Continuity From Below

Why Increasing Unions Have a Limit Measure

Suppose measurable sets grow with time: every point in one set remains in all later sets. Their union records all points that eventually appear. A basic question is whether the measure of that union can be obtained as the limit of the measures of the individual sets. The answer is yes, including when that limit is infinite. The key is to separate each set into the points already present at the preceding stage and the points newly added.

Throughout, let \((X,\mathcal{M},\mu)\) be a measure space. No assumption that \(\mu\) is finite, sigma-finite, or complete is needed. A sequence \((A_n)_{n\ge1}\) of sets in \(\mathcal{M}\) is increasing if \(A_n\subseteq A_{n+1}\) for every \(n\ge1\). The measure \(\mu(A_n)\) may be any value in \([0,\infty]\), so it is important to allow infinite values in the conclusion.

Disjoint Successive Differences

For an increasing sequence, define its successive increments by

D_1=A_1,\qquad D_n=A_n\setminus A_{n-1}\quad(n\ge2).

The set \(D_n\) consists of the points that have appeared by stage \(n\) but were not present at stage \(n-1\). Each \(D_n\) is measurable: \(D_1\in\mathcal{M}\), and for \(n\ge2\), \(A_n\setminus A_{n-1}=A_n\cap A_{n-1}^c\), which belongs to \(\mathcal{M}\). These increments are pairwise disjoint. Indeed, if \(j<n\), then \(D_j\subseteq A_j\subseteq A_{n-1}\), while \(D_n\) is disjoint from \(A_{n-1}\).

The first \(N\) increments make up exactly \(A_N\): points in \(A_N\) have a first stage at which they appear, and belong to the increment for that stage. Thus \(A_N=\bigcup_{n=1}^{N}D_n\). The full union is \(\bigcup_{n=1}^{\infty}A_n=\bigcup_{n=1}^{\infty}D_n\). These identities turn the desired limit into an application of countable additivity.

Theorem (Continuity From Below): Let \((X,\mathcal{M},\mu)\) be any measure space, and let \((A_n)_{n\ge1}\) be an increasing sequence in \(\mathcal{M}\). Then \(\displaystyle \mu\!\left(\bigcup_{n=1}^{\infty}A_n\right)=\lim_{n\to\infty}\mu(A_n)\). The limit and the measure of the union are allowed to equal \(+\infty\).

Proof. Form the measurable, pairwise disjoint sets \(D_1=A_1\) and \(D_n=A_n\setminus A_{n-1}\) for \(n\ge2\). For every \(N\ge1\), the finite union \(\bigcup_{n=1}^{N}D_n\) equals \(A_N\). Finite additivity, which follows from countable additivity, therefore gives \(\mu(A_N)=\sum_{n=1}^{N}\mu(D_n)\). The sets \(D_n\) are pairwise disjoint and their union is \(\bigcup_{n=1}^{\infty}A_n\). Countable additivity gives \(\mu(\bigcup_{n=1}^{\infty}A_n)=\sum_{n=1}^{\infty}\mu(D_n)\), where the sum is allowed to be infinite. By the definition of a series of nonnegative extended real numbers, its value is the limit of its partial sums. Those partial sums are precisely \(\sum_{n=1}^{N}\mu(D_n)=\mu(A_N)\). Hence \(\mu(\bigcup_{n=1}^{\infty}A_n)=\lim_{N\to\infty}\mu(A_N)\). This also covers empty sets and zero-measure increments; if \(X\) is empty, all the sets and both sides are empty or zero as appropriate. \(\square\)

The increasing condition also ensures that the sequence \(\mu(A_n)\) is nondecreasing, by monotonicity of a measure. Its limit thus exists in \([0,\infty]\). The proof gives more than existence of the limit: it identifies each step’s increase in measure with the measure of the corresponding increment.

Corollary (Measure as a Sum of Increments): Under the hypotheses of continuity from below, with \(D_1=A_1\) and \(D_n=A_n\setminus A_{n-1}\) for \(n\ge2\), \(\displaystyle \mu\!\left(\bigcup_{n=1}^{\infty}A_n\right)=\sum_{n=1}^{\infty}\mu(D_n)\).

Proof. The sets \(D_n\) are measurable, pairwise disjoint, and have union \(\bigcup_{n=1}^{\infty}A_n\), as established in the theorem’s proof. Countable additivity of \(\mu\) gives the asserted equality, with \(+\infty\) allowed. \(\square\)

Worked Example: Growing Finite Sets Under Counting Measure

Let \(X=\mathbb{N}=\{1,2,\ldots\}\), let \(\mathcal{M}\) be its power set, and let \(\mu\) be counting measure, so \(\mu(E)\) is the number of elements of \(E\), or \(+\infty\) if \(E\) is infinite. Set \(A_n=\{1,\ldots,n\}\). These measurable sets are increasing, and \(D_1=\{1\}\), while \(D_n=\{n\}\) for \(n\ge2\). Every increment has measure \(1\), so

\mu(A_N)=\sum_{n=1}^{N}\mu(D_n)=\sum_{n=1}^{N}1=N, \qquad \sum_{n=1}^{\infty}\mu(D_n)=\sum_{n=1}^{\infty}1=+\infty.

The union is \(\mathbb{N}\), whose counting measure is \(+\infty\). Continuity from below gives \(\lim_{N\to\infty}\mu(A_N)=+\infty=\mu(\mathbb{N})\). Check: directly, the first \(N\) natural numbers are \(N\) points, and the union contains at least \(N\) points for every \(N\); therefore its counting measure is infinite, in agreement with the increment sum.

Worked Example: Increasing Intervals in a Finite-Measure Space

Take \(X=[0,1]\), \(\mathcal{M}\) to be the Lebesgue sigma-algebra on \(X\), and \(\mu\) to be Lebesgue measure on \(X\). For \(n\ge1\), let \(A_n=[0,1-2^{-n}]\). These intervals increase, and their union is \([0,1)\): every \(x\in[0,1)\) eventually satisfies \(x\le1-2^{-n}\), while \(1\) belongs to none of the \(A_n\). Their measures are

\mu(A_n)=1-2^{-n}.

The first increment is \(D_1=[0,1/2]\), with measure \(1/2\). For \(n\ge2\), \(D_n=(1-2^{-(n-1)},\,1-2^{-n}]\), whose length is \((1-2^{-n})-(1-2^{-(n-1)})=2^{-n}\). Consequently, the total increment measure is

\sum_{n=1}^{\infty}\mu(D_n) =\frac12+\sum_{n=2}^{\infty}2^{-n} =\frac12+\frac{1/4}{1-1/2} =1.

The limit of the measures is \(\lim_{n\to\infty}(1-2^{-n})=1\). Independently, \([0,1)\) has Lebesgue measure equal to its length, \(1\). Thus the limit, the union’s measure, and the sum of the increments all agree. The omitted endpoint \(1\) does not change the measure: a single point has Lebesgue measure zero.

Worked Example: A Dirac Measure and Zero-Measure Increments

Let \(X=\mathbb{R}\), \(\mathcal{M}=\mathcal{B}_{\mathbb{R}}\), and \(\mu=\delta_0\), the Dirac measure at \(0\), defined by \(\delta_0(E)=1\) if \(0\in E\) and \(\delta_0(E)=0\) otherwise. This is a finite measure. Put \(A_n=(-n,n)\). The sets increase and their union is \(\mathbb{R}\). Every \(A_n\) contains \(0\), so \(\delta_0(A_n)=1\) for every \(n\), and \(\delta_0(\mathbb{R})=1\).

Here \(D_1=(-1,1)\), which contains \(0\) and has \(\delta_0\)-measure \(1\). For \(n\ge2\), \(D_n=(-n,-(n-1)]\cup[n-1,n)\), which does not contain \(0\), so \(\delta_0(D_n)=0\). The increment sum is \(1+\sum_{n=2}^{\infty}0=1\). This agrees with the limit of the measures and with the direct calculation \(\delta_0(\mathbb{R})=1\). The example shows that later increments can be nonempty while having measure zero.

When the Union Has Finite Measure

Corollary: Let \((X,\mathcal{M},\mu)\) be a measure space, and let \((A_n)_{n\ge1}\) be an increasing sequence in \(\mathcal{M}\). The union \(\bigcup_{n=1}^{\infty}A_n\) has finite measure if and only if the numbers \(\mu(A_n)\) are bounded above by a finite constant.

Proof. If \(\mu(\bigcup_n A_n)\) is finite, then \(A_n\subseteq\bigcup_k A_k\) for every \(n\). Monotonicity gives \(\mu(A_n)\le\mu(\bigcup_k A_k)\), so the sequence is bounded above by a finite number. Conversely, suppose there is a finite constant \(C\) such that \(\mu(A_n)\le C\) for every \(n\). Continuity from below gives \(\mu(\bigcup_nA_n)=\lim_n\mu(A_n)\le C\), so the union has finite measure. \(\square\)

This conclusion does not require each increment to have positive measure. For example, in the Dirac-measure calculation above, the union has finite measure even though it contains infinitely many points and all increments after the first have measure zero.

A Common Mistake: Forgetting the Increasing Hypothesis

Continuity from below concerns sequences that satisfy \(A_n\subseteq A_{n+1}\) at every step; it is not a rule for arbitrary sequences of measurable sets. On \(\mathbb{R}\) with its Borel sigma-algebra and Lebesgue measure \(m\), let \(A_n=(n,n+1)\) for \(n\ge1\). Each set has measure \(1\), so \(\lim_n m(A_n)=1\). But the sequence is not increasing: the intervals are disjoint, and \(A_n\) is not a subset of \(A_{n+1}\). Their union has infinite measure, since it is a countable disjoint union of intervals each of measure \(1\). Thus the incorrect calculation \(m(\bigcup_n A_n)=\lim_n m(A_n)=1\) applies the theorem without its hypothesis; the actual union measure is \(+\infty\).

The disjoint-increment argument also pinpoints why the hypothesis matters. For an increasing sequence, the sets \(A_n\setminus A_{n-1}\) are disjoint and account for the union exactly once. Without inclusion, those differences need not form a disjoint decomposition of the union, so their partial sums need not equal the measures of the original sets.

Check Your Understanding

Use the increasing-set hypothesis and the disjoint-increment decomposition when deciding whether continuity from below applies.

For an increasing sequence \((A_n)\), define \(D_1\) and \(D_n\) for \(n\ge2\). Explain why the \(D_n\) are pairwise disjoint.
On \(\mathbb{N}\) with counting measure, take \(A_n=\{2,4,\ldots,2n\}\). Find \(\mu(A_n)\), the union, and its measure. Which increments contribute to the increment sum?
Give the precise hypothesis that fails for \(A_n=(n,n+1)\), and explain why the measures of these sets do not determine the measure of their union by continuity from below.
Suppose \((A_n)\) is increasing and \(\mu(A_n)\le 7\) for every \(n\). What can you conclude about \(\mu(\bigcup_n A_n)\), and which result justifies the conclusion?
In the Dirac-measure example, why does a nonempty increment \(D_n\) for \(n\ge2\) have measure zero?

What You'll Learn