Simple Integral Additivity
Why
If we stack two rectangles, with equal base lengths but different heights, on top of each other, the additivity principle says that the area of the so-formed rectangle is the sum of the areas of the stacked rectangles. Our definition of integral for simple functions has this property, as it ought to.
Result
Let $(X, \mathcal{A} , \mu )$be a measure space. Let $\SimpleF_+(X)$ denote the non-negative real-valued simple functions on $X$. Define $s: \SimpleF_+(X) \to [0, \infty]$ by $s(f) = \int f d\mu $ for $f \in \SimpleF_+{X}$.
In this notation, we want to show that$s(f+g)
= s(f) + s(g)$ for all $f, g \in
\SimpleF_+(X)$.
Toward this end, let $f,g \in \SimpleF_+(X)$
with the simple partitions:
\[
\set{A_i}_{i = 1}^{m},\set{B_j}_{j = 1}^{n} \subset
\mathcal{A} \quad \text{and} \quad\set{a_i}_{i =
1}^{m},\set{b_j}_{j = 1}^{n} \subset [0, \infty].
\]
We consider the refinement of the two partitions. $\set{A_i \cap B_j}_{i, j = 1}^{i = m, j = n}$.
First, let $\alpha \in (0, \infty)$. Then $\alpha f \in \SimpleF_+(X)$,with the simple partition$\set{A_n} \subset \mathcal{A} $ and $\set{\alpha a_n} \subset [0, \infty]$.
\[ \begin{aligned} s(\alpha f) &= \sum_{i = 1}^{n} \alpha a_n \mu (A_i) \\ &= \alpha \sum_{i = 1}^{n} a_n \mu (A_i) \\ &= \alpha s(f). \end{aligned} \]
If $\alpha = 0$, then $\alpha f$ is uniformly zero; it is the non-negative simple with partition $\set{X}$ and $\set{0}$. Regardless of the measure of $X$, this non-negative simple function is zero. Recall that we define $0\cdot \infty = \infty\cdot 0 = 0$.