Indicator Integrals and Increasing Sets

Indicators Turn Measure into an Integral

An indicator function records membership in a set: it is \(1\) on the set and \(0\) elsewhere. Its integral therefore measures the size of the set, including when that size is infinite. This simple identity also connects the integral of indicators of increasing sets to continuity from below of measure. The argument uses only countable additivity and the definition of the nonnegative integral; it does not require the measure space to be finite or sigma-finite.

Throughout, let \((X,\mathcal{M},\mu)\) be any measure space; no finiteness, sigma-finiteness, or completeness assumption is imposed. For \(E\in\mathcal{M}\), the function \(\chi_E:X\to\{0,1\}\) is measurable and real-valued. Its nonnegative Lebesgue integral is allowed to take the value \(+\infty\). A sequence \((E_n)\) is increasing if \(E_n\subseteq E_{n+1}\) for every \(n\); equality between successive sets is permitted.

Proposition (Integral of an indicator): For every \(E\in\mathcal{M}\), \(\displaystyle\int_X\chi_E\,d\mu=\mu(E)\), with equality in \([0,\infty]\). In particular, the integral is \(0\) if \(E\) is null and \(+\infty\) if \(\mu(E)=+\infty\).

Proof. The function \(\chi_E=1\chi_E\) is a nonnegative simple function. By the definition of the integral of a simple function, its integral is its sole nonzero value, \(1\), multiplied by the measure of the set where that value occurs. Thus \(\int_X\chi_E\,d\mu=1\cdot\mu(E)=\mu(E)\). This includes \(E=\varnothing\), for which both sides are \(0\), and \(\mu(E)=+\infty\), for which \(1\cdot\mu(E)=+\infty\). \(\square\)

The proposition also gives the integral over any measurable region \(F\in\mathcal{M}\). Since \(\chi_E\chi_F=\chi_{E\cap F}\) pointwise,

\int_F\chi_E\,d\mu =\int_X\chi_E\chi_F\,d\mu =\mu(E\cap F).

This identity concerns the measure of the intersection, not necessarily the measure of all of \(E\). If \(E\cap F\) is null, the integral over \(F\) is zero even if \(E\) itself has positive or infinite measure.

Continuity from Below

For an increasing sequence of measurable sets, every point in the union eventually belongs to one of the sets and hence to all subsequent sets. Countable additivity makes this pointwise observation into a precise statement about measures.

Theorem (Continuity from Below): Let \((X,\mathcal{M},\mu)\) be a measure space, and let \(E_1\subseteq E_2\subseteq\cdots\) be measurable sets. Then \(E=\bigcup_{n=1}^{\infty}E_n\) is measurable and \(\displaystyle\mu(E)=\lim_{n\to\infty}\mu(E_n)\). The limit and the measure may equal \(+\infty\).

Proof. A sigma-algebra is closed under countable unions, so \(E\in\mathcal{M}\). Define \(D_1=E_1\), and for \(n\ge2\) define \(D_n=E_n\setminus E_{n-1}\). Each \(D_n\) is measurable. The sets \(D_n\) are pairwise disjoint: if \(j<n\), then \(D_j\subseteq E_j\subseteq E_{n-1}\), while \(D_n\cap E_{n-1}=\varnothing\). For each \(N\), their first \(N\) pieces have union \(E_N\), and all the pieces together have union \(E\). Countable additivity therefore gives \(\mu(E)=\sum_{n=1}^{\infty}\mu(D_n)\) and \(\mu(E_N)=\sum_{n=1}^{N}\mu(D_n)\). The partial sums of a series of nonnegative extended real numbers increase to its sum, possibly \(+\infty\). Hence \(\lim_{N\to\infty}\mu(E_N)=\mu(E)\), as claimed. This proof does not subtract measures, so it remains valid if some \(E_n\) has infinite measure. \(\square\)

Combining the indicator proposition with continuity from below gives the integral version. The indicators are nonnegative, real-valued, and measurable, and their pointwise limit can be identified exactly: for each \(x\in X\), if \(x\in\bigcup_n E_n\), then \(x\in E_N\) for some \(N\), so \(\chi_{E_n}(x)=1\) for every \(n\ge N\). If \(x\notin\bigcup_nE_n\), every indicator is \(0\). Thus \(\chi_{E_n}\) increases pointwise everywhere to \(\chi_E\).

Corollary (Increasing indicators): If \(E_n\in\mathcal{M}\) and \(E_n\subseteq E_{n+1}\) on a measure space \((X,\mathcal{M},\mu)\), then \(\displaystyle\int_X\chi_{\cup_n E_n}\,d\mu=\lim_{n\to\infty}\int_X\chi_{E_n}\,d\mu\). Both sides may be \(+\infty\), and the indicators converge pointwise everywhere.

Proof. The indicator proposition identifies the left side with \(\mu(\bigcup_nE_n)\) and each integral on the right with \(\mu(E_n)\). Continuity from below identifies the measure of the union with the limit of those measures. Therefore the two sides are equal, including when their common value is infinite. \(\square\)

Worked Examples

Worked Example: Increasing Intervals with a Finite Union Measure

On \(X=\mathbb{R}\) with its Lebesgue sigma-algebra \(\mathcal{L}\) and Lebesgue measure \(m\), let \(E_n=[0,1-1/n]\) for integers \(n\ge2\). Find the pointwise limit of \(\chi_{E_n}\) and the limit of its integrals.

The endpoints \(1-1/n\) increase as \(n\) increases, so \(E_n\subseteq E_{n+1}\). Every \(E_n\) lies in \([0,1)\). Conversely, if \(0\le x<1\), choose \(n\ge2\) large enough that \(1/n<1-x\). Then \(x<1-1/n\), so \(x\in E_n\). It follows that \(\bigcup_{n\ge2}E_n=[0,1)\), and \(\chi_{E_n}\) increases pointwise everywhere to \(\chi_{[0,1)}\).

Each interval has length \(1-1/n\), so the indicator proposition gives \(\int_{\mathbb{R}}\chi_{E_n}\,dm=1-1/n\). Therefore

\lim_{n\to\infty}\int_{\mathbb{R}}\chi_{E_n}\,dm =\lim_{n\to\infty}\left(1-\frac1n\right) =1 =\int_{\mathbb{R}}\chi_{[0,1)}\,dm.

Check: at \(n=2,3,4\), the integrals are \(1-1/2=1/2\), \(1-1/3=2/3\), and \(1-1/4=3/4\). Directly, the union \([0,1)\) has Lebesgue measure \(1-0=1\), agreeing with the limit.

Worked Example: Increasing Sets Whose Measures Tend to Infinity

On \((\mathbb{R},\mathcal{L},m)\), let \(E_n=[-n,n]\) for integers \(n\ge1\). Compute the integral of each indicator and the integral of the indicator of the union.

The sets increase, and every real number belongs to \([-n,n]\) once \(n\ge |x|\). Hence \(\bigcup_{n\ge1}E_n=\mathbb{R}\), and \(\chi_{E_n}\) increases pointwise everywhere to \(\chi_{\mathbb{R}}\). The length of \([-n,n]\) is \(n-(-n)=2n\), so

\int_{\mathbb{R}}\chi_{E_n}\,dm=m([-n,n])=2n \quad\text{and}\quad \lim_{n\to\infty}\int_{\mathbb{R}}\chi_{E_n}\,dm=+\infty.

The union has measure \(m(\mathbb{R})=+\infty\), so \(\int_{\mathbb{R}}\chi_{\mathbb{R}}\,dm=+\infty\), in agreement with the limit. Check: for \(n=1,2,3\), the measures are \(2\), \(4\), and \(6\), respectively; they increase without a finite upper bound. This example shows why continuity from below must allow infinite values.

Worked Example: Increasing Finite Sets in a Finite Atomic Measure Space

Let \(X=\mathbb{N}=\{1,2,\ldots\}\), \(\mathcal{M}=\mathcal{P}(\mathbb{N})\), and \(\mu(A)=\sum_{k\in A}2^{-k}\). This is a finite measure: the nonnegative geometric series has total \(\sum_{k=1}^{\infty}2^{-k}=1\). For \(E_n=\{1,\ldots,n\}\), compute the integrals and their limit.

The sets increase and their union is \(\mathbb{N}\). Every subset is measurable, and the indicator proposition applies. The finite geometric sum gives \(\int_X\chi_{E_n}\,d\mu=\mu(E_n)=\sum_{k=1}^n2^{-k}=1-2^{-n}\). Thus the integral limit is \(1\). On the union, \(\int_X\chi_{\mathbb{N}}\,d\mu=\mu(\mathbb{N})=\sum_{k=1}^{\infty}2^{-k}=1\).

Check: for \(n=1,2,3\), the finite sums are \(2^{-1}=1/2\), \(2^{-1}+2^{-2}=1/2+1/4=3/4\), and \(2^{-1}+2^{-2}+2^{-3}=1/2+1/4+1/8=7/8\). The formula gives \(1-2^{-1}=1/2\), \(1-2^{-2}=3/4\), and \(1-2^{-3}=7/8\), respectively. The omitted tail after \(n\) is \(\sum_{k=n+1}^{\infty}2^{-k}=2^{-n}\), so the partial sums tend to the full measure \(1\).

Why the Infinite-Measure Case Matters

A common mistake is to assume that continuity from below needs \(\mu(E_1)<\infty\), or that the measures of the sets must remain bounded. Neither condition is needed. For the intervals \(E_n=[-n,n]\), a mistaken argument might say that the measures cannot converge because \(m(E_n)=2n\) has no finite limit. The correction is that convergence is in the extended interval \([0,\infty]\): \(2n\to+\infty\), exactly matching \(m(\bigcup_nE_n)=m(\mathbb{R})=+\infty\).

There is a related proof error to avoid when sets have infinite measure. One should not write \(\mu(E_n\setminus E_{n-1})=\mu(E_n)-\mu(E_{n-1})\) without knowing the terms are finite: the expression \(+\infty-(+\infty)\) is undefined. The proof instead partitions the union into disjoint differences and applies countable additivity directly. This is why the disjoint-piece argument handles both finite and infinite measures without an extra hypothesis.

The result is useful whenever a measurable set is built as an increasing union of simpler measurable pieces. For indicators, it says exactly that the integrals of those pieces approach the integral over the full union. It is a set-level continuity property, derived directly from countable additivity; the corresponding general theorem for increasing nonnegative functions requires additional development.

Check Your Understanding

In each question, specify whether the relevant equality is pointwise or almost everywhere, and allow for infinite values when appropriate.

For a measurable set \(E\) with \(\mu(E)=0\), what is \(\int_X\chi_E\,d\mu\)? What if \(\mu(E)=+\infty\)?
Let \(E_n=[0,n)\) in \((\mathbb{R},\mathcal{L},m)\). Find \(\bigcup_nE_n\) and compute \(\lim_n\int_{\mathbb{R}}\chi_{E_n}\,dm\).
In the proof of continuity from below, why are the sets \(D_n=E_n\setminus E_{n-1}\) pairwise disjoint?
Give an increasing sequence of measurable sets whose measures tend to a finite value, and verify that the limit equals the measure of the union.
Explain why the expression \(\mu(E_n)-\mu(E_{n-1})\) may not be used when both measures equal \(+\infty\), and describe the disjoint-set argument that avoids it.