Indicator Functions

From Sets to Functions

A measurable set can be recorded by a function that assigns $1$ to its points and $0$ to all other points. This simple encoding turns questions about set membership into algebraic identities between functions. In the previous tutorial, simple functions were represented as finite sums of constants times indicators. Here we examine the indicator functions themselves: how their algebra reflects set operations, and what pointwise convergence of indicators says about membership in a sequence of sets.

Let $(X,\mathcal{F})$ be a measurable space. For a subset $A\subseteq X$, its indicator function is $\mathbf{1}_A:X\to\mathbb{R}$, defined by $\mathbf{1}_A(x)=1$ if $x\in A$, and $\mathbf{1}_A(x)=0$ if $x\notin A$. The Measurability of an Indicator Function Theorem, established earlier in the course, says that $\mathbf{1}_A$ is measurable exactly when $A\in\mathcal{F}$. We will use this result without repeating its proof.

At every point, an indicator has only two possible values. Consequently, any identity involving indicators can be checked by considering whether the point belongs to the sets involved. That pointwise check is the bridge between set operations and function operations.

Set Operations as Indicator Identities

Theorem (Indicator Identities for Set Operations): For any subsets $A,B\subseteq X$, $$ \mathbf{1}_{X\setminus A}=1-\mathbf{1}_A,\qquad \mathbf{1}_{A\cap B}=\mathbf{1}_A\mathbf{1}_B,\qquad \mathbf{1}_{A\cup B}=\mathbf{1}_A+\mathbf{1}_B-\mathbf{1}_A\mathbf{1}_B, $$ and $$ \mathbf{1}_{A\setminus B}=\mathbf{1}_A(1-\mathbf{1}_B). $$ If $A$ and $B$ are disjoint, then $\mathbf{1}_{A\cup B}=\mathbf{1}_A+\mathbf{1}_B$.

Proof. Fix $x\in X$. If $x\in A$, then $\mathbf{1}_A(x)=1$; if $x\notin A$, then $\mathbf{1}_A(x)=0$. Thus $1-\mathbf{1}_A(x)$ equals $0$ on $A$ and $1$ outside $A$, which is exactly $\mathbf{1}_{X\setminus A}(x)$.

For the intersection identity, if $x\in A\cap B$, both indicators equal $1$, so their product is $1$. If $x\notin A\cap B$, at least one indicator is $0$, so the product is $0$. This proves $\mathbf{1}_{A\cap B}(x)=\mathbf{1}_A(x)\mathbf{1}_B(x)$.

For the union identity, there are four membership possibilities. If $x$ belongs to neither set, the expression is $0+0-0=0$. If $x\in A$ and $x\notin B$, it is $1+0-0=1$. If $x\notin A$ and $x\in B$, it is $0+1-0=1$. If $x\in A\cap B$, it is $1+1-1=1$. In each case this agrees with $\mathbf{1}_{A\cup B}(x)$. For the difference identity, the product is $1$ exactly when $\mathbf{1}_A(x)=1$ and $\mathbf{1}_B(x)=0$, which is exactly when $x\in A\setminus B$. Finally, if $A\cap B=\varnothing$, the product $\mathbf{1}_A\mathbf{1}_B$ is zero everywhere, giving the disjoint-union formula. Since $x$ was arbitrary, all the identities hold on $X$. $\square$

Worked Example: Intersections, Unions, and Differences

Take $X=\mathbb{R}$, $A=(-3,2]$, and $B=[0,4)$, which are Borel sets. Their intersection is $[0,2]$, their union is $(-3,4)$, and their difference $A\setminus B$ is $(-3,0)$. The identities give

\mathbf{1}_{A\cap B}=\mathbf{1}_A\mathbf{1}_B,\qquad \mathbf{1}_{A\cup B}=\mathbf{1}_A+\mathbf{1}_B-\mathbf{1}_A\mathbf{1}_B,\qquad \mathbf{1}_{A\setminus B}=\mathbf{1}_A(1-\mathbf{1}_B).

For $x=1$, both indicators equal $1$, so the intersection formula gives $1\cdot1=1$, while the union formula gives $1+1-1=1$; indeed, $1$ belongs to both sets. For $x=-1$, the indicators are $\mathbf{1}_A(-1)=1$ and $\mathbf{1}_B(-1)=0$. The union formula gives $1+0-0=1$, and the difference formula gives $1(1-0)=1$; indeed, $-1\in A\setminus B$. For $x=3$, the indicators are $0$ and $1$, so the union formula gives $0+1-0=1$, whereas the difference formula gives $0(1-1)=0$. These values agree with the sets described above.

The product in the union identity prevents a point in both sets from being counted twice. This is the same correction that appears in the addition formula for two events in probability. In contrast, when the sets are disjoint, no correction is needed: their indicators add directly.

Disjoint Unions and Partitions

The same reasoning extends to a countable disjoint family. At any given point, at most one of the indicators in the family can equal $1$. Thus the infinite sum is pointwise well-defined: its partial sums are eventually constant at that point.

Theorem (Indicator of a Countable Disjoint Union): If $A_1,A_2,\ldots\subseteq X$ are pairwise disjoint, then, pointwise on $X$, $$ \mathbf{1}_{\bigcup_{n=1}^{\infty}A_n}=\sum_{n=1}^{\infty}\mathbf{1}_{A_n}. $$ If each $A_n$ is measurable, the union is measurable and both sides are measurable functions.

Proof. Fix $x\in X$. If $x\notin\bigcup_{n=1}^{\infty}A_n$, then $x\notin A_n$ for every $n$, so every term $\mathbf{1}_{A_n}(x)$ is zero. The sum is zero, as is the indicator of the union. If $x\in\bigcup_{n=1}^{\infty}A_n$, then $x\in A_k$ for some $k$. Pairwise disjointness ensures that $x\notin A_n$ for every $n\neq k$. Therefore the $k$-th term is $1$, all other terms are zero, and the sum is $1$, again equal to the indicator of the union. This proves the pointwise identity. If all $A_n$ are measurable, their union is measurable by closure of a sigma-algebra under countable unions. The indicator measurability theorem then gives measurability of the left side; the identity also expresses it as a pointwise limit of measurable partial sums. $\square$

Worked Example: Indicators of a Measurable Partition

Let $X=\mathbb{R}$ with its Borel sigma-algebra, and partition it into $A_1=(-\infty,-2)$, $A_2=[-2,1)$, and $A_3=[1,\infty)$. These three Borel sets are disjoint and cover $\mathbb{R}$. At each real number exactly one of their indicators is $1$, so

\mathbf{1}_{A_1}+\mathbf{1}_{A_2}+\mathbf{1}_{A_3}=1.

For $x=-3$, the three indicator values are $1,0,0$, whose sum is $1$. For $x=0$, they are $0,1,0$, again summing to $1$. For $x=2$, they are $0,0,1$, also summing to $1$. The endpoints are assigned consistently: $-2\in A_2$, and $1\in A_3$.

Assigning the respective values $-1$, $4$, and $7$ to these three pieces gives the function $$ s=-\mathbf{1}_{A_1}+4\mathbf{1}_{A_2}+7\mathbf{1}_{A_3}. $$ At $x=-3$, this formula gives $-1(1)+4(0)+7(0)=-1$; at $x=0$, it gives $-1(0)+4(1)+7(0)=4$; and at $x=2$, it gives $-1(0)+4(0)+7(1)=7$. This is a finite partition representation of a simple function, as in the previous tutorial.

Disjointness matters in the countable-sum identity. If the sets overlap, a point in two of them contributes $1+1=2$ to the sum, whereas the indicator of their union is still $1$. For a finite family that is not disjoint, the union formula requires correction terms; the two-set identity above displays that correction explicitly.

When Do Indicators Converge Pointwise?

A sequence of indicators can converge at a point only if membership in the corresponding sets eventually settles down at that point. Alternating membership produces alternating values $0$ and $1$, which cannot converge. The next result makes this observation exact and identifies the limit set.

Theorem (Pointwise Convergence Criterion for Indicators): Let $E_n\subseteq X$ for each positive integer $n$. The sequence $\mathbf{1}_{E_n}$ converges pointwise on $X$ if and only if, for every $x\in X$, membership of $x$ in $E_n$ is eventually constant: either $x\in E_n$ for every sufficiently large $n$, or $x\notin E_n$ for every sufficiently large $n$. In that case, $$ \lim_{n\to\infty}\mathbf{1}_{E_n}(x)=\mathbf{1}_{\liminf_{n\to\infty}E_n}(x). $$

Proof. First suppose that $\mathbf{1}_{E_n}(x)$ converges for a fixed $x$, with limit $L$. Each term is either $0$ or $1$. The limit must also be either $0$ or $1$: if $L\notin\{0,1\}$, then $d=\min\{|L|,|L-1|\}>0$. Every term has distance at least $d$ from $L$, contradicting convergence, which requires the terms eventually to have distance less than $d/2$ from $L$.

If $L=0$, convergence gives an $N$ such that $|\mathbf{1}_{E_n}(x)-0|<1/2$ for every $n\geq N$. A value of $1$ has distance $1$ from $0$, so every such term must equal $0$. If $L=1$, convergence gives an $N$ such that $|\mathbf{1}_{E_n}(x)-1|<1/2$ for every $n\geq N$. A value of $0$ has distance $1$ from $1$, so every such term must equal $1$. Thus membership is eventually constant at $x$. The threshold is $1/2$ in both cases: it is a fixed distance separating the two possible indicator values, including when the limit is zero.

Conversely, if membership of $x$ is eventually constant, then $\mathbf{1}_{E_n}(x)$ is eventually constantly $0$ or eventually constantly $1$, and hence converges. It is eventually $1$ exactly when $x$ belongs to every $E_n$ from some index onward. By definition, that is the condition $x\in\liminf_{n\to\infty}E_n$. Therefore the pointwise limit equals the indicator of this set. Applying the argument at each $x$ proves the criterion on all of $X$. If all $E_n$ are measurable, the Measurability of Set Limsup and Liminf Theorem from earlier in the course shows that this limit set is measurable; its indicator is measurable by the earlier indicator measurability theorem. $\square$

Worked Example: Shrinking Intervals and a Pointwise Limit

On $\mathbb{R}$, let $E_n=(-1/n,1/n)$. The point $0$ belongs to every $E_n$, so $\mathbf{1}_{E_n}(0)=1$ for every $n$. If $x\neq0$, choose $N$ so large that $1/N\leq |x|$. For every $n\geq N$, $1/n\leq1/N\leq |x|$, and therefore $x\notin(-1/n,1/n)$. Thus $\mathbf{1}_{E_n}(x)=0$ for all $n\geq N$. Membership stabilizes at every point, and

\lim_{n\to\infty}\mathbf{1}_{(-1/n,1/n)}=\mathbf{1}_{\{0\}}

pointwise. The limit is $1$ at $0$ and $0$ at every nonzero real number. Since $\{0\}$ is Borel, this limit is measurable.

Worked Example: A Sequence with No Pointwise Limit

Let $C=[2,5)\subseteq\mathbb{R}$, and set $E_n=C$ for even $n$ and $E_n=\mathbb{R}\setminus C$ for odd $n$. At $x=3$, the value $\mathbf{1}_{E_n}(3)$ is $0$ for odd $n$ and $1$ for even $n$. Both values occur arbitrarily far along the sequence, so membership does not become constant and the numerical sequence does not converge. Hence $\mathbf{1}_{E_n}$ does not converge pointwise on $\mathbb{R}$. By contrast, for any $x\notin C$, its indicator sequence is $1$ for odd $n$ and $0$ for even $n$, so it also fails to converge there. This example shows why convergence requires eventual stabilization, not merely that the sets are all measurable.

Why Indicator Algebra Is Useful

Indicator functions let us carry set information into calculations with functions. The product identity turns intersection into multiplication; subtraction from $1$ turns a set into its complement; and disjoint unions become sums. These formulas are useful in measure theory because they let us express simple functions in terms of measurable pieces and reason about those pieces through pointwise identities. The convergence criterion also warns that pointwise limits of indicators are restrictive: a point cannot keep switching between membership and nonmembership if the numerical values are to converge.

A common pitfall is to use $\mathbf{1}_{A\cup B}=\mathbf{1}_A+\mathbf{1}_B$ without checking disjointness. On $A\cap B$, the right side is $2$, not $1$, so the formula fails there. Another is to confuse pointwise convergence with a vague notion of sets becoming close: the criterion is pointwise and asks whether each individual point's membership eventually stops changing. When the sets are measurable and convergence holds, the resulting set is the set liminf, which is measurable by an earlier theorem.

Key takeaway: Indicator functions encode membership using the values $0$ and $1$. Their products and complements reflect intersections and complements, disjoint unions correspond to sums, and a sequence of indicators converges pointwise exactly when membership eventually stabilizes at every point.

Check Your Understanding

Use the identities and convergence criterion to answer the following questions.

Why does $\mathbf{1}_A\mathbf{1}_B$ equal $1$ exactly on $A\cap B$?
Why does the formula for $\mathbf{1}_{A\cup B}$ include the product term $-\mathbf{1}_A\mathbf{1}_B$?
What does pairwise disjointness ensure in the indicator formula for a countable union?
Why does convergence of a sequence taking only the values $0$ and $1$ force it eventually to have one fixed value?
For $E_n=(-1/n,1/n)$, why is the pointwise limit the indicator of $\{0\}$?

What You'll Learn