Integral of Positive Measurable Function Extends Integral of Positive Simple Function

Theorem
Let $\struct {X, \Sigma, \mu}$ be a measure space.

Let $f: X \to \R, f \in \EE^+$ be a positive simple function.

Then:
 * $\ds \int f \rd \mu = \map {I_\mu} f$

where:


 * $\ds \int \cdot \rd \mu$ denotes the $\mu$-integral of positive measurable functions
 * $I_\mu$ denotes the $\mu$-integral of positive simple functions.

That is:
 * $\ds \int \cdot \rd \mu \restriction_{\EE^+} = I_\mu$

using the notion of restriction, $\restriction$.

Proof
From the definition of the integral of a positive measure function, we have:


 * $\ds \int f \rd \mu = \sup \set {\map {I_\mu} g: g \le f, g \in \EE^+}$

Let $g \in \EE^+$ be such that $g \le f$.

Then, from Integral of Positive Simple Function is Increasing, we have:


 * $\map {I_\mu} g \le \map {I_\mu} f$

So $\map {I_\mu} f$ is an upper bound of $\set {\map {I_\mu} g: g \le f, g \in \EE^+}$.

Since $f \in \EE^+$, and $f \le f$, we have:


 * $\map {I_\mu} f \in \set {\map {I_\mu} g: g \le f, g \in \EE^+}$

So:


 * $\map {I_\mu} f$ is the greatest element of $\set {\map {I_\mu} g: g \le f, g \in \EE^+}$.

so, from Greatest Element is Supremum:


 * $\map {I_\mu} f = \sup \set {\map {I_\mu} g: g \le f, g \in \EE^+}$

So:


 * $\ds \int f \rd \mu = \map {I_\mu} f$