The Monotonicity Property of the Lebesgue Int. of Leb. Int. Functs.
The Monotonicity Property of the Lebesgue Integral of Lebesgue Integrable Functions
| Lemma 1: Let $f$ be a Lebesgue integrable function defined on a Lebesgue measurable set $E$. If $f(x) \geq 0$ for all $x \in E$ then $\displaystyle{\int_E f \geq 0}$. |
- Proof: Since $f(x) \geq 0$ we have that $f$ is a nonnegative Lebesgue measurable function defined on $E$ and so we already have that $\displaystyle{\int_E f \geq 0}$ since the Lebesgue integral for nonnegative Lebesgue measurable functions is the same as the Lebesgue integral for a Lebesgue measurable function $f$ that is nonnegative. $\blacksquare$
| Theorem 2 (Monotonicity of the Lebesgue Integral for Lebesgue Integrable Functions): Let $f$ and $g$ be Lebesgue integrable functions defined on a Lebesgue measurable set $E$. If $f(x) \leq g(x)$ for all $x \in E$ then $\displaystyle{\int_E f \leq \int_E g}$. |
- Proof: Since $f(x) \leq g(x)$ for all $x \in E$ we have that $0 \leq g(x) - f(x)$ for all $x \in E$ and by Lemma 1 we have that:
\begin{align} \quad \int_E (g - f) \geq 0 \end{align}
- By linearity of the Lebesgue integral for 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 Lebesgue integrable function defined on a Lebesgue measurable set $E$. Then $\displaystyle{\biggr \lvert \int_E f \biggr \rvert \leq \int_E |f|}$. |
- Proof: For all $x \in E$ we have that $-|f(x)| \leq f(x) \leq |f(x)|$ and so by Theorem 2 we have that:
\begin{align} \quad \int_E -|f| \leq \int_E f \leq \int_E |f| \end{align}
- By the linearity of the Lebesgue integral for 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 < \int_E |f| \quad \blacksquare \end{align}