16.1 Integration of Simple Functions

Definition 16.1.1 (Space of Simple Functions). Let $(X, \cm)$ be a measurable space, then $\Sigma(X, \cm) = \Sigma(X, \cm; \complex)$ is the space of $\complex$-valued simple functions on $(X, \cm)$, and $\Sigma^{+}(X, \cm)$ is the space of non-negative simple functions.

Definition 16.1.2 (Integral of Non-Negative Simple Functions). Let $(X, \cm, \mu)$ be a measure space and $f = \sum_{y \in f(X)}y \cdot \one_{\bracs{f = y}}\in \Sigma^{+}(X, \cm)$ be a non-negative simple function in standard form, thenWith the convention that $0 \cdot \infty = 0$.

\[\int f d\mu = \int f(x) \mu(dx) = \sum_{y \in f(X)}y \cdot \mu(\bracs{f = y})\]

is the Lebesgue integral of $f$.

Proposition 16.1.3. Let $(X, \cm, \mu)$ be a measure space and $f, g \in \Sigma^{*}(X, \cm)$, then:

For any $\alpha \ge 0$, $\int \alpha f d\mu = \alpha \int f d\mu$.
$\int f + g d\mu = \int f d\mu + \int g d\mu$.
If $f \le g$, then $\int f \le \int g$.
The mapping $A \mapsto \int \one_{A} \cdot f d\mu$ is a measure on $(X, \cm)$.

Proof. (1): If $\alpha = 0$, then $\int \alpha f d\mu = \int 0 d\mu = 0 = 0 \cdot \int f d\mu$. Otherwise, the mapping $y \mapsto \alpha y$ is a bijection. Hence

\[\alpha f = \sum_{y \in f(X)}(\alpha y) \cdot \one_{\bracs{f = y}}\]

is the standard form of $\alpha f$. Therefore

\[\int \alpha f d\mu = \sum_{y \in f(X)}(\alpha y) \cdot \mu({\bracs{f = y}}) = \alpha \int f d\mu\]

(2): Since $\mu$ is finitely additive,

\begin{align*}\int f d\mu + \int g d\mu&= \sum_{y \in f(X)}y \cdot \mu(\bracs{f = y}) + \sum_{y \in g(X)}y \cdot \mu(\bracs{g = y}) \\&= \sum_{y \in f(X)}\sum_{z \in g(X)}(y + z) \cdot \mu(\bracs{f = y, g = z}) \\&= \sum_{y \in (f + g)(X)}\sum_{{z \in f(X) \atop {z' \in g(X) \atop z + z' = y}}}(z + z') \cdot \mu(\bracs{f = z, g = z'}) \\&= \sum_{y \in (f + g)(X)}y \cdot \mu(\bracs{f + g = y}) = \int f + g d\mu\end{align*}

(3): Since $\mu$ is finitely additive,

\begin{align*}\int f d\mu&= \sum_{y \in f(X)}y \cdot \mu(\bracs{f = y}) = \sum_{y \in f(X)}\sum_{z \in g(X)}y \cdot \mu(\bracs{f = y, g = z}) \\&\le \sum_{y \in f(X)}\sum_{z \in g(X)}z \cdot \mu(\bracs{f = y, g = z}) = \sum_{z \in g(X)}z \cdot \mu\paren{\bracs{g = z}}= \int g d\mu\end{align*}

(4): Firstly, for any $A \in \cm$,

\[\one_{A} \cdot f = \sum_{y \in f(X)}y \cdot \one_{A \cap \bracs{f = y}}\]

so by (1),

\[\int \one_{A} \cdot f d\mu = \sum_{y \in f(X)}y \cdot \mu(A \cap \bracs{f = y})\]

Since $\mu$ is countably additive, for any $\seq{E_n}\subset \cm$ pairwise disjoint with $E = \bigcup_{n \in \natp}E_{n}$,

\[\int \one_{E} \cdot f d\mu = \sum_{y \in f(X)}y \cdot \one_{E \cap \bracs{f = y}}= \sum_{y \in f(X)}\sum_{n \in \natp}y \cdot \one_{E_n \cap \bracs{f = y}}= \sum_{n \in \natp}\int \one_{E_n}\cdot f d\mu\]

$\square$

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16.1 Integration of Simple Functions

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