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:

\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:

\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:

\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:

\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:

\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}