The Lebesgue Integral of a Simp. Funct. Defined on a Uni. of Disj. Sets

# The Lebesgue Integral of a Simple Function Defined on a Union of Disjoint Sets

We have already discussed the linearity and monotonicity properties of the Lebesgue integral of simple functions on the following pages:

We now state another result which says that if $\varphi$ is a simple function defined on a Lebesgue measurable set $A \cup B$ with $m(A \cup B) < \infty$ and $A$ and $B$ are mutually disjoint, then the Lebesgue integral of $\varphi$ of $A \cup B$ is the sum of the Lebesgue integrals of $\varphi$ on $A$ and on $B$.

 Theorem 1: Let $\varphi$ be a simple function defined on a Lebesegue measurable set $A \cup B$ with $m(A \cup B) < \infty$ where $A$ and $B$ are disjoint then $\displaystyle{\int_{A \cup B} \varphi = \int_A \varphi + \int_B \varphi}$.
• Proof: Let $\varphi$ be a simple function defined on a Lebesegue measurable set $A \cup B$ with $m(A \cup B) < \infty$ and let $\varphi(x)$ have the following canonical representation:
(1)
\begin{align} \quad \varphi(x) = \sum_{k=1}^{n} a_k \chi_{E_k} (x) \end{align}
• Consider the sets $A \cap E_1, A \cap E_2, ..., A \cap E_n, B \cap E_1, B \cap E_2, ..., B \cap E_n$. Note that these sets are mutually disjoint. Then:
(2)
\begin{align} \quad \varphi(x) &= \sum_{k=1}^{n} a_k \chi_{A \cap E_k} (x) \quad \mathrm{for \: all \:} x \in A \\ \quad \varphi(x) &= \sum_{k=1}^{n} a_k \chi_{B \cap E_k} (x) \quad \mathrm{for \: all \:} x \in B \end{align}
• Therefore:
(3)
\begin{align} \quad \int_{A \cup B} \varphi &= \sum_{k=1}^{n} a_k m(E_k) \\ &= \sum_{k=1}^{n} a_k [m(A \cap E_k) + m(B \cap E_k)] \\ &= \sum_{k=1}^{n} a_km(A \cap E_k) + \sum_{k=1}^{n} a_km(B \cap E_k) \\ &= \int_A \varphi + \int_B \varphi \quad \blacksquare \end{align}

The following corollary generalizes Theorem 1 above and can be proven with mathematical induction.

 Corollary 2: Let $\varphi$ be a simple function defined on a Lebesegue measurable set $\displaystyle{A = \bigcup_{k=1}^{n} A_k}$ with $m(A) < \infty$ where $A_1, A_2, ..., A_n$ are mutually disjoint then $\displaystyle{\int_{A} \varphi =\sum_{k=1}^{n} \int_{A_k} \varphi}$.