Simple Integral Homogeneity

Why

If we stack a rectangle on top of itself we have a rectangle twice the height. 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.

Result

The simple non-negative integral operator is homogenous over non-negative real values.

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 $s(\alpha f) = \alpha s(f)$ for all $\alpha \in [0, \infty)$ and $f \in \SimpleF_+(X)$. Toward this end, let $f \in \SimpleF_+(X)$ with the simple partition $\set{A_n} \subset \mathcal{A} $ and $\set{a_n} \subset [0, \infty]$.

First, let $\alpha \in (0, \infty)$. Then $\alpha f \in \SimpleF_+(X)$, with the simple partition $\set{A_n} \subset \mathcal{A} $ and $\set{\alpha a_n} \subset [0, \infty]$.

\[ s(\alpha f) = \sum_{i = 1}^{n} \alpha a_n \mu (A_i) = \alpha \sum_{i = 1}^{n} a_n \mu (A_i) = \alpha s(f). \]

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 $0 \cdot \infty = \infty \cdot 0 = 0$.