Indicator Functions and Jordan Measurability

Why the Boundary Determines Integrability

An indicator function takes only the values zero and one, but it can still oscillate at many points. The key question is whether the points where that oscillation occurs form a set of Lebesgue measure zero. For bounded sets, this question has a precise answer: the indicator is Riemann integrable exactly when the boundary of the set is null. Such a set is called Jordan measurable.

Fix a compact interval $[a,b]$, with $a<b$, and a set $A\subseteq[a,b]$. The function under consideration is the real-valued function $\chi_A:[a,b]\to\{0,1\}$, defined by $\chi_A(x)=1$ for $x\in A$ and $\chi_A(x)=0$ for $x\notin A$. Boundaries in this tutorial are the ordinary topological boundaries in $\mathbb{R}$, denoted $\partial A$. Lebesgue measure on $\mathbb{R}$ is $m$. The endpoint convention will not affect the criterion: adding or removing the two endpoints changes a set by at most a null set.

Definition: A bounded set $A\subseteq\mathbb{R}$ is Jordan measurable if $m(\partial A)=0$. This concerns the boundary, not merely whether $A$ is Lebesgue measurable: a Lebesgue measurable set can have a boundary of positive measure.

We use two results from the earlier material on comparing Riemann and Lebesgue integrals. Lebesgue’s criterion for Riemann integrability states that a bounded real-valued function on a compact interval is Riemann integrable if and only if its set of discontinuities has Lebesgue measure zero. Equality of Riemann and Lebesgue integrals for bounded functions states that if such a function is Riemann integrable, its Lebesgue integral exists and equals its Riemann integral. These results apply to the bounded indicator functions considered here.

The Indicator’s Discontinuities

Proposition: For every $x\in(a,b)$, $\chi_A$ is discontinuous at $x$ if and only if $x\in\partial A$.

Proof. If $x\notin\partial A$, then some open neighborhood of $x$ lies entirely in $A$ or entirely in $A^c$. On that neighborhood $\chi_A$ is constant, so it is continuous at $x$. If $x\in\partial A$, every open neighborhood of $x$ meets both $A$ and $A^c$. In every such neighborhood, $\chi_A$ therefore takes both values zero and one. In particular, its values do not approach $\chi_A(x)$ as the neighborhood shrinks, so it is discontinuous at $x$. $\square$

At $a$ and $b$, continuity is tested relative to the domain $[a,b]$, so it need not agree with continuity on all of $\mathbb{R}$. This causes no problem for measure: the discontinuity set in $(a,b)$ is exactly $\partial A\cap(a,b)$, and any discrepancy between the boundary and the discontinuity set is contained in $\{a,b\}$, a null set.

Theorem: Let $A\subseteq[a,b]$, where $a<b$, and let $\chi_A$ be its indicator on $[a,b]$. Then $\chi_A$ is Riemann integrable if and only if $m(\partial A)=0$.

Proof. The indicator is bounded and real-valued, so Lebesgue’s criterion applies. If $m(\partial A)=0$, the proposition shows that its discontinuity set inside $(a,b)$ is null; any additional discontinuities can only be at the two endpoints. Thus the full discontinuity set is null, and the criterion implies that $\chi_A$ is Riemann integrable.

Conversely, if $\chi_A$ is Riemann integrable, Lebesgue’s criterion says its discontinuity set is null. By the proposition, $\partial A\cap(a,b)$ is contained in that set. Since $A\subseteq[a,b]$, its boundary is contained in $[a,b]$, possibly including $a$ or $b$. The boundary outside $(a,b)$ is therefore contained in $\{a,b\}$. Adding these at most two points to a null set still gives a null set, so $m(\partial A)=0$. $\square$

The Integral Is the Measure of the Set

Corollary: If $A\subseteq[a,b]$ and $m(\partial A)=0$, then $A$ is Lebesgue measurable and $$ \int_a^b \chi_A(x)\,dx=m(A). $$

Proof. The interior $\operatorname{int}(A)$ is open and hence Borel measurable. Also $A\triangle\operatorname{int}(A)\subseteq\partial A$: a point of $A$ that is not interior is a boundary point, and every point of $\operatorname{int}(A)$ belongs to $A$. Since Lebesgue measure is complete and $\partial A$ is null, $A$ is Lebesgue measurable. The Lebesgue integral of the indicator of a measurable set is its measure, so $\int_{[a,b]}\chi_A\,dm=m(A)$. The theorem gives Riemann integrability, and equality of Riemann and Lebesgue integrals for bounded Riemann-integrable functions identifies this value with the Riemann integral. $\square$

The value is finite because $A\subseteq[a,b]$ and $m(A)\le b-a$. The same statement applies to any bounded $A\subseteq\mathbb{R}$: choose a compact interval containing $A$, and integrate its indicator on that interval. The integral does not depend on enlarging the interval, because the indicator is zero outside $A$.

Worked Example: An Indicator of an Interval

Let $A=[\tfrac14,\tfrac34]\subseteq[0,1]$. Its boundary is $\{\tfrac14,\tfrac34\}$, a two-point set, so $m(\partial A)=0$. The theorem gives Riemann integrability, and the corollary gives $$ \int_0^1\chi_A(x)\,dx=m(A)=\frac34-\frac14=\frac12. $$

One can also verify the integral directly using Darboux sums. For $0<\delta<\tfrac14$, take a partition that includes $\tfrac14-\delta,\tfrac14+\delta,\tfrac34-\delta,\tfrac34+\delta$, as well as $0$ and $1$. The function is one on $[\tfrac14+\delta,\tfrac34-\delta]$, so the lower sum is at least $$ \left(\frac34-\delta\right)-\left(\frac14+\delta\right)=\frac12-2\delta. $$ It vanishes outside $[\tfrac14-\delta,\tfrac34+\delta]$, so the upper sum is at most $$ \left(\frac34+\delta\right)-\left(\frac14-\delta\right)=\frac12+2\delta. $$ The difference between these bounds is $4\delta$, which can be made arbitrarily small. Both bounds approach $\tfrac12$. Check: the interval length is $\tfrac34-\tfrac14=\tfrac12$, matching both the Lebesgue measure and the Riemann integral.

Worked Example: The Rationals in the Unit Interval

Let $A=\mathbb{Q}\cap[0,1]$. Every open interval in $\mathbb{R}$ contains both a rational and an irrational number. Thus every point of $[0,1]$ is a boundary point of $A$, and $\partial A=[0,1]$. Its measure is $m(\partial A)=1$, so the indicator is not Riemann integrable by the theorem.

The Darboux sums confirm this. Every subinterval of positive length contains rational and irrational points, so the infimum of $\chi_A$ on each such subinterval is zero and its supremum is one. For any partition of $[0,1]$, the lower sum is therefore $0$, while the upper sum is the sum of the subinterval lengths, namely $1$. The upper and lower sums never approach one another.

This example also separates two notions: $A$ is Lebesgue measurable because it is countable, and $m(A)=0$, so its Lebesgue integral is zero. But Lebesgue measurability alone does not imply Riemann integrability; the boundary is too large. Check: the Darboux lower and upper sums are $0$ and $1$, respectively, in agreement with the boundary criterion.

Worked Example: The Middle-Thirds Cantor Set

Let $C\subseteq[0,1]$ be the middle-thirds Cantor set, obtained by repeatedly removing the open middle third of each remaining interval. At stage $n$, the remaining set is covered by $2^n$ closed intervals, each of length $3^{-n}$. The sum of their lengths is $$ 2^n3^{-n}=\left(\frac23\right)^n. $$ Since $C$ is contained in the stage-$n$ set for every $n$, $m(C)\le(\tfrac23)^n$ for every $n$. Letting $n$ tend to infinity gives $m(C)=0$.

The set $C$ is closed and has empty interior: at stage $n$ it is contained in intervals of length $3^{-n}$, which tend to zero, so it cannot contain any interval of positive length. Hence $\partial C=C$, and $m(\partial C)=0$. The theorem shows that $\chi_C$ is Riemann integrable, and its integral is $m(C)=0$. Check: the Lebesgue integral of $\chi_C$ is also $m(C)=0$, as required by the equality of integrals.

A Boundary-Containment Pitfall

A common mistake is to assume that if $A\subseteq(a,b)$, then $\partial A\subseteq(a,b)$. This is false: boundary points can lie at the endpoints of the containing interval. For example, for $A=(a,\tfrac{a+b}{2})$, the boundary is $$ \partial A=\left\{a,\frac{a+b}{2}\right\}, $$ so $a\in\partial A$, even though $a\notin A$. The correct containment is $\partial A\subseteq[a,b]$. In the integrability criterion, this endpoint issue is harmless because a finite set has measure zero, but the containment itself must be stated correctly.

A related error is to infer Riemann integrability merely from the fact that $A$ is Lebesgue measurable. The rational example is a counterexample: it is measurable and null, but its boundary has measure one and its indicator is not Riemann integrable. For indicators, the decisive condition is not just measurability of the set; it is nullity of its boundary.

Check Your Understanding

Use the boundary criterion and the distinction between the two integral notions to answer these questions.

For $A=[0,\tfrac13)\cup[\tfrac23,1]$ on $[0,1]$, find $\partial A$, decide whether $\chi_A$ is Riemann integrable, and compute its integral.
Explain why the indicator of $\mathbb{Q}\cap[0,1]$ has Lebesgue integral zero but is not Riemann integrable.
If $A\subseteq[a,b]$ and $\partial A$ is finite, what does the theorem imply about $\chi_A$ and its integral?
Why does $A\triangle\operatorname{int}(A)\subseteq\partial A$ imply that $A$ is Lebesgue measurable when $m(\partial A)=0$?
Give an example of a bounded set whose indicator is Riemann integrable but whose boundary is nonempty. Explain why nonemptiness does not obstruct integrability.

Indicator Functions and Jordan Measurability

What You'll Learn