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}