Sum of Integrals on Complementary Sets

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\left({X, \Sigma, \mu}\right)$ be a measure space.

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

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


Then

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


Proof

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

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

$A \cap \left({E \setminus A}\right) = \varnothing$

By Characteristic Function of Union:

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


Integrating $f$ over $E$ gives:

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

$\blacksquare$


Also see


Sources