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 $f(X)=\{c_{1},\dots ,c_{n}\}$.

Examples

Consider the functions $f,g:\mathbb {R} \rightarrow \mathbb {R}$ determined by $f(x)=2$ and $g(x)=x$. The function $f(x)$ is a simple function and can be expressed in the following manner:

$$f(x)=1\cdot 1_{\mathbb {R} }+1\cdot 1_{\mathbb {R} }=2\cdot 1_{(-\infty ,0]}+2\cdot 1_{[0,\infty )}=2\cdot 1_{\mathbb {R} }.$$

The last of these representations is the standard representation of $f(x)$. The function $g(x)$ is not a simple function as $\{g(x):x\in \mathbb {R} \}=\mathbb {R}$.

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 we can rewrite $E_{i}$ and $F_{j}$ as unions of disjoint sets as follows

$E_{i}=\cup _{j=1}^{m}(E_{i}\cap F_{j})$ and $F_{j}=\cup _{i=1}^{n}(F_{j}\cap E_{i}).$

Then

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

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

which by countable additivity,

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

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

$=\int (f+g).$

It is worth noting that this may not be the standard representation for the integral of $f+g$.

As for the third statement, if $f\leq g$, then whenever $E_{i}\cap F_{j}\neq \emptyset$, it must be that $a_{i}\leq b_{j}$, implying that

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

Finally, we show the last statement. Define $\nu (A)=\int _{A}fd\mu$. Now we show $\nu$ satisfies all the measure properties. Notice that $\nu$ is a nonnegative function on ${\mathcal {M}}$. Then compute

$\nu (\emptyset )=\int _{\emptyset }fd\mu =\int f1_{\emptyset }d\mu =\int f\cdot 0d\mu =0.$

Consider a disjoint sequence of sets $\{A_{k}\}_{k\in \mathbb {N} }$ and let $A$ be its union. Then

$\nu (A)=\int _{A}fd\mu =\int f1_{A}d\mu =\sum _{i=1}^{n}a_{i}\mu (E_{i}\cap A)=\sum _{i=1}^{n}a_{i}\mu (E_{i}\cap (\cup _{k=1}^{\infty }A_{k})),$

which by countable additivity is equal to

$=\sum _{i=1}^{n}\sum _{k=1}^{\infty }a_{i}\mu (E_{i}\cap A_{k})=\sum _{k=1}^{\infty }\sum _{i=1}^{n}a_{i}\mu (E_{i}\cap A_{k})=\sum _{k=1}^{\infty }\int f1_{A_{k}}=\sum _{k=1}^{\infty }\int _{A_{k}}f=\sum _{k=1}^{\infty }\nu (A_{k}).$