Simple Function

The simplest functions you will ever integrate, hence the name.

Definition

Let $(X,{\mathcal {M}},\mu )$ be a measure space. A measurable function $f:X\rightarrow \mathbb {R}$ is a simple function if $f(X)$ is a finite subset of $\mathbb {R}$. The standard representation for a simple function is given by

$f(x)=\sum _{i=1}^{n}c_{i}1_{E_{i}}(x)$,

where $1_{E_{i}}(x)$ is the indicator function on the disjoint sets $E_{i}=f^{-1}(\{c_{i}\})\in {\mathcal {M}}$ that partition $X$, where $f(X)=\{c_{1},\dots ,c_{n}\}$.

Integration of Simple Functions

These functions earn their name from the simplicity in which their integrals are defined. Let $L^{+}$ be the space of all measurable functions from $X$ to $[0,+\infty ].$ Then

$\int _{X}f(x)d\mu (x)=\sum _{i=1}^{n}c_{i}\mu (E_{i}),$

where by convention, we let $0\cdot \infty =0$. Note that $d\mu (x)$ is equivalent to $\mu (dx)$ and that some arguments may be omitted when there is no confusion.

Furthermore, for any $A\in {\mathcal {M}}$, we define

$\int _{A}f(x)d\mu (x)=\int _{X}f(x)1_{A}(x)d\mu (x)=\sum _{i=1}^{n}c_{i}\mu (E_{i}\cap A).$

Properties of Simple Functions

Given simple functions $f,g\in L^{+}$, the following are true:

• if $c\geq 0,\int cf=c\int f$;
• $\int (f+g)=\int f+\int g$;
• if $f\leq g$, then $\int f\leq \int g$;
• the function $A\mapsto \int _{A}f$ is a measure on ${\mathcal {M}}$.

Proof

Let $f=\sum _{i=1}^{n}a_{i}1_{E_{i}}$ and $g=\sum _{j=1}^{m}b_{j}1_{F_{j}}$ be simple functions with their corresponding standard representations.

We show the first claim. Suppose $c=0$. Then $cf=0$, implying $\int cf=0$. Similarly, $c\int f=0$. Thus, the first statement holds for this case.

Suppose $c\neq 0$. Then

$c\int f=c\sum _{i=1}^{n}a_{i}1_{E_{i}}=\sum _{i=1}^{n}ca_{i}1_{E_{i}}=\int cf$.

Next, we show the second statement. Notice that

Failed to parse (unknown function "\math"): {\displaystyle E_i = \cup_{j=1}^m (E_i \cap F_j)<\math> and <math>F_j = \cup_{i=1}^n (F_j \cap E_i).}

Then

$\int f+\int g=\sum _{i=1}^{n}a_{i}\mu (E_{i})+\sum _{j=1}^{m}b_{j}\mu (F_{j})$

${\displaystyle }$