Integral of Positive Measurable Function is Monotone/Corollary

Corollary to Integral of Positive Measurable Function is Monotone

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

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

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

From the definition of $\mu$-integration over $A$, we have:

$\ds \int_A f \rd \mu = \int \paren {\chi_A \times f} \rd \mu$

and:

$\ds \int_A g \rd \mu = \int \paren {\chi_A \times g} \rd \mu$

We show that:

$f \times \chi_A \le g \times \chi_A$

If $x \in A$, we have:

\(\ds \map {\paren {f \times \chi_A} } x\)

\(=\)

\(\ds \map f x \map {\chi_A} x\)

Definition of Pointwise Multiplication of Mappings

\(\ds \)

\(=\)

\(\ds \map f x\)

Definition of Characteristic Function of Set

and:

\(\ds \map {\paren {g \times \chi_A} } x\)

\(=\)

\(\ds \map g x \map {\chi_A} x\)

Definition of Pointwise Multiplication of Mappings

\(\ds \)

\(=\)

\(\ds \map g x\)

Definition of Characteristic Function of Set

So:

$\map {\paren {f \times \chi_A} } x \le \map {\paren {g \times \chi_A} } x$ for all $x \in A$.

Now take $x \in X \setminus A$.

We have:

\(\ds \map {\paren {f \times \chi_A} } x\)

\(=\)

\(\ds \map f x \map {\chi_A} x\)

Definition of Pointwise Multiplication of Mappings

\(\ds \)

\(=\)

\(\ds 0\)

Definition of Characteristic Function of Set

and:

\(\ds \map {\paren {g \times \chi_A} } x\)

\(=\)

\(\ds \map g x \map {\chi_A} x\)

Definition of Pointwise Multiplication of Mappings

\(\ds \)

\(=\)

\(\ds 0\)

Definition of Characteristic Function of Set

So:

$\map {\paren {f \times \chi_A} } x \le \map {\paren {g \times \chi_A} } x$ for all $x \in X \setminus A$

giving:

$f \times \chi_A \le g \times \chi_A$

From Integral of Positive Measurable Function is Monotone, we therefore have:

$\ds \int \paren {f \times \chi_A} \rd \mu \le \int \paren {g \times \chi_A} \rd \mu$

From the definition of $\mu$-integration over $A$, we have:

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

$\blacksquare$