Simple Function

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

Definition

Let ${\displaystyle (X,{\mathcal {M}},\mu )}$ be a measure space. A measurable function ${\displaystyle f:X\rightarrow \mathbb {R} }$ is a simple function[1] if ${\displaystyle f(X)}$ is a finite subset of ${\displaystyle \mathbb {R} }$. The standard representation[1] for a simple function is given by

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

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

Examples

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

${\displaystyle f(x)=1\cdot 1_{\mathbb {R} }+1\cdot 1_{\mathbb {R} }=2\cdot 1_{(-\infty ,0]}+2\cdot 1_{[0,\infty )}=2\cdot 1_{\mathbb {R} }.}$
The last of these representations is the standard representation of ${\displaystyle f(x)}$. The function ${\displaystyle g(x)}$ is not a simple function as ${\displaystyle \{g(x):x\in \mathbb {R} \}=\mathbb {R} }$.

Integration of Simple Functions

These functions earn their name from the simplicity in which their integrals are defined[3]. Let ${\displaystyle L^{+}}$ be the space of all measurable functions from ${\displaystyle X}$ to ${\displaystyle [0,+\infty ].}$ Then

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

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

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

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

Properties of Simple Functions

Given simple functions ${\displaystyle f,g\in L^{+}}$, the following are true[3]:

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

Proof[4]

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

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

Suppose ${\displaystyle c\neq 0}$. Then

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

Next, we show the second statement. Notice that we can rewrite ${\displaystyle E_{i}}$ and ${\displaystyle F_{j}}$ as unions of disjoint sets as follows

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

Then

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

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

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

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

${\displaystyle =\int (f+g).}$

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

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

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

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

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

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

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

which by countable additivity is equal to

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

References

1. Craig, Katy. MATH 201A Lecture 11. UC Santa Barbara, Fall 2020.
2. Craig, Katy. MATH 201A Lecture 12. UC Santa Barbara, Fall 2020.
3. Folland, Gerald B. (1999). Real Analysis: Modern Techniques and Their Applications, John Wiley and Sons, ISBN 0471317160, Second edition.
4. Craig, Katy. MATH 201A Lectures 12-13. UC Santa Barbara, Fall 2020.