Simple Functions

Finite-Valued Measurable Functions

In the previous tutorial, measurability was characterized by preimages of threshold sets. Here we focus on a particularly manageable class of measurable functions: those that take only finitely many values. Such functions can be described by listing their values and the measurable sets on which each value occurs. This description is useful throughout measure theory, where finite-valued functions provide simple building blocks for more general functions.

Let $(X,\mathcal{F})$ be a measurable space. For $A\subseteq X$, write $\mathbf{1}_A$ for the indicator function, which equals $1$ on $A$ and $0$ outside $A$. The measurability of indicator functions was established earlier in the course.

Definition: A function $s:X\to\mathbb{R}$ is a simple function if it is measurable and its range $s(X)$ is finite. Equivalently, it takes only finitely many distinct real values. The zero function is a simple function.

If $s$ takes the values $c_1,\ldots,c_m$, its level sets are the sets $A_i=\{x\in X:s(x)=c_i\}$. These sets are pairwise disjoint and cover $X$. Since each singleton $\{c_i\}$ is Borel, measurability of $s$ ensures that every $A_i=s^{-1}(\{c_i\})$ belongs to $\mathcal{F}$. Thus each point of $X$ belongs to exactly one of the level sets, and the value of $s$ on that set is fixed.

The Partition Representation

Theorem (Partition Representation of a Simple Function): A function $s:X\to\mathbb{R}$ is simple if and only if there are finitely many pairwise disjoint measurable sets $A_1,\ldots,A_m$ whose union is $X$, and real numbers $c_1,\ldots,c_m$, such that $$ s=\sum_{i=1}^{m}c_i\mathbf{1}_{A_i}. $$ The sets may be empty if a redundant value is included. When $X$ is empty, the representation is the empty sum.

Proof. Suppose first that $s$ is simple and $X$ is nonempty. List its distinct values as $c_1,\ldots,c_m$, and define $A_i=s^{-1}(\{c_i\})$. Each $A_i$ is measurable because $\{c_i\}$ is Borel and $s$ is measurable. Distinct level sets are disjoint: a point cannot have two different values under $s$. They cover $X$, since every $s(x)$ is one of the listed values. For any $x\in A_i$, the indicator $\mathbf{1}_{A_i}(x)$ equals $1$, while all the other indicators equal $0$. Therefore the sum at $x$ is $c_i=s(x)$, proving the representation.

Conversely, suppose a finite measurable partition $A_1,\ldots,A_m$ and real numbers $c_1,\ldots,c_m$ are given, and let $s=\sum_{i=1}^{m}c_i\mathbf{1}_{A_i}$. Each $x\in X$ lies in exactly one set $A_i$, so $s(x)=c_i$ there. In particular, the range of $s$ is finite. For any Borel set $B\subseteq\mathbb{R}$,

s^{-1}(B)=\bigcup_{\{i:c_i\in B\}}A_i.

This is a finite union of measurable sets, and hence belongs to $\mathcal{F}$. Thus $s$ is measurable and simple. If $X$ is empty, its unique real-valued function has empty range and is measurable; the empty-sum convention gives the stated representation. $\square$

The theorem uses a partition: the sets are disjoint and cover the entire domain. A finite sum of indicators of measurable sets need not initially be written on a partition, because its sets may overlap. But it can always be rewritten in partition form by recording which of the finitely many sets contain each point.

Worked Example: Reading Values from a Partition

Let $A,B\in\mathcal{F}$ be disjoint and suppose $A\cup B=X$. Define $s=4\mathbf{1}_A-2\mathbf{1}_B$. If $x\in A$, then $\mathbf{1}_A(x)=1$ and $\mathbf{1}_B(x)=0$, so $s(x)=4(1)-2(0)=4$. If $x\in B$, then $s(x)=4(0)-2(1)=-2$. Consequently $s$ has range contained in $\{4,-2\}$, and the partition representation theorem shows it is simple. Its level sets are exactly $A$ and $B$ if both sets are nonempty.

Worked Example: Refining Overlapping Sets

Suppose $A,B\in\mathcal{F}$, with no assumption that they are disjoint, and define $h=2\mathbf{1}_A+3\mathbf{1}_B$. The four measurable sets

A\cap B,\qquad A\setminus B,\qquad B\setminus A,\qquad X\setminus(A\cup B)

are pairwise disjoint and cover $X$. On $A\cap B$, both indicators equal $1$, so $h=2+3=5$. On $A\setminus B$, the indicators are $1$ and $0$, so $h=2$. On $B\setminus A$, they are $0$ and $1$, so $h=3$. Outside $A\cup B$, both equal $0$, so $h=0$. Thus

h=5\mathbf{1}_{A\cap B} +2\mathbf{1}_{A\setminus B} +3\mathbf{1}_{B\setminus A} +0\mathbf{1}_{X\setminus(A\cup B)}.

All four sets are measurable by closure of a sigma-algebra under intersections, complements, and relative differences. Empty pieces cause no problem. This example shows explicitly how overlaps change the value and how to produce a disjoint partition that records every possibility.

Nonnegative Simple Functions

A simple function is called nonnegative when $s(x)\geq 0$ for every $x\in X$. For such a function, it is often convenient to omit the zero level set and list only its positive values. This gives a representation with positive coefficients on disjoint measurable sets.

Corollary (Positive-Level Representation): If $s$ is a nonnegative simple function, then there are distinct positive values $a_1,\ldots,a_k$ and pairwise disjoint measurable sets $E_1,\ldots,E_k$ such that $$ s=\sum_{j=1}^{k}a_j\mathbf{1}_{E_j}. $$ The union of the $E_j$ is $\{x:s(x)>0\}$. If $s$ is identically zero, the sum is empty.

Proof. List the distinct positive values of $s$ as $a_1,\ldots,a_k$, and set $E_j=s^{-1}(\{a_j\})$. Each $E_j$ is measurable, and distinct level sets are disjoint. At a point where $s(x)>0$, exactly one of these sets contains $x$, and the corresponding coefficient is $s(x)$. At a point where $s(x)=0$, none contains $x$, so the displayed sum is zero. This proves the identity and the claim about the union. If there are no positive values, nonnegativity forces $s(x)=0$ everywhere. $\square$

Worked Example: A Nonnegative Step Function

On $\mathbb{R}$ with its Borel sigma-algebra, define $s$ to be $0$ on $(-\infty,-1)$, $2$ on $[-1,2)$, and $5$ on $[2,\infty)$. These three intervals are Borel, disjoint, and cover $\mathbb{R}$. The function is therefore simple, and its positive-level representation is

s=2\mathbf{1}_{[-1,2)}+5\mathbf{1}_{[2,\infty)}.

For $x=-2$, both indicators are zero and the formula gives $s(-2)=0$. For $x=0$, the first indicator is $1$ and the second is $0$, giving $s(0)=2$. For $x=3$, the first is $0$ and the second is $1$, giving $s(3)=5$. The endpoint $2$ belongs to $[2,\infty)$, so the formula gives $s(2)=5$, as required by the definition.

Arithmetic Operations Preserve Simplicity

Finite range is stable under basic arithmetic: combining two functions that each take only finitely many values cannot produce infinitely many possible sums or products. Measurability is preserved as well. The partition representation makes both facts transparent without requiring a separate general theorem about operations on measurable functions.

Theorem (Arithmetic Closure of Simple Functions): If $f,g:X\to\mathbb{R}$ are simple and $\alpha\in\mathbb{R}$, then $\alpha f$, $f+g$, and $fg$ are simple.

Proof. Write $f=\sum_{i=1}^{m}a_i\mathbf{1}_{A_i}$ and $g=\sum_{j=1}^{n}b_j\mathbf{1}_{B_j}$, where each family is a finite measurable partition of $X$. The sets $A_i\cap B_j$ are measurable, pairwise disjoint as $(i,j)$ varies, and cover $X$. On $A_i\cap B_j$, the functions $f$ and $g$ have the constant values $a_i$ and $b_j$. Therefore $\alpha f$, $f+g$, and $fg$ have the respective constant values $\alpha a_i$, $a_i+b_j$, and $a_ib_j$ on these finitely many sets. Each operation consequently gives a function with finite range and a measurable partition representation. By the partition representation theorem, each resulting function is simple. $\square$

Worked Example: Adding and Multiplying Simple Functions

Let $A\in\mathcal{F}$, and define $f=1+2\mathbf{1}_A$ and $g=3-\mathbf{1}_A$. On $A$, the indicator equals $1$, so $f=3$ and $g=2$. On $X\setminus A$, it equals $0$, so $f=1$ and $g=3$. Hence the sum and product are

f+g= \begin{cases} 5,&x\in A,\\ 4,&x\in X\setminus A, \end{cases} \qquad fg= \begin{cases} 6,&x\in A,\\ 3,&x\in X\setminus A. \end{cases}

For example, on $A$, the calculations are $3+2=5$ and $3\cdot2=6$; outside $A$, they are $1+3=4$ and $1\cdot3=3$. Since $A$ and its complement form a measurable partition, both functions are simple.

Why the Finite Partition Matters

The key feature of a simple function is not merely that it has a formula involving indicators. Its values are controlled by finitely many measurable pieces. This makes it possible to check identities one piece at a time, as in the arithmetic-closure proof. In later measure theory, simple functions serve as approximations from which more general measurable functions and their integrals can be developed.

A common pitfall is to treat overlapping sets as if they formed a partition. In the expression $2\mathbf{1}_A+3\mathbf{1}_B$, a point in $A\cap B$ contributes both $2$ and $3$, so the value is $5$, not either coefficient separately. Another pitfall is to infer measurability from finite range alone: a finite-range function need not be measurable unless its level sets are measurable. The definition requires both finite range and measurability; the partition representation packages these requirements together.

Key takeaway: A simple function is a measurable finite-range function. Its measurable level sets form a finite partition, giving a representation as a finite sum of constants times indicators. This representation also makes nonnegative simple functions and basic arithmetic operations easy to handle.

Check Your Understanding

Use the partition representation and the examples above to answer the following questions.

Why is the level set $\{x:s(x)=c\}$ measurable when $s$ is a measurable real-valued function?
How does a finite measurable partition show that its indicator representation has finite range?
For $h=2\mathbf{1}_A+3\mathbf{1}_B$, what value does $h$ take on $A\cap B$, and why?
How is the positive-level representation of a nonnegative simple function constructed?
In the arithmetic-closure proof, why do the sets $A_i\cap B_j$ form a finite measurable partition?

What You'll Learn