📂Analysis

Integrable Functions and Absolute Values

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

Theorem¹

Let function $f$ be Riemann(-Stieltjes) integrable over the interval $[a,b]$. Then

• (a) $\left|f\right|$ is also integrable over $[a,b]$.

• (b) Furthermore, the following inequality holds:

  $$\left|\int_{a}^{b}fd\alpha \right| \le \int_{a}^{b}\left| f\right| d\alpha$$

Proof

(a)

Integrability is defined for bounded functions. Hence, assuming that $f$ is integrable implies that $f$ is bounded. Let’s consider $M, m$ as the upper and lower bounds.

$$M=\sup_{[a,b]} f \quad \text{and} \quad m= \inf_{[a,b]}f$$

Let’s say $\phi (t)=\left| t \right|$. Then $\phi$ is a function that is continuous in $[m,M]$. Moreover, the following holds:

$$\phi \circ f=\left| f\right|$$

Since the composition with a continuous function preserves integrability, $|f|$ is integrable over $[a,b]$.

■

(b)

Let there be a positive number $\varepsilon >0$. Then, according to the necessary and sufficient condition for integrability, there exists a division $P=\left\{ a=x_{0},\cdots,x_{n}=b \right\}$ of $[a,b]$ that satisfies the following equation:

$$U(P,\left| f\right|,\alpha) - L(P,\left| f\right|,\alpha) < \varepsilon$$

Moreover, the below inequality holds:

$$U(P,\left| f \right|,\alpha) < \int_{a}^{b}\left| f \right| d\alpha +\varepsilon$$

Then, by the definition of integration and upper sum, the following equation holds:

$$\int_{a}^{b} f d\alpha \le U(P,f,\alpha) \le U(P,\left| f \right|,\alpha ) <\int_{a}^{b}\left| f \right| d\alpha +\varepsilon$$

Furthermore, if $f$ is integrable then $-f$ is also integrable, thus the following equation holds:

$$-\int_{a}^{b} f d\alpha=\int_{a}^{b}(-f)d\alpha \le U(P,-f,\alpha) \le U(P,\left| f \right|,\alpha ) <\int_{a}^{b}\left| f \right| d\alpha +\varepsilon$$

Then, since $\varepsilon$ is any positive number, the below two equations hold:

$$\begin{align*} \int_{a}^{b}f d\alpha &\le \int _{a}^{b} \left| f \right| d\alpha \\ -\int_{a}^{b}f d\alpha &\le \int _{a}^{b} \left| f \right| d\alpha \end{align*}$$

Therefore, we obtain the following:

$$\left| \int_{a}^{b}fd\alpha \right| \le \int_{a}^{b}\left| f \right| d\alpha$$

■