Level Sets and Disjoint Forms of Simple Functions

Fibers Record the Values of a Simple Function

For a finite-valued function, the sets on which it takes each value are more than a convenient way to describe the function: they form a partition of its domain and give a canonical indicator representation. This is also the right way to interpret a sum of indicators when the sets overlap. On an overlap, all the corresponding coefficients contribute to the function value; disjoint level sets record their combined contribution.

Throughout this tutorial, let $(X,\mathcal{M},\mu)$ be any measure space, with no assumption that $\mu$ is finite, complete, or sigma-finite. The results below depend only on the measurable space $(X,\mathcal{M})$, not on the choice of $\mu$. A function $f:X\to\mathbb{R}$ is simple here if it is measurable and its range $f(X)=\{f(x):x\in X\}$ is finite. For $a\in\mathbb{R}$, its fiber at $a$ is the set $f^{-1}(\{a\})=\{x\in X:f(x)=a\}$. A partition of $X$ means a family of pairwise disjoint subsets whose union is $X$; empty sets may be omitted.

Proposition: If $f:X\to\mathbb{R}$ has finite range $R=f(X)$, then $f$ is $\mathcal{M}$-measurable if and only if $f^{-1}(\{a\})\in\mathcal{M}$ for every $a\in R$. When these fibers are measurable, they are a finite measurable partition of $X$, and $$ f=\sum_{a\in R}a\,\chi_{f^{-1}(\{a\})} $$ at every point of $X$.

Proof. Suppose first that $f$ is measurable. Each singleton $\{a\}$ is Borel in $\mathbb{R}$, so $f^{-1}(\{a\})\in\mathcal{M}$ by the definition of measurability. Conversely, suppose each fiber for $a\in R$ is measurable. For any Borel set $B\subseteq\mathbb{R}$, the finite range gives $f^{-1}(B)=\bigcup_{a\in R\cap B}f^{-1}(\{a\})$. This is a finite union of measurable sets, so $f$ is measurable. Every $x\in X$ has exactly one value $a=f(x)\in R$, hence belongs to exactly one of these fibers. Thus the fibers cover $X$ and are pairwise disjoint. At a point $x$, all indicators in the displayed sum are zero except the one for $a=f(x)$, which equals one; the sum therefore equals $f(x)$. $\square$

The formula includes the zero level set if $0\in R$, where its coefficient is zero. If $0\notin R$, the fibers for values in $R$ still cover all of $X$, and the formula remains pointwise valid. This representation is canonical in the sense that it uses the distinct values actually taken by $f$, rather than a possibly redundant list of coefficients.

Worked Example: Writing a Piecewise Function by Its Fibers

Let $X=[0,4]$, with the trace of the Lebesgue sigma-algebra and Lebesgue measure restricted to this interval. Define $f:X\to\mathbb{R}$ by $f(x)=-2$ on $[0,1)$, $f(x)=0$ on $[1,3)$, and $f(x)=3$ on $[3,4]$. Find its fibers and its disjoint indicator representation.

The three fibers are $E_{-2}=[0,1)$, $E_0=[1,3)$, and $E_3=[3,4]$. Each is measurable in the trace sigma-algebra. They are disjoint: the first ends before $1$, the second includes $1$ but excludes $3$, and the third includes $3$. Their union is $[0,1)\cup[1,3)\cup[3,4]=[0,4]$, including both endpoints of $X$. The fiber formula gives

f=-2\chi_{[0,1)}+0\chi_{[1,3)}+3\chi_{[3,4]} =-2\chi_{[0,1)}+3\chi_{[3,4]}.

Check: at $x=0$, the two nonzero indicators are $1$ and $0$, so the formula gives $-2$; at $x=2$, both are $0$, so it gives $0$; at $x=3$, the first is $0$ and the second is $1$, so it gives $3$. These values agree with the definition, including the boundary point $3$.

Turning Overlapping Indicators into Disjoint Level Sets

A simple function is often first presented as a finite indicator sum. The sets in that sum need not be disjoint, so the coefficient on an overlap is the sum of the coefficients of every set containing that point. To make this calculation systematic, record each point’s membership pattern.

Theorem: Let $(X,\mathcal{M},\mu)$ be any measure space, let $A_1,\ldots,A_n\in\mathcal{M}$, and let $c_1,\ldots,c_n\in\mathbb{R}$, where $n$ is a positive integer. Define $f=\sum_{i=1}^n c_i\chi_{A_i}$. For each subset $S\subseteq\{1,\ldots,n\}$, define $$ D_S=\left(\bigcap_{i\in S}A_i\right)\cap\left(\bigcap_{i\notin S}A_i^c\right), \qquad v_S=\sum_{i\in S}c_i, $$ using $X$ for an intersection over no indices and $0$ for a sum over no indices. Then the sets $D_S$ form a finite measurable partition of $X$, and $$ f=\sum_{S\subseteq\{1,\ldots,n\}}v_S\chi_{D_S} $$ everywhere. Grouping together all $D_S$ with the same value $v_S$ gives the disjoint level-set representation of $f$.

Proof. Each $D_S$ is a finite intersection of measurable sets, because $A_i\in\mathcal{M}$ and $A_i^c\in\mathcal{M}$; hence $D_S\in\mathcal{M}$. Given any $x\in X$, let $S_x=\{i:x\in A_i\}$. For each $i$, $x$ belongs to $A_i$ exactly when $i\in S_x$, and otherwise belongs to $A_i^c$. Therefore $x\in D_{S_x}$. If $S\ne T$, choose an index belonging to exactly one of them. The corresponding sets $D_S,D_T$ require $x$ to belong to $A_i$ and $A_i^c$, respectively, so they are disjoint. Thus the $D_S$ partition $X$. For $x\in D_S$, $\chi_{A_i}(x)=1$ if $i\in S$ and is $0$ otherwise. Consequently $f(x)=\sum_{i\in S}c_i=v_S$, which is also the value of the displayed partition sum at $x$. Finally, for each value $v$ among the finitely many $v_S$, the union $E_v=\bigcup_{S:v_S=v}D_S$ is measurable, and these unions are disjoint and cover $X$. On $E_v$, $f=v$, so grouping the terms gives exactly the level-set representation. $\square$

Worked Example: Overlapping Intervals and Their Level Sets

On $X=[0,4]$ with restricted Lebesgue measure, take $A_1=[0,2]$, $A_2=(1,3]$, and $A_3=[2,4]$, and define $f=2\chi_{A_1}-\chi_{A_2}+3\chi_{A_3}$. Find the disjoint level sets and check the values at the shared endpoints.

The membership patterns and their coefficient sums are as follows. Points in $[0,1]$ belong only to $A_1$, giving value $2$. Points in $(1,2)$ belong to $A_1,A_2$, giving $2-1=1$. The point $2$ belongs to all three sets, giving $2-1+3=4$. Points in $(2,3]$ belong to $A_2,A_3$, giving $-1+3=2$. Points in $(3,4]$ belong only to $A_3$, giving $3$. These cases are disjoint and cover $X$. The value $2$ occurs on two separate pieces, which must be combined into one fiber:

E_1=(1,2),\qquad E_2=[0,1]\cup(2,3],\qquad E_3=(3,4],\qquad E_4=\{2\}.

These sets are measurable, pairwise disjoint, and their union is $[0,4]$. Thus $f=1\chi_{E_1}+2\chi_{E_2}+3\chi_{E_3}+4\chi_{E_4}$. Check: at $x=1$, only $A_1$ contains the point, so the original formula gives $2$; $1\in E_2$, so the new formula gives $2$. At $x=2$, all three indicators equal $1$, so the original gives $2-1+3=4$; $2\in E_4$, so the new formula gives $4$. At $x=3$, $A_2,A_3$ contain the point, giving $-1+3=2$; $3\in E_2$, again giving $2$.

A Finite-Set Calculation

The same procedure works without intervals or a geometric picture. The following example makes explicit that one level set can combine several distinct membership patterns. We use counting measure, although the representation itself does not depend on measure.

Worked Example: Grouping Equal Values from Different Patterns

Let $X=\{a,b,c,d\}$ with its power-set sigma-algebra and counting measure. Set $A_1=\{a,b\}$, $A_2=\{b,c\}$, $A_3=\{c,d\}$, and $f=2\chi_{A_1}-2\chi_{A_2}+\chi_{A_3}$. Determine $f$ pointwise and write its level-set form.

At $a$, membership is in $A_1$ only, so $f(a)=2$. At $b$, membership is in $A_1,A_2$, so $f(b)=2-2=0$. At $c$, membership is in $A_2,A_3$, so $f(c)=-2+1=-1$. At $d$, membership is in $A_3$ only, so $f(d)=1$. Therefore the fibers are $E_2=\{a\}$, $E_0=\{b\}$, $E_{-1}=\{c\}$, and $E_1=\{d\}$, and

f=2\chi_{\{a\}}+0\chi_{\{b\}}-\chi_{\{c\}}+\chi_{\{d\}}.

The zero term may be omitted without changing the function. Check: at $b$, the original two nonzero contributions are $2$ and $-2$, whose sum is $0$; the level-set formula also gives $0$ there. At each other point exactly one displayed level-set indicator is $1$, giving the value computed above.

Why Disjointification Must Preserve the Overlap Values

A common mistake is to make the sets disjoint by removing earlier sets, then keep the original coefficients unchanged. That procedure can discard contributions on overlaps. For example, on $X=\{u,v\}$ with the power-set sigma-algebra, let $A_1=\{u\}$, $A_2=\{u,v\}$, and $f=2\chi_{A_1}+3\chi_{A_2}$. The original function satisfies $f(u)=2+3=5$ and $f(v)=3$.

If one replaces $A_2$ by $A_2\setminus A_1=\{v\}$ and writes $2\chi_{A_1}+3\chi_{A_2\setminus A_1}$, the resulting value at $u$ is $2$, not $5$. The correction is to assign the overlap its combined coefficient: the fibers are $\{u\}$ at value $5$ and $\{v\}$ at value $3$, so $f=5\chi_{\{u\}}+3\chi_{\{v\}}$. This is precisely what the membership-pattern construction guarantees: it partitions the domain first, then computes the function value on every piece.

Disjoint level sets are useful because they expose the actual values and where they occur. In particular, they make equality of two finite-valued functions a question about their fibers and values, rather than about a potentially redundant indicator formula. They also prevent errors when several terms contribute at the same point. The key checks are always: the sets cover the whole domain, no point lies in two different pieces, and the coefficient assigned to each piece equals the sum of the original coefficients active there.

Check Your Understanding

For each question, keep track of both the values and the exact membership of boundary or overlap points.

For a measurable finite-valued function $f:X\to\mathbb{R}$, explain why the fibers $f^{-1}(\{a\})$, for $a\in f(X)$, cover $X$ and are pairwise disjoint.
Let $f=3\chi_A-2\chi_B$, where $A,B\in\mathcal{M}$. Write the four membership-pattern sets and the value of $f$ on each.
In the interval example above, why does the point $2$ belong to all three original sets? What is the value of the original function there?
Give an overlapping indicator sum for which two different membership patterns yield the same value, and identify the level set that combines them.
For $f=2\chi_A+3\chi_B$, explain why replacing $B$ by $B\setminus A$ while retaining coefficient $3$ need not preserve $f$.

Level Sets and Disjoint Forms of Simple Functions

What You'll Learn