The Integral of Nonnegative Simple Functions

# The Integral of Nonnegative Simple Functions

Recall from The Lebesgue Integral of Simple Functions page that if $\varphi$ is a simple function defined on a Lebesgue measurable set $E$ with $m(E) < \infty$ and with canonical representation:

(1)\begin{align} \quad \varphi(x) = \sum_{k=1}^{n} a_k \chi_{E_k}(x) \end{align}

Then we defined the Lebesgue integral of $\varphi$ on $E$ to be:

(2)\begin{align} \quad \int_E \varphi(x) \: dx = \sum_{k=1}^{n} a_k \mu (E_k) \end{align}

We now give a similar definition for NONNEGATIVE simple functions from a general measure space.

Definition: Let $(X, \mathcal A, \mu)$ be a complete measure space and let $\varphi$ be a nonnegative simple function defined on a measurable set $E$. We define the Integral of the Zero Function $\psi(x) = 0$ to be $\displaystyle{\int_E \psi(x) \: d \mu = 0}$. If $a_1, a_2, ..., a_n > 0$ are the POSITIVE values in the range of the simple $\varphi$ function and $\varphi(x) = a_k$ if and only if $x \in E_k \subseteq E$, then we defined the Integral of the Nonnegative Simple Function $\varphi$ on $E$ to be $\displaystyle{\int_E \varphi(x) \: d \mu = \sum_{k=1}^{n} a_k \mu (E_k)}$. |

We now state some important results regarding the integrals of nonnegative simple functions.

Theorem 1 (Linearity of the Integral of Nonnegative Simple Functions): Let $(X, \mathcal A, \mu)$ be a complete measure space and let $\varphi$ and $\psi$ be simple functions defined on a measurable set $E$. Then for all $\alpha, beta \in \mathbb{R}$ with $\alpha, \beta \geq 0$ we have that $\displaystyle{\int_E (\alpha \varphi(x) + \beta \psi(x)) \: d \mu = \alpha \int_E \varphi(x) \: d \mu + \beta \int_E \psi(x) \: d \mu}$. |

Theorem 2 (Monotonicity of the Integral of Nonnegative Simple Functions): Let $(X, \mathcal A, \mu)$ be a complete measure space and let $\varphi$ and $\psi$ be simple functions defined on a measurable set $E$. If $\varphi(x) \leq \psi(x)$ on $E$ then $\displaystyle{\int_E \varphi(x) \: d \mu \leq \int_E \psi(x) \: d \mu}$. |

Theorem 3: Let $(X, \mathcal A, \mu)$ be a complete measure space and let $\varphi$ be a simple function defined on the measurable sets $A$ and $B$. If $A$ and $B$ are mutually disjoint then $\displaystyle{\int_{A \cup B} \varphi(x) \: d \mu = \int_A \varphi(x) \: d \mu + \int_B \varphi(x) \: d \mu}$. |