The Linearity Property of the Lebesgue Integral of Leb. Int. Functs.

# The Linearity Property of the Lebesgue Integral of Lebesgue Integrable Functions

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

(1)
\begin{align} \quad \int_E f = \int_E f^+ - \int_E f^- \end{align}

We said that $f$ is Lebesgue integrable on $E$ if $|f|$ is Lebesgue integrable on $E$, that is, $\displaystyle{\int_E |f| < \infty}$, and we proved that:

(2)
\begin{align} \quad f \: \mathrm{is \: Lebesgue \: measurable} \: \mathrm{and} \: \int_E |f| < \infty \quad \Leftrightarrow \quad \int_E f^+ < \infty \: \mathrm{and} \: \int_E f^- < \infty \end{align}

We now show that the Lebesgue integral for Lebesgue measurable functions has the linearity property.

 Theorem 1 (Linearity of the Lebesgue Integral of Lebesgue Measurable Functions): Let $f$ and $g$ be Lebesgue integrable functions defined on a Lebesgue measurable set $E$. Then for all $\alpha, \beta \in \mathbb{R}$, the function $\alpha f + \beta g$ is Lebesgue integrable on $E$ and $\displaystyle{\int_E (\alpha f + \beta g) = \alpha \int_E f + \beta \int_E g}$.
• Proof: Since $f$ and $g$ are Lebesgue integrable functions they are also Lebesgue measurable functions and we have that for all $\alpha, \beta \in \mathbb{R}$ that $\alpha f + \beta g$ is a Lebesgue measurable function. Furthermore we have that:
(3)
\begin{align} \quad |\alpha f + \beta g| \leq | \alpha | |f| + |\beta ||g| \end{align}
• Since $f$ is Lebesgue measurable and Lebesgue integrable we have that $\displaystyle{\int_E |f| < \infty}$. Similarly, since $g$ is Lebesgue measurable and Lebesgue integrable we have that $\displaystyle{\int_E |g| < \infty}$. Therefore:
(4)
\begin{align} \quad \int_E |\alpha f + \beta g| \leq |\alpha| \int_E |f| + |\beta| \int_E |g| < \infty \end{align}
• So $\alpha f + \beta g$ is a Lebesgue measurable function such that $\displaystyle{\int_E |\alpha f + \beta g| < \infty}$. So $\alpha f + \beta g$ is Lebesgue integrable on $E$.
• We now show that the linearity property holds. First, we show that for all $\alpha \in \mathbb{R}$ that $\displaystyle{\int_E \alpha f = \alpha \int_E}$. We break this up into three cases.
• Case 1: If $\alpha > 0$ then:
(5)
\begin{align} \quad \int_E \alpha f &= \int_E (\alpha f)^+ - \int_E (\alpha f)^- \\ &= \int_E \max \{ \alpha f, 0 \} - \int_E \max \{ -\alpha f, 0 \} \\ &= \int_E \alpha \max \{ f, 0 \} - \int_E \alpha \max \{ -f, 0 \} \\ &= \alpha \int_E \max \{ f, 0 \} - \alpha \int_E \max \{ -f, 0 \} \\ &= \alpha \left ( \int_E \max \{ f, 0\} - \int_E \max \{ -f, 0 \} \right ) \\ &= \alpha \left ( \int_E f^+ - \int_E f^- \right ) \\ &= \alpha \int_E f \end{align}
• Case 2: If $\alpha < 0$ then:
(6)
\begin{align} \quad \int_E \alpha f &= \int_E (\alpha f)^+ - \int_E (\alpha f)^- \\ &= \int_E \max \{ \alpha f, 0 \} - \int_E \max \{ -\alpha f, 0 \} \\ &= \int_E -\alpha \max \{ -f, 0 \} - \int_E -\alpha \max \{ f, 0 \} \\ &= - \alpha \int_E \max \{ -f, 0 \} + \alpha \int_E \max \{ f, 0 \} \\ &= \alpha \left ( -\int_E \max \{ -f, 0 \} + \int_E \max \{ f, 0 \} \right ) \\ &= \alpha \left ( -\int_E f^- + \int_E f^+ \right ) \\ &= \alpha \int_E f \end{align}
• Case 3: If $\alpha = 0$ then $\displaystyle{\int_E \alpha f = 0}$ so trivially $\displaystyle{\int_E \alpha f = \alpha \int_E f}$.
• We now show that the additivity property holds. We first prove an important result:
• If $f_1$, $f_2$, $g_1$, and $g_2$ are nonnegative Lebesgue integrable functions defined on a Lebesgue measurable set $E$ and $f_1 - f_2 = g_1 - g_2$ then:
(7)
\begin{align} \quad \int_E f_1 - \int_E f_2 = \int_E g_1 - \int_E g_2 \end{align}
• To prove this, we note that since $f_1 - f_2 = g_1 - g_2$ then $f_1 + g_2 = g_1 + f_2$. We have by the additivity of the Lebesgue integral for nonnegative Lebesgue measurable functions that:
(8)
\begin{align} \quad \int_E (f_1 + g_2) &= \int_E (g_1 + f_2) \\ \quad \int_E f_1 + \int_E g_2 &= \int_E g_1 + \int_E f_2 \\ \quad \int_E f_1 - \int_E f_2 &= \int_E g_1 - \int_E g_2 \end{align}
• Now let $f = f^+ - f^-$ and $g = g^+ - g^-$. Then $(f + g) = (f^+ + g^+) - (f^- + g^-)$. But also $(f + g) = (f + g)^+ - (f + g)^-$. Therefore:
(9)
\begin{align} \quad \int_E (f + g) &= \int_E (f + g)^+ - \int_E (f + g)^- \\ &= \int_E (f^+ + g^+) - \int_E (f^- + g^-) \\ &= \left ( \int_E f^+ - \int_E f^- \right ) + \left ( \int_E g^+ - \int_E g^- \right )\\ &= \int_E f + \int_E g \quad \blacksquare \end{align}