that divides the interval $[a,b]$ evenly. Now let’s set the following:

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

Since $f$ is a monotonically increasing function, the following holds true:

$$M_{i}=f(x_{i}) \quad \text{and} \quad m_{i}=f(x_{i-1})\quad (i=1,\cdots,n)$$

Then, for sufficiently large $n$, the following equation holds:

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

Since this is a necessary and sufficient condition for integrability, $f$ is integrable.

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About Stieltjes Integration²

If the function $f$ is monotonic over $[a,b]$, and the function $\alpha$ is also monotonic and continuous over $[a,b]$, then $f$ is Riemann-Stieltjes integrable.

Proof

Assume that $f$ is a monotonically increasing function¹. Let $\epsilon >0$ be given. Consider a partition $P= \left\{ x_{i} : a=x_{0} < x_{1} < x_{2} < \cdots < x_{n}=b \right\}$ of the interval $[a,b]$ that satisfies the following for any natural number $n$:

$$\Delta \alpha_{i} = \dfrac{\alpha (b) -\alpha (a) }{n},\quad (i=1,2,\dots,n)$$

To put it another way, $P$ is a partition that divides the values of function $\alpha$ evenly. This is possible due to the continuity of $\alpha$. Now, let’s set the following:

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

Since $f$ is a monotonically increasing function, the following holds true:

$$M_{i}=f(x_{i}) \quad \text{and} \quad m_{i}=f(x_{i-1})\quad (i=1,\cdots,n)$$

Then, for sufficiently large $n$, the following equation holds:

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

Since this is a necessary and sufficient condition for integrability, $f$ is integrable.

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