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:
(2)
\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:
(3)
\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:
(4)
\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:
(5)
\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:
(6)
\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}