The Monotonicity Prop. of the Leb. Int. of Bound. Leb. Meas. Functs.

The Monotonicity Property of the Lebesgue Integral of Bounded Lebesgue Measurable Functions

Recall from The Linearity Property of the Lebesgue Integral of Bounded, Lebesgue Measurable Functions page that if $f$ and $g$ are bounded Lebesgue measurable functions defined on a Lebesgue measurable set $E$ with $m(E) < \infty$ then for all $\alpha, \beta \in \mathbb{R}$ we have that:

(1)
\begin{align} \quad \int_E (\alpha f + \beta g) = \alpha \int_E f + \beta \int_E g \end{align}

We now show that the Lebesgue integral for bounded Lebesgue measurable functions defined on Lebesgue measurable sets $E$ with $m(E) < \infty$ have the monotonicity property.

 Lemma 1: Let $f$ be a bounded Lebesgue measurable function defined on a Lebesgue measurable set $E$ with $m(E) < \infty$. If $f(x) \geq 0$ for all $x \in E$ then $\displaystyle{\int_E f \geq 0}$.
• Proof: Let $f$ be a bounded Lebesgue measurable function on a Lebesgue measurable set $E$ with $m(E) < \infty$. Since $f(x) \geq 0$ for all $x \in E$, we see that for the simple function $\varphi_0(x) = 0$ for all $x \in E$ that $\varphi(x) \leq f(x)$ for all $x \in E$. So $\displaystyle{0 = \int_E \varphi_0 \in \left \{ \int_E \varphi : \varphi \: \mathrm{is \: a \: simple \: function}, \: \varphi(x) \leq f(x) \: \mathrm{on} \: E \right \}}$ and hence:
(2)
\begin{align} \quad \int_E f = \underline{\int_E} f = \sup \left \{ \int_E \varphi : \varphi \: \mathrm{is \: a \: simple \: function}, \: \varphi(x) \leq f(x) \: \mathrm{on} \: E \right \} \geq 0 \quad \blacksquare \end{align}
 Theorem 2 (Monotonicity of the Lebesgue Integral for Bounded, Lebesgue Measurable Functions): Let $f$ and $g$ be bounded Lebesgue measurable functions defined on a Lebesgue measurable set $E$ with $m(E) < \infty$. If $f(x) \leq g(x)$ for all $x \in E$ then $\displaystyle{\int_E f \leq \int_E g}$.
• Proof: Let $f$ and $g$ be bounded, Lebesgue measurable functions on a Lebesgue measurable set $E$ with $m(E) < \infty$. Since $f(x) \leq g(x)$ for all $x \in E$ we have that $0 \leq g(x) - f(x)$ for all $x \in E$. By Lemma 1 we have that:
(3)
\begin{align} \quad \int_E (g - f) \geq 0 \end{align}
• And by the linearity of the Lebesgue integral for bounded, Lebesgue measurable functions we have that:
(4)
\begin{align} \quad \int_E f \leq \int_E g \quad \blacksquare \end{align}
 **Theorem 3: Let $f$ be a bounded Lebesgue measurable function defined on a Lebesgue measurable set $E$ with $m(E) < \infty$. Then $\displaystyle{\biggr \lvert \int_E f \biggr \rvert \leq \int_E |f|}$.
• Proof: Note that since $f$ is a bounded Lebesgue measurable function, so is $|f|$. Furthermore, $-|f(x)| \leq f(x) \leq |f(x)|$ for all $x \in E$. So by Theorem 2 we have that:
(5)
\begin{align} \quad \int_E -|f| \leq \int_E f \leq \int_E |f| \end{align}
• And by the monotonicity of the Lebesgue integral for bounded, Lebesgue measurable functions we have that:
(6)
\begin{align} \quad - \int_E |f| \leq \int_E f \leq \int_E |f| \quad \Leftrightarrow \quad \biggr \lvert \int_E f \biggr \rvert \leq \int_E |f| \quad \blacksquare \end{align}