Additivity over Dom. of Int. of the Leb. Int. of Leb. Int. Functs.

# Additivity over Domains of Integration of the Lebesgue Integral of Lebesgue Integrable Functions

Theorem 1: Let $f$ be a Lebesgue integrable function defined on the Lebesgue measurable set $E$. If $A, B \subseteq E$ are disjoint Lebesgue measurable subsets of $E$ then $f$ is Lebesgue integrable on $A$ and $B$ and $\displaystyle{\int_{A \cup B} f = \int_A f + \int_B f}$. |

**Proof:**Let $f$ be Lebesgue integrable on a Lebesgue measurable set $E$ and let $A, B \subseteq E$ be Lebesgue measurable with $A \cap B = \emptyset$. Let $\chi_A$ and $\chi_B$ denote the characteristic functions of $A$ and $B$ respectively, on $E$.

- Note that for all $x \in A$ and for all $x \in B$ we respectively have that:

\begin{align} \quad | \chi_A(x) f(x) | \leq |f(x)| \quad \mathrm{and} \quad | \chi_B(x) f(x) | \leq |f(x)| \end{align}

- Since $f$ is Lebesgue integrable on $E$ and hence on $A$ and $B$ we have that $|f|$ is Lebesgue integrable on $E$ and on $A$ and $B$. So by The Comparison Test for Lebesgue Integrability we have that [$\chi_A(x) f(x)$ and $\chi_B(x) f(x)$ are Lebesgue integrable on $A$ and $B$ respectively.

- Now note that $f = \chi_A f + \chi_B f$ on $A \cup B$. Therefore by the linearity property of the Lebesgue integral for Lebesgue integrable functions we have that:

\begin{align} \quad \int_{A \cup B} f &= \int_{A \cup B} (\chi_A f + \chi_B f) \\ &= \int_{A \cup B} \chi_A f + \int_{A \cup B} \chi_B f \\ &= \int_A \chi_A f + \int_B \chi_B f \\ &= \int_A f + \int_B f \end{align}