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}