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:

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

$\blacksquare$

Sources

• 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $9.5$