Sum of Integrals on Complementary Sets

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Theorem

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

Let $A, E \in \Sigma$ with $A \subseteq E$.

Let $f$ be a $\mu$-integrable function on $X$.


Then

$\ds \int_E f \rd \mu = \int_A f \rd \mu + \int_{E \mathop \setminus A} f \rd \mu$


Proof

Let $\chi_E$ be the characteristic function of $E$.

Because $A$ and $E \setminus A$ are disjoint:

$A \cap \paren {E \setminus A} = \O$

By Characteristic Function of Union:

$\chi_E = \chi_A + \chi_{E \mathop \setminus A}$


Integrating $f$ over $E$ gives:

\(\ds \int_E f \rd \mu\) \(=\) \(\ds \int \chi_E \cdot f \rd \mu\) Definition of Integral of Integrable Function over Measurable Set
\(\ds \) \(=\) \(\ds \int \paren {\chi_A + \chi_{E \mathop \setminus A} } \cdot f \rd \mu\) by the above argument
\(\ds \) \(=\) \(\ds \int \paren {\chi_A \cdot f + \chi_{E \mathop \setminus A} \cdot f} \rd \mu\) Pointwise Operation on Distributive Structure is Distributive
\(\ds \) \(=\) \(\ds \int \chi_A \cdot f \rd \mu + \int \chi_{E \mathop \setminus A} \cdot f \rd \mu\) Integral of Integrable Function is Additive
\(\ds \) \(=\) \(\ds \int_A f \rd \mu + \int_{E \mathop \setminus A} f \rd \mu\) Definition of Integral of Integrable Function over Measurable Set

$\blacksquare$


Also see


Sources

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