Integral of Positive Measurable Function is Additive/Corollary
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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 |
$\blacksquare$