# Sum of Integrals on Complementary Sets

## 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

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

## Proof

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

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

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

Integrating $f$ over $E$ gives:

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

$\blacksquare$