The Part of a Decreasing Set That Disappears
Suppose \(A_1\supseteq A_2\supseteq\cdots\) is a decreasing sequence of measurable sets. Its intersection \(A=\bigcap_{n=1}^{\infty}A_n\) consists of the points that remain in every set. The difference \(A_n\setminus A\) is the part of \(A_n\) that will eventually disappear: each of its points is absent from at least one later set. When the first set has finite measure, continuity from above implies that this disappearing part has measure tending to zero.
Throughout, let \((X,\mathcal{M},\mu)\) be a measure space. No completeness or sigma-finiteness assumption is needed. Sets in the sequence are required to belong to \(\mathcal{M}\), and “decreasing” means \(A_{n+1}\subseteq A_n\) for every \(n\). The excess set \(A_n\setminus A\) is measurable because sigma-algebras are closed under countable intersections and differences.
Indeed, \(D_{n+1}\subseteq D_n\) follows from \(A_{n+1}\subseteq A_n\). If \(x\) belonged to every \(D_n\), it would belong to every \(A_n\), hence to \(A\), while also being outside \(A\). That contradiction proves the intersection is empty.
The Measure of the Excess Tends to Zero
We use continuity from above for measures: if \((A_n)\) is a decreasing sequence of measurable sets and \(\mu(A_1)<\infty\), then \(\mu(\bigcap_n A_n)=\lim_{n\to\infty}\mu(A_n)\). This result requires finite measure of the initial set; it is not a consequence of countable additivity alone when that measure is infinite.
Proof. Since \(A\subseteq A_n\), the sets \(A\) and \(A_n\setminus A\) are disjoint and their union is \(A_n\). Countable additivity implies finite additivity, so \[ \mu(A_n)=\mu(A)+\mu(A_n\setminus A). \] Monotonicity gives \(\mu(A)\leq\mu(A_1)<\infty\) and \(\mu(A_n)\leq\mu(A_1)<\infty\). Thus all terms in this equality are finite, and subtracting \(\mu(A)\) is valid: \[ \mu(A_n\setminus A)=\mu(A_n)-\mu(A). \] By continuity from above, \(\mu(A_n)\to\mu(A)\). Therefore the right-hand side tends to zero, proving the assertion. \(\square\)
The finite-measure hypothesis makes the subtraction meaningful: it prevents an expression of the form \(+\infty-(+\infty)\). It also ensures that continuity from above applies. The theorem gives a quantitative statement about the sequence: not only do the measures \(\mu(A_n)\) approach the measure of the intersection, but the exact part of \(A_n\) outside that intersection has measure approaching zero.
Proof. For every \(n\), the disjoint decomposition \(A_n=A\cup(A_n\setminus A)\) and finite additivity give \(\mu(A_n)=\mu(A)+\mu(A_n\setminus A)\). If \(\mu(A_n)\to\mu(A)\), subtracting the finite constant \(\mu(A)\) shows that the excess measures tend to zero. Conversely, if the excess measures tend to zero, the same equality shows that \(\mu(A_n)\to\mu(A)\). Both implications use only this equality and the finiteness guaranteed by \(\mu(A_1)<\infty\). \(\square\)
In epsilon language, the theorem says that for every \(\varepsilon>0\), there exists an integer \(N\) such that \(n\geq N\) implies \(\mu(A_n\setminus A)<\varepsilon\). The choice of \(N\) may depend on the sequence and on \(\varepsilon\); the assertion is not a uniform bound over all decreasing sequences.
Worked Example: Nested Intervals in Lebesgue Measure
Take \(X=\mathbb{R}\), \(\mathcal{M}=\mathcal{L}\), and \(\mu=m\), Lebesgue measure. For \(n\geq1\), let \(A_n=[0,1+1/n]\). These are closed, Lebesgue measurable intervals. Since \(1+1/(n+1)<1+1/n\), the sequence is decreasing, and its intersection is \(A=[0,1]\).
The excess set is \[ A_n\setminus A=(1,1+1/n]. \] The left endpoint is excluded because it belongs to \(A\); the right endpoint is included because it belongs to \(A_n\). Its Lebesgue measure is its interval length: \[ m(A_n\setminus A)=(1+1/n)-1=1/n. \] Thus the excess measure tends to zero. To check by subtraction, \(m(A_n)=1+1/n\) and \(m(A)=1\), so \(m(A_n)-m(A)=(1+1/n)-1=1/n\), in agreement with the direct interval calculation. At \(n=1\), the excess is \((1,2]\) and has measure \(1\); at \(n=4\), it is \((1,5/4]\) and has measure \(1/4\).
Worked Example: A Decreasing Sequence for Counting Measure
Let \(X=\{a,b,c,d,e\}\), let \(\mathcal{M}\) be its power set, and let \(\mu\) be counting measure, so \(\mu(E)\) is the number of elements of \(E\). Define \(A_1=\{a,b,c,d,e\}\), \(A_2=\{a,b,c,d\}\), \(A_3=\{a,b,c\}\), \(A_4=\{a,b\}\), and \(A_n=\{a\}\) for \(n\geq5\). This is a decreasing sequence of measurable sets with \(\mu(A_1)=5<\infty\), and its intersection is \(A=\{a\}\).
For \(n=1,2,3,4\), respectively, the excess sets are \(\{b,c,d,e\}\), \(\{b,c,d\}\), \(\{b,c\}\), and \(\{b\}\). For every \(n\geq5\), the excess is empty. Their measures are therefore \(4,3,2,1,0,0,\ldots\), which tend to zero. The subtraction check gives \(\mu(A_n)-\mu(A)=5-1=4\) at \(n=1\), \(4-1=3\) at \(n=2\), \(3-1=2\) at \(n=3\), \(2-1=1\) at \(n=4\), and \(1-1=0\) for \(n\geq5\). These values match the direct counts in each case.
Worked Example: Weighted Tails on the Natural Numbers
Let \(X=\mathbb{N}=\{1,2,\ldots\}\), \(\mathcal{M}\) be the power set of \(\mathbb{N}\), and define \(\mu(E)=\sum_{k\in E}2^{-k}\). This is a measure; it assigns each singleton \(\{k\}\) the weight \(2^{-k}\). Set \(A_n=\{k\in\mathbb{N}:k\geq n\}\). The sets are decreasing and their intersection is empty: for any fixed \(k\), every \(n>k\) has \(k\notin A_n\). Also, \(\mu(A_1)=\sum_{k=1}^{\infty}2^{-k}=1\), so the initial measure is finite.
Because the intersection is empty, \(A_n\setminus A=A_n\). The geometric series gives \[ \mu(A_n\setminus A)=\sum_{k=n}^{\infty}2^{-k} =2^{-n}\sum_{j=0}^{\infty}2^{-j} =2^{-n}\cdot2 =2^{1-n}. \] For instance, this is \(1\) at \(n=1\), \(1/2\) at \(n=2\), and \(1/8\) at \(n=4\); the formula tends to zero as \(n\to\infty\). A check using the geometric-tail formula gives \(\mu(A_n)=2^{1-n}\) and \(\mu(A)=0\), so \(\mu(A_n)-\mu(A)=2^{1-n}\), the same value as the direct sum.
Why Infinite Initial Measure Is a Pitfall
The condition \(\mu(A_1)<\infty\) cannot simply be dropped. In \((\mathbb{R},\mathcal{L},m)\), take \(A_n=[n,\infty)\) for \(n\geq1\). These sets are measurable and decreasing, their intersection is empty, and \(m(A_1)=\infty\). For every \(n\), \(A_n\setminus\bigcap_k A_k=A_n\), and \(m(A_n)=\infty\). The excess measures do not tend to zero.
A common incorrect calculation is to write \(m(A_n\setminus A)=m(A_n)-m(A)=\infty-\infty=0\), because \(A=\varnothing\) is mistakenly assigned infinite measure, or because an informal subtraction is performed without checking finiteness. Here \(m(A)=m(\varnothing)=0\), but the actual difference is \(\infty-0=\infty\), not zero. More generally, subtraction from an equality of measures is justified in this argument only when the quantities being subtracted are finite. The finite initial measure hypothesis guarantees that for every \(n\), both \(m(A_n)\) and \(m(A)\) are finite.
The result is useful whenever a decreasing family describes what remains after successive restrictions: it controls the measure of points still present at stage \(n\) but absent from the limiting set. The conclusion concerns measure, not necessarily the number of points in the excess. In the weighted-tail example, every \(A_n\setminus A\) contains infinitely many points, even though its measure tends to zero.
Check Your Understanding
For each question, identify the measure space and verify whether the finite-initial-measure hypothesis is satisfied before applying the result.
- For \(A_n=[0,2+1/n]\) in \(\mathbb{R}\) with Lebesgue measure, find the intersection and compute \(m(A_n\setminus\bigcap_k A_k)\).
- Explain why the decomposition \(A_n=A\cup(A_n\setminus A)\) is disjoint, and state which measure property gives the corresponding sum of measures.
- For counting measure on \(\mathbb{N}\), let \(A_n=\{1,\ldots,n\}\). Does the theorem apply? Explain the role of the measure of \(A_1\), and find the intersection.
- For \(A_n=[n,\infty)\) with Lebesgue measure, compute the measure of the excess and identify the hypothesis that fails.
- In a finite measure space, prove directly from \(A_n=A\cup(A_n\setminus A)\) that \(\mu(A_n)\to\mu(A)\) whenever \(\mu(A_n\setminus A)\to0\).