The Linearity Property of the Leb. Int. of Bounded, Leb. Meas. Functs.

# The Linearity Property of the Lebesgue Integral of Bounded, Lebesgue Measurable Functions

 Theorem 1 (Linearity of the Lebesgue Integral of Bounded, Lebesgue Measurable Functions): Let $f$ and $g$ be bounded, Lebesgue measurable functions defined on a Lebesgue measurable set $E$ with $m(E) < \infty$. Then for all $\alpha, \beta \in \mathbb{R}$, $\displaystyle{\int_E (\alpha f + \beta g) = \alpha \int_E f + \beta \int_E g}$.
• Proof: Since $f$ and $g$ are bounded, Lebesgue measurable functions defined on a Lebesgue measurable set $E$ with $m(E) < \infty$ we have that for all $\alpha, \beta \in \mathbb{R}$, $\alpha f + \beta g$ is a bounded, Lebesgue measurable function defined on $E$.
• We first show that for all $\alpha \in \mathbb{R}$ that $\displaystyle{\int_E \alpha f = \alpha \int_E f}$. We break this into three cases. We use the linearity of the Lebesgue integral for simple functions in the first two cases.
• Case 1 ($\alpha > 0$): If $\alpha > 0$ then:
(1)
\begin{align} \quad \int_E \alpha f &= \sup \left \{ \int_E \varphi : \varphi \: \mathrm{is \: a \: simple \: function}, \: \varphi(x) \leq \alpha f(x) \: \mathrm{on} \: E \right \} \\ &= \sup \left \{ \int_E \varphi : \varphi \: \mathrm{is \: a \: simple \: function}, \: \frac{\varphi(x)}{\alpha} \leq f(x) \: \mathrm{on} \: E \right \} \\ &= \sup \left \{ \int_E \alpha \varphi' : \varphi' \: \mathrm{is \: a \: simple \: function}, \: \varphi'(x) \leq f(x) \: \mathrm{on} \: E \right \} \\ &= a \sup \left \{ \int_E \varphi' : \varphi' \: \mathrm{is \: a \: simple \: function}, \: \varphi'(x) \leq f(x) \: \mathrm{on} \: E \right \} \\ &= a \underline{\int_E} f \\ &= a \int_E f \end{align}
• Case 2 ($\alpha < 0$): If $\alpha < 0$ then:
(2)
\begin{align} \quad \int_E \alpha f &= \sup \left \{ \int_E \varphi : \varphi \: \mathrm{is \: a \: simple \: function}, \: \varphi(x) \leq \alpha f(x) \: \mathrm{on} \: E \right \} \\ &= \sup \left \{ \int_E \varphi : \varphi \: \mathrm{is \: a \: simple \: function}, \: \frac{\varphi(x)}{\alpha} \geq f(x) \: \mathrm{on} \: E \right \} \\ &= \sup \left \{ \int_E a \varphi' : \varphi' \: \mathrm{is \: a \: simple \: function}, \: \varphi'(x) \geq f(x) \: \mathrm{on} \: E \right \} \\ &= \sup \left \{ a \int_E \varphi' : \varphi' \: \mathrm{is \: a \: simple \: function}, \: \varphi'(x) \geq f(x) \: \mathrm{on} \: E \right \} \\ &= a \inf \left \{ a \int_E \varphi' : \varphi' \: \mathrm{is \: a \: simple \: function}, \: \varphi'(x) \geq f(x) \: \mathrm{on} \: E \right \} \\ &= a \overline{\int_E} f \\ &= a \int_E f \end{align}
• Case 3 ($\alpha = 0$): If $\alpha = 0$ then:
(3)
\begin{align} \quad 0 = \int_E 0 \cdot f = 0 \int_E f = 0 \end{align}
• We now show that $\displaystyle{\int_E (f + g) = \int_E f + \int_E g}$.
• Let $\varphi$ and $\varphi'$ be simple functions such that $\varphi(x) \leq f(x)$ and $\varphi'(x) \leq g(x)$ on $E$. Then $\varphi(x) + \varphi'(x) \leq f(x) + g(x)$ on $E$. Furthermore, the sum of two simple functions is a simple function, so $\varphi + \varphi'$ is a simple function. By the linearity of the Lebesgue integral of simple functions we have that:
(4)
\begin{align} \quad \int_E \varphi + \int_E \varphi' = \int_E (\varphi + \varphi') \leq \int_E (f + g) \end{align}
• Therefore:
(5)
\begin{align} \quad \int_E (f + g) & \geq \sup \left \{ \int_E \varphi + \int_E \varphi' : \varphi, \varphi' \: \mathrm{are \: simple \: functions}, \: \varphi(x) \leq f(x), \: \varphi'(x) \leq g(x) \: \mathrm{on} \: E \right \} \\ & \geq \sup \left \{ \int_E \varphi : \varphi \: \mathrm{is \: a \: simple \: function}, \: \varphi(x) \leq f(x) \: \mathrm{on} \: E \right \} + \sup \left \{ \int_E \varphi' : \varphi' \: \mathrm{is \: a \: simple \: function}, \: \varphi(x) \leq g(x) \: \mathrm{on} \: E \right \} \\ & \geq \int_E f + \int_E g \quad (*) \end{align}
• Note that since $f$ and $g$ are bounded, Lebesgue measurable functions, we have that $-f$ and $-g$ are also bounded, Lebesgue measurable functions. Therefore:
(6)
\begin{align} \quad \int_E (-f - g) \geq \int_E -f + \int_E -g \quad \Leftrightarrow -\int_E (f + g) \geq -\int_E f - \int_E g \quad \Leftrightarrow \int_E (f + g) \leq \int_E f + \int_E g \quad (**) \end{align}
• Combining $(*)$ and $(**)$ yields:
(7)
\begin{align} \quad \int_E (f + g) = \int_E f + \int_E g \quad \blacksquare \end{align}