The Integral of Measurable Functions

Recall that if $f$ is a function and $f^+ = \max \{ f, 0 \}$ and $f^- = \max \{ -f, 0\}$ then:

(1)

\begin{align} \quad f = f^+ - f^- \end{align}

And:

(2)

\begin{align} \quad |f| = f^+ + f^- \end{align}

We have already defined the integral for a nonnegative measurable function. From this definition, we can define the integral for any measurable function from a measure space $(X, \mathcal A, \mu)$.

Definition: Let $(X, \mathcal A, \mu)$ be a complete measure space and let $f$ be a measurable function defined on a measurable set $E$. Then the Integral of $f$ on $E$ is defined as $\displaystyle{\int_E f(x) \: d \mu = \int_E f^+(x) \: d\mu - \int_E f^-(x) \: d \mu}$ where $f^+(x) = \max \{ f(x), 0 \}$ and $f^-(x) = \max \{ -f(x), 0 \}$.

Definition: Let $(X, \mathcal A, \mu)$ be a complete measure space and let $f$ be a measurable function defined on a measurable set $E$. Then $f$ is said to be Integrable on $E$ if $\displaystyle{\int_E f^+(x) \: d \mu < \infty}$ and $\displaystyle{\int_E f^-(x) \: d \mu < \infty}$.

We now state an equivalent statement for a function $f$ to be integrable.

Theorem 1: Let $(X, \mathcal A, \mu)$ be a complete measure space and let $E$ be a measurable set. Then $f$ is integrable on $E$ if and only if $f$ is a measurable function on $E$ and $\displaystyle{\int_E |f(x)| \: d \mu < \infty}$.

• Proof: $\Rightarrow$ If $f$ is integrable then by definition, $f$ is a measurable function and:

(3)

\begin{align} \quad \int_E f^+(x) \: d \mu < \infty, \quad \mathrm{and} \quad \int_E f^-(x) \: d \mu < \infty \end{align}

• Therefore:

(4)

\begin{align} \quad \int_E |f(x)| \: d \mu = \int_E f^+(x) \: d \mu + \int_E f^-(x) \: d \mu < \infty \end{align}

• Proof: $\Leftarrow$ Suppose that $f$ is a measurable function on $E$ and $\displaystyle{\int_E |f(x)| \: d \mu < \infty}$. Then:

(5)

\begin{align} \quad \int_E f^+(x) \: d \mu \leq \int_E |f(x)| \: d \mu < \infty \quad \mathrm{and} \quad \int_E f^-(x) \: d \mu \leq \int_E |f(x)| \: d \mu < \infty \end{align}

• So $f$ is integrable on $E$. $\blacksquare$

We now state some basic theorems about integrable and measurable functions.

Theorem 1 (The Linearity of Integrable Functions): Let $(X, \mathcal A, \mu)$ be a complete measure space and let $f$ and $g$ be INTEGRABLE functions defined on a measurable set $E$. Then for all $\alpha, \beta \in \mathbb{R}$, $\alpha f + \beta g$ is integrable on $E$ and $\displaystyle{\int_E [\alpha f(x) + \beta g(x)] \: d \mu = \alpha \int_E f(x) \: d \mu + \beta \int_E g(x) \: d \mu}$.

Theorem 2 (The Monotonicity Property of Measurable Functions): Let $(X, \mathcal A, \mu)$ be a complete measure space and let $f$ and $g$ be measurable functions defined on a measurable set $E$ such that $f(x) \leq g(x)$ on $E$. Then $\displaystyle{\int_E f(x) \: d \mu \leq \int_E g(x) \: d \mu}$.

Theorem 3 (The Finite Additivity Over Domains Property of Integrable Functions): Let $(X, \mathcal A, \mu)$ be a complete measure space and let $f$ be a measurable function defined on the measurable sets $A$ and $B$ where $A$ and $B$ are mutually disjoint. Then $\displaystyle{\int_{A \cup B} f(x) \: d \mu = \int_A f(x) \: d \mu + \int_B f(x) \: d \mu}$.

Theorem 4 (The Countable Additivity Over Domains Property of Integrable Functions): Let $(X, \mathcal A, \mu)$ be a complete measure space and let $f$ be a measurable function defined on the measurable sets $(E_n)_{n=1}^{\infty}$ where $(E_n)_{n=1}^{\infty}$ are mutually disjoint. Then $\displaystyle{\int_{\bigcup_{n=1}^{\infty} E_n} f(x) \: d \mu = \sum_{n=1}^{\infty} \int_{E_n} f(x) \: d \mu}$.