Integral of Positive Measurable Function is Monotone

Theorem

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

Let $f, g: X \to \overline \R$ be positive $\Sigma$-measurable functions.

Suppose that $f \le g$, where $\le$ denotes pointwise inequality.

Then:

$\ds \int f \rd \mu \le \int g \rd \mu$

where the integral sign denotes $\mu$-integration.

This can be summarized by saying that $\ds \int \cdot \rd \mu$ is monotone.

Corollary

Let $A \in \Sigma$.

Suppose that $f \le g$, where $\le$ denotes pointwise inequality.

Then:

$\ds \int_A f \rd \mu \le \int_A g \rd \mu$

where the integral sign denotes $\mu$-integration over $A$.

This can be summarized by saying that $\ds \int_A \cdot \rd \mu$ is monotone.

Proof

By the definition of $\mu$-integration, we have:

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

and:

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

where:

$\EE^+$ denotes the space of positive simple functions

$\map {I_\mu} g$ denotes the $\mu$-integral of the positive simple function $g$.

Since $f \le g$:

if $h \in \EE^+$ satisfies $h \le f$ then it satisfies $h \le g$.

So:

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

From Supremum of Subset, we therefore have:

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

That is:

$\ds \int f \rd \mu \le \int g \rd \mu$

$\blacksquare$

Sources

• 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $9.8 \ \text{(iv)}$, $\S 9$: Problem $2$