The Monotonicity Property of the Lebesgue Integral of Simple Functions

# The Monotonicity Property of the Lebesgue Integral of Simple Functions

Recall from The Linearity Property of the Lebesgue Integral of Simple Functions page that if $\varphi$ and $\psi$ are simple functions defined on a Lebesgue measurable set $E$ with $m(E) < \infty$ then for any $\alpha, \beta \in \mathbb{R}$ we have that:

(1)\begin{align} \quad \int_E (\alpha \varphi + \beta \psi) = \alpha \int_E \varphi + \beta \int_E \psi \end{align}

We will now show that the Lebesgue integral of simple functions also has a monotonicity property by first proving an important lemma.

Lemma 1: Let $\varphi$ be a simple function defined on a Lebesgue measurable set $E$ with $m(E) < \infty$. If $\varphi(x) \geq 0$ for all $x \in E$ then $\displaystyle{\int_E \varphi \geq 0}$. |

**Proof:**Let $\varphi$ be a simple function defined on a Lebesgue measurable set $E$ with $m(E) < \infty$ and let $\varphi(x)$ have canonical representation $\displaystyle{\varphi(x) = \sum_{k=1}^{n} a_k \chi_{E_k}(x)}$. Since $\varphi(x) \geq 0$ for all $x \in E$ we must have that $a_k \geq 0$ for each $k \in \{ 1, 2, ..., n \}$. Noting that the Lebesgue measurable of a set is always nonnegative we have that:

\begin{align} \quad \int_E \varphi = \sum_{k=1}^{n} a_k m(E_k) \geq 0 \quad \blacksquare \end{align}

Theorem 2 (Monotonicity of the Lebesgue Integral for Simple Functions): Let $\varphi$ and $\psi$ be simple functions defined on a Lebesgue measurable set $E$ with $m(E) < \infty$. If $\varphi (x) \leq \psi(x)$ for all $x \in E$ then $\displaystyle{\int_E \varphi \leq \int_E \psi}$. |

- Let $\varphi$ and $\psi$ be simple functions defined on a Lebesgue measurable set $E$ with $m(E) < \infty$. Since $\varphi(x) \leq \psi(x)$ for all $x \in E$ we have that $\psi(x) - \varphi(x) \geq 0$ for all $x \in E$. By Lemma 1 this means that:

\begin{align} \quad \int_E (\psi - \varphi) \geq 0 \end{align}

- And by the linearity of the Lebesgue integral of simple functions we have that:

\begin{align} \quad 0 \leq \int_E (\psi - \varphi) = \int_E \psi - \int_E \varphi \end{align}

- Hence $\displaystyle{\int_E \varphi \leq \int_E \psi}$. $\blacksquare$

Theorem 3: Let $\varphi$ be a simple function defined on a Lebesgue measurable set $E$ with $m(E) < \infty$. Then $\displaystyle{\biggr \lvert \int_E \varphi \biggr \rvert \leq \int_E |\varphi|}$. |

**Proof:**Let $\varphi$ be a simple function defined on a Lebesgue measurable set $E$ with $m(E) < \infty$. Then clearly $| \varphi |$ and $-|\varphi|$ are simple functions defined on $E$. Furthermore, $-| \varphi(x) | \leq \varphi(x) \leq | \varphi(x) |$ for all $x \in E$. So by Theorem 2:

\begin{align} \quad \int_E -|\varphi| \leq \int_E \varphi \leq \int_E | \varphi | \end{align}

- By the linearity property of the Lebesgue integral of simple functions we have that:

\begin{align} \quad -\int_E |\varphi| \leq \int_E \varphi \leq \int_E | \varphi | \quad \Leftrightarrow \quad \biggr \lvert \int_E \varphi \biggr \rvert \leq \int_E |\varphi| \quad \blacksquare \end{align}