Integral of Positive Measurable Function is Additive/Corollary

Theorem

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

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

Let $A \in \Sigma$.

Then:

$\ds \int_A \paren {f + g} \rd \mu = \int_A f \rd \mu + \int_A g \rd \mu$

where:

$f + g$ is the pointwise sum of $f$ and $g$

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

This can be summarized by saying that $\ds \int_A \cdot \rd \mu$ is (conventionally) additive.

Proof

We have:

\(\ds \int_A \paren {f + g} \rd \mu\)

\(=\)

\(\ds \int \paren {f + g} \times \chi_A \rd \mu\)

Definition of Integral of Positive Measurable Function over Measurable Set

\(\ds \)

\(=\)

\(\ds \int \paren {f \times \chi_A + g \times \chi_A} \rd \mu\)

\(\ds \)

\(=\)

\(\ds \int \paren {f \times \chi_A} \rd \mu + \int \paren {g \times \chi_A} \rd \mu\)

Integral of Positive Measurable Function is Additive

\(\ds \)

\(=\)

\(\ds \int_A f \rd \mu + \int_A g \rd \mu\)

Definition of Integral of Positive Measurable Function over Measurable Set

$\blacksquare$