The Lin. Prop. of the Leb. Int. of Nonneg. Lebes. Meas. Functs.

The Linearity Property of the Lebesgue Integral of Nonnegative Lebesgue Measurable Functions

Recall from The Lebesgue Integral of Nonnegative Lebesgue Measurable Functions page that if $f$ is a nonnegative Lebesgue measurable function defined on a Lebesgue measurable set $E$ then we defined the Lebesgue integral of $f$ on $E$ to be:

(1)
\begin{align} \quad \int_E f = \sup \left \{ \int_{E_{\varphi}} \varphi : \varphi \: \mathrm{is \: bounded \: and \: Lebesgue \: measurable}, 0 \leq \varphi(x) \leq f(x) \: \mathrm{on} \: E, \: E_{\varphi} \subseteq E \: \mathrm{with} \: m(E_{\varphi}) < \infty, \: \varphi \: \mathrm{vanishes \: outside \: of \:} E_{\varphi} \right \} \end{align}

We now show that such Lebesgue integrals have the linearity property when $\alpha, \beta \geq 0$

 Theorem 1: Let $f$ and $g$ be a nonnegative Lebesgue measurable functions defined on a Lebesgue measurable set $E$. Then for all $\alpha, \beta \in \mathbb{R}$, $\alpha, \beta \geq 0$ we have that $\displaystyle{\int_E (\alpha f + \beta g) = \alpha \int_E f + \beta \int_E g}$.
• Proof: We break this proof into two parts.
• Part 1: We first show that for all $\alpha \in \mathbb{R}$ that $\displaystyle{\int_E \alpha f = \alpha \int_E f}$. We break this up into two cases.
• Case 1: If $\alpha > 0$ then:
(2)
\begin{align} \quad \int_E \alpha f &= \sup \left \{ \int_{E_{\varphi}} \varphi : \varphi \: \mathrm{is \: bdd. \: and \: Leb. \: meas.}, 0 \leq \varphi(x) \leq \alpha f(x) \: \mathrm{on} \: E, \: E_{\varphi} \subseteq E \: \mathrm{with} \: m(E_{\varphi}) < \infty, \: \varphi \: \mathrm{van. \: out. \: of \:} E_{\varphi} \right \} \\ &= \sup \left \{ \int_{E_{\varphi}} \varphi : \varphi \: \mathrm{is \: bdd. \: and \: Leb. \: meas.}, 0 \leq \frac{1}{\alpha} \varphi(x) \leq f(x) \: \mathrm{on} \: E, \: E_{\varphi} \subseteq E \: \mathrm{with} \: m(E_{\varphi}) < \infty, \: \varphi \: \mathrm{van. \: out. \: of \:} E_{\varphi} \right \} \\ &= \sup \left \{ \int_{E_{\varphi}} \alpha \varphi : \varphi \: \mathrm{is \: bdd. \: and \: Leb. \: meas.}, 0 \leq \varphi(x) \leq f(x) \: \mathrm{on} \: E, \: E_{\varphi} \subseteq E \: \mathrm{with} \: m(E_{\varphi}) < \infty, \: \varphi \: \mathrm{van. \: out. \: of \:} E_{\varphi} \right \} \\ &= \sup \left \{ \alpha \int_{E_{\varphi}} \varphi : \varphi \: \mathrm{is \: bdd. \: and \: Leb. \: meas.}, 0 \leq \varphi(x) \leq f(x) \: \mathrm{on} \: E, \: E_{\varphi} \subseteq E \: \mathrm{with} \: m(E_{\varphi}) < \infty, \: \varphi \: \mathrm{van. \: out. \: of \:} E_{\varphi} \right \} \\ &= \alpha \sup \left \{ \int_{E_{\varphi}} \alpha \varphi : \varphi \: \mathrm{is \: bdd. \: and \: Leb. \: meas.}, 0 \leq \varphi(x) \leq f(x) \: \mathrm{on} \: E, \: E_{\varphi} \subseteq E \: \mathrm{with} \: m(E_{\varphi}) < \infty, \: \varphi \: \mathrm{van. \: out. \: of \:} E_{\varphi} \right \} \\ &= \alpha \int_E f \end{align}
• Case 2: If $\alpha = 0$ then:
(3)
\begin{align} \int_E \alpha f &= \int_E 0 \\ &= \sup \left \{ \int_{E_{\varphi}} \alpha \varphi : \varphi \: \mathrm{is \: bdd. \: and \: Leb. \: meas.}, 0 \leq \varphi(x) \leq 0 \: \mathrm{on} \: E, \: E_{\varphi} \subseteq E \: \mathrm{with} \: m(E_{\varphi}) < \infty, \: \varphi \: \mathrm{van. \: out. \: of \:} E_{\varphi} \right \} \\ &= 0 \end{align}
• Therefore $\displaystyle{\int_E \alpha f = \int_E 0 f = 0 = 0 \int_E f = \alpha \int_E f}$.
• Part 2: We now show if $f$ and $g$ are nonnegative Lebesgue measurable functions defined on a Lebesgue measurable set $E$ then $\displaystyle{\int_E (f + g) = \int_E f + \int_E g}$.
• Now let $\varphi(x) \leq f(x)$ be a bounded, Lebesgue measurable function that vanishes outside of $E_{\varphi}$ with $m(E_{\varphi}) < \epsilon$ and let $\psi(x) \leq g(x)$ be a bounded, Lebesgue measurable function that vanishes outside of $E_{\psi}$ with $m(E_{\psi}) < \infty$. Then $\varphi(x) + \psi(x) \leq f(x) + g(x)$ and $\varphi + \psi$ is a Lebesgue measurable function that vanishes outside of $E_{\varphi} \cup E_{\psi}$. Therefore:
(4)
\begin{align} \quad \int_E (f + g) & \geq \int_{E_{\varphi} \cup E_{\psi}} (\varphi + \psi) \\ & \geq \int_{E_{\varphi} \cup E_{\psi}} \varphi + \int_{E_{\varphi} \cup E_{\psi}} \psi \\ & \geq \int_{E_{\varphi}} \varphi + \underbrace{\int_{(E_{\varphi} \cup E_{\psi}) \setminus E_{\varphi}} \varphi}_{=0} + \int_{E_{\psi}} \psi + \underbrace{\int_{(E_{\varphi} \cup E_{\psi}) \setminus E_{\psi}} \psi}_{=0} \\ & \geq \int_{E_{\varphi}} \varphi + \int_{E_{\psi}} \psi \\ & \geq \sup \left \{ \int_{E_{\varphi}} \varphi + \int_{E_{\psi}} \psi : ... \right \} \\ & \geq \sup \left \{ \int_{E_{\varphi}} \varphi : ... \right \} + \sup \left \{ \int_{E_{\psi}} \psi : ... \right \} \\ & \geq \int_E f + \int_E g \quad (*) \end{align}
• Now let $\varphi$ be a bounded Lebesgue measurable function with $0 \leq \varphi(x) \leq f(x) + g(x)$ for all $x \in E$ that vanishes outside of $E_{\varphi}$ with $m(E_{\varphi}) < \infty$. Define two new functions as follows:
(5)
\begin{align} \quad \varphi_1 &= \min \{ \varphi, f \} \\ \quad \varphi_2 &= \varphi - \varphi_1 \end{align}
• Then $\varphi_1$ and $\varphi_2$ are both bounded Lebesgue measurable sets that vanish outside a set of finite measure. Let $x \in E$. If $\varphi(x) \leq f(x)$ then $\varphi_1(x) = \varphi(x)$. So $\varphi_2(x) = 0$. If $\varphi(x) \geq f(x)$ then $\varphi_1(x) = f(x)$ and $\varphi_2(x) = \varphi(x) - f(x) \leq g(x)$. In both cases we see that $\varphi_2(x) \leq g(x)$ for all $x \in E$.
• So $\varphi_1(x) \leq f(x)$ for all $x \in E$ and $\varphi_2(x) \leq g(x)$ for all $x \in E$. Furthermore, $\varphi(x) = \varphi_1(x) + \varphi_2(x)$. We have that:
(6)
\begin{align} \quad \int_{E_{\varphi}} \varphi = \int_{E_{\varphi}} (\varphi_1 + \varphi_2) = \int_{E_{\varphi}} \varphi_1 + \int_{E_{\varphi}} \varphi_2 = \int_{E_{\varphi_1}} \varphi_1 + \int_{E_{\varphi_2}} \varphi_2 \leq \int_E f + \int_E g \end{align}
• Therefore:
(7)
\begin{align} \quad \int_E (f + g) = \sup \left \{ \int_{E_{\varphi}} \varphi : ... \right \} \leq \int_E f + \int_E g \quad (**) \end{align}
• From $(*)$ and $(**)$ we conclude that:
(8)
\begin{align} \quad \int_E (f + g) = \int_E f + \int_E g \quad \blacksquare \end{align}