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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{M}} .

Proof

Let Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f = \sum_{i=1}^n a_i 1_{E_i}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c=0} . Then Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle cf = 0} , implying Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int cf = 0} . Similarly, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c \int f = 0} . Thus, the first statement holds for this case.

Suppose Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c \neq 0} . Then

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_i} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F_j} as unions of disjoint sets as follows

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_i = \cup_{j=1}^m (E_i \cap F_j)} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F_j = \cup_{i=1}^n (F_j \cap E_i).}

Then

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int f + \int g = \sum_{i=1}^n a_i \mu(E_i) + \sum_{j=1}^m b_j \mu(F_j)}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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))}

which by countable additivity,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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)}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = \sum_{i,j=1}^{n,m} (a_i+b_j) \mu(E_i \cap F_j)}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = \int (f + g). }

It is worth noting that this may not be the standard representation for the integral of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f+g} .

As for the third statement, if Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f \leq g} , then whenever Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_i \cap F_j \neq \emptyset} , it must be that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \nu(A) = \int_A f d\mu} . Now we show Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \nu} satisfies all the measure properties. Notice that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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

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