The Issue: Decreasing Sets Can Lose Mass at Infinity
For a decreasing sequence of measurable sets, it is tempting to expect that the measures decrease to the measure of the intersection. That conclusion is guaranteed when the first set has finite measure. Without that hypothesis, measure can remain infinite at every finite stage even when the intersection is empty. Counting measure on the natural numbers gives a direct computation of this failure.
Let \((X,\mathcal{M},\mu)\) be a measure space. A sequence \((A_n)\) in \(\mathcal{M}\) is decreasing if \(A_{n+1}\subseteq A_n\) for every \(n\ge1\). Its intersection consists of the points that belong to every \(A_n\). The continuity-from-above theorem says that if \(\mu(A_1)<\infty\), then \(\lim_{n\to\infty}\mu(A_n)=\mu(\bigcap_{n=1}^{\infty}A_n)\). The finiteness assumption concerns the measure of the first set, not merely whether each \(A_n\) is measurable.
Here \(\#E\) denotes the number of elements of a finite set \(E\). Since every subset of \(\mathbb{N}\) is measurable, there is no measurability issue in the examples below. The distinction between finite and sigma-finite matters: sigma-finiteness gives a countable cover by sets of finite measure, but it does not say that the particular set \(A_1\) in a decreasing sequence has finite measure.
Worked Calculations on the Natural Numbers
Worked Example: Decreasing Tails with Empty Intersection
For each integer \(n\ge1\), define \(A_n=\{k\in\mathbb{N}:k\ge n\}\). Compute \(\bigcap_{n=1}^{\infty}A_n\) and \(\lim_n\mu(A_n)\).
The sets are decreasing: if \(k\in A_{n+1}\), then \(k\ge n+1\), so \(k\ge n\) and \(k\in A_n\). For any fixed \(n\), the distinct integers \(n,n+1,n+2,\ldots\) all belong to \(A_n\). Thus \(A_n\) is infinite and \(\mu(A_n)=\infty\).
No natural number belongs to every tail. Indeed, for a fixed \(k\), the choice \(n=k+1\) gives \(k<n\), so \(k\notin A_{k+1}\). Therefore \(\bigcap_{n=1}^{\infty}A_n=\varnothing\), and \(\mu(\bigcap_n A_n)=\mu(\varnothing)=0\). On the other hand, the sequence of measures is \(\infty,\infty,\infty,\ldots\), so its limit in the extended nonnegative reals is \(\infty\).
Check: For \(n=1\), \(A_1=\mathbb{N}\) and \(\mu(A_1)=\infty\); for \(n=2\), \(A_2=\{2,3,\ldots\}\) and \(\mu(A_2)=\infty\). These computed values agree with the general count for every tail. The hypotheses for continuity from above fail exactly at \(\mu(A_1)<\infty\).
Proof. If \(k\in A_{n+1}\), then \(k\ge n+1\ge n\), so \(k\in A_n\); hence the sequence is decreasing. For every \(n\), all integers \(n+j\), \(j=0,1,2,\ldots\), are distinct members of \(A_n\), so \(A_n\) is infinite and \(\mu(A_n)=\infty\). If \(k\in\mathbb{N}\), then \(k\notin A_{k+1}\), so \(k\) is not in the intersection. The intersection is therefore empty and has measure zero. Consequently, \(\lim_n\mu(A_n)=\infty\ne0=\mu(\bigcap_n A_n)\), as claimed. \(\square\)
Worked Example: A Finite Intersection but Infinite Measures
For \(n\ge1\), let \(D_n=\{1\}\cup\{k\in\mathbb{N}:k\ge n+1\}\). Determine the intersection and the measure of each set.
The point \(1\) belongs to every \(D_n\). If \(k\ge2\), choose \(n=k\); then \(k\not\ge n+1=k+1\), and \(k\ne1\), so \(k\notin D_k\). Thus \(\bigcap_nD_n=\{1\}\), whose counting measure is \(1\).
For every \(n\), \(D_n\) contains all integers \(n+1,n+2,\ldots\), so \(D_n\) is infinite and \(\mu(D_n)=\infty\). The sequence is decreasing because increasing \(n\) raises the threshold \(n+1\), while the singleton \(\{1\}\) stays fixed. Consequently, \(\lim_n\mu(D_n)=\infty\), whereas \(\mu(\bigcap_nD_n)=1\).
Check: \(D_1=\{1\}\cup\{2,3,\ldots\}=\mathbb{N}\), so \(\mu(D_1)=\infty\); \(D_2=\{1\}\cup\{3,4,\ldots\}\) is also infinite. The computed intersection contains exactly \(1\), not the whole tail. This example shows that even a finite-measure intersection does not repair the failure when the first set has infinite measure.
Worked Example: A Decreasing Sequence with an Infinite Intersection
Define \(C_n=\{2,4,6,\ldots\}\cup\{k\in\mathbb{N}:k\text{ is odd and }k\ge2n+1\}\). Find \(C_1\), the intersection, and the measures.
The even positive integers are in every \(C_n\), and the odd-number threshold increases with \(n\). Thus \(C_{n+1}\subseteq C_n\). At \(n=1\), \(C_1\) contains all even integers and all odd integers at least \(3\). Therefore \(C_1=\mathbb{N}\setminus\{1\}\), not all of \(\mathbb{N}\).
Every \(C_n\) contains all positive even integers, so it is infinite and \(\mu(C_n)=\infty\). Every odd integer \(k\) is eventually excluded: choose an integer \(n\ge1\) with \(2n+1>k\); then \(k\) fails the odd-integer condition, and it is not even. Hence the intersection consists exactly of the positive even integers. It is infinite, so \(\mu(\bigcap_nC_n)=\infty\).
Check: At \(n=1\), the even integers together with odd integers at least \(3\) omit precisely \(1\). At \(n=2\), the odd integers included are \(5,7,9,\ldots\), so \(3\) has been removed while every even integer remains. The measures are infinite at each stage and in the intersection, so equality holds here in the extended-real sense, \(\infty=\infty\). Infinite measure permits failure; it does not force failure for every sequence.
What the Finite-Measure Hypothesis Controls
The finite-measure condition prevents the sequence from carrying an unbounded amount of measure from arbitrarily far out in the first set. In the tail example, each finite stage still contains infinitely many integers, even though any particular integer is eventually removed. The total measure at each stage is therefore infinite, and the numerical sequence \((\mu(A_n))\) does not decrease to the measure of the intersection.
There is a simple complementary result for counting measure when the first set is finite. It also handles the degenerate case \(A_1=\varnothing\).
Proof. Every \(A_n\) is a subset of the finite set \(A_1\). Whenever \(A_{n+1}\) is a proper subset of \(A_n\), at least one element of \(A_n\) is removed. There can be at most \(\#A_1\) such removals, because no removed element can return in a decreasing sequence. Hence only finitely many indices can satisfy \(A_{n+1}\subsetneq A_n\). After the last such index, the sets are constant; call their common value \(A_N\). If there are no proper removals, take \(N=1\). The intersection of a sequence that equals \(A_N\) for all \(n\ge N\) is \(A_N\), since earlier sets contain it. Also, \(\mu(A_n)=\#A_N\) for all \(n\ge N\). Thus both sides equal \(\#A_N\). If \(A_1=\varnothing\), every set is empty and the same conclusion holds with \(N=1\). \(\square\)
Worked Example: Restricting the Tails to a Finite Set
Let \(F=\{1,2,3,4,5,6\}\) and \(E_n=F\cap A_n\), where \(A_n=\{k\in\mathbb{N}:k\ge n\}\). List the measures and compute the intersection.
Membership requires both \(1\le k\le6\) and \(k\ge n\). Thus \(E_1=\{1,2,3,4,5,6\}\), \(E_2=\{2,3,4,5,6\}\), \(E_3=\{3,4,5,6\}\), \(E_4=\{4,5,6\}\), \(E_5=\{5,6\}\), \(E_6=\{6\}\), and \(E_n=\varnothing\) for \(n\ge7\). Their measures are, respectively, \(6,5,4,3,2,1,0,\ldots\).
For every \(k\in F\), the choice \(n=k+1\) excludes \(k\) from \(E_n\). Therefore \(\bigcap_nE_n=\varnothing\), and \(\mu(\bigcap_nE_n)=0\). The displayed measure sequence is eventually zero, so its limit is also zero.
Check: At \(n=6\), the membership conditions leave exactly \(\{6\}\), with measure \(1\); at \(n=7\), no element of \(F\) is at least \(7\), so the set is empty and its measure is \(0\). This agrees with the finite-set proposition.
A Common Mistake and Its Correction
A common error is to apply continuity from above to any decreasing sequence and write \(\mu(\bigcap_nA_n)=\lim_n\mu(A_n)\) without checking \(\mu(A_1)<\infty\). For the tails \(A_n=\{k:k\ge n\}\), that calculation would give \(0=\infty\): the left side is \(0\) because the intersection is empty, while the right side is \(\infty\) because every tail is infinite. The correction is not to alter either computation. It is to recognize that the theorem's finite-measure hypothesis is absent.
A related mistake is to treat sigma-finiteness as if it supplied that hypothesis. Counting measure on \(\mathbb{N}\) is sigma-finite because \(\mathbb{N}=\bigcup_{k=1}^{\infty}\{k\}\) and each singleton has measure \(1\). But the tail sequence begins with \(A_1=\mathbb{N}\), which has infinite measure. Sigma-finiteness of the whole space does not turn this first set into a finite-measure set.
The central lesson is specific: continuity from above is guaranteed when the first set has finite measure, and the hypothesis cannot simply be dropped. The tail computation provides a counterexample even in a sigma-finite measure space with a very simple measure. At the same time, examples such as \(C_n\) show why the right conclusion is that the theorem is no longer guaranteed, not that equality must fail in every infinite-measure case.
Check Your Understanding
For each question, identify the decreasing sets and check the finite-measure hypothesis before applying continuity from above.
- For \(A_n=\{k\in\mathbb{N}:k\ge n\}\), why is the intersection empty, and what is \(\mu(A_n)\) for each \(n\)?
- For \(D_n=\{1\}\cup\{k\in\mathbb{N}:k\ge n+1\}\), compute the intersection and compare its measure with \(\lim_n\mu(D_n)\).
- Why is counting measure on \(\mathbb{N}\) sigma-finite but not finite? Explain why sigma-finiteness does not make the tail example satisfy the finite-measure hypothesis.
- For a decreasing sequence of subsets of a finite set \(A_1\), explain why only finitely many strict decreases can occur.
- For the sets \(C_n\) in the worked example, what is \(C_1\), and why does the intersection contain every positive even integer but no odd integer?