From Decreasing Sets to Their Intersection
For an increasing sequence of measurable sets, continuity from below identifies the measure of the union with the limit of the measures. The corresponding statement for decreasing sets concerns their intersection. Its proof is a duality argument: inside the first set, take complements relative to that set. Those relative complements form an increasing sequence, so continuity from below applies. The finiteness of the first set ensures that the subtraction in this argument is valid.
Throughout, \((X,\mathcal{M},\mu)\) is a measure space: \(\mathcal{M}\) is a sigma-algebra of subsets of \(X\), and \(\mu:\mathcal{M}\to[0,\infty]\) is a measure. The measure space need not be complete, finite, or sigma-finite. A sequence \((A_n)\) in \(\mathcal{M}\) is decreasing if \(A_{n+1}\subseteq A_n\) for every \(n\ge1\). All set inclusions and identities below hold exactly, and measures may generally take the value \(+\infty\).
We use continuity from below, established in the earlier material on that result: if \((B_n)\) is an increasing sequence in \(\mathcal{M}\), then \(\bigcup_{n=1}^{\infty}B_n\in\mathcal{M}\) and \(\mu(\bigcup_n B_n)=\lim_{n\to\infty}\mu(B_n)\). The limit exists in \([0,\infty]\) because monotonicity of measure makes \((\mu(B_n))\) nondecreasing.
The Continuity-From-Above Theorem
Proof. Write \(A=\bigcap_{n=1}^{\infty}A_n\). Because \(\mathcal{M}\) is a sigma-algebra, \(A\in\mathcal{M}\). Since \(A_n\subseteq A_1\), monotonicity gives \(0\le\mu(A_n)\le\mu(A_1)<\infty\) for every \(n\). Also \(A_{n+1}\subseteq A_n\), so \(\mu(A_{n+1})\le\mu(A_n)\). Thus the real sequence \((\mu(A_n))\) is nonincreasing and bounded below by zero, and it has a finite limit.
For each \(n\), define the relative complement \(B_n=A_1\setminus A_n\). These sets are measurable because both \(A_1\) and \(A_n\) are measurable. If \(x\in B_n\), then \(x\notin A_n\); since \(A_{n+1}\subseteq A_n\), this implies \(x\notin A_{n+1}\), so \(x\in B_{n+1}\). Hence \((B_n)\) is increasing. De Morgan's law, together with \(A_n\subseteq A_1\), gives $$ \bigcup_{n=1}^{\infty}B_n =\bigcup_{n=1}^{\infty}(A_1\setminus A_n) =A_1\setminus\bigcap_{n=1}^{\infty}A_n =A_1\setminus A. $$
For every \(n\), the disjoint union \(A_1=A_n\cup B_n\) and finite additivity give \(\mu(A_1)=\mu(A_n)+\mu(B_n)\). All these measures are finite because \(B_n\subseteq A_1\). Therefore \(\mu(A_n)=\mu(A_1)-\mu(B_n)\). Applying continuity from below to \((B_n)\) yields $$ \lim_{n\to\infty}\mu(A_n) =\mu(A_1)-\lim_{n\to\infty}\mu(B_n) =\mu(A_1)-\mu(A_1\setminus A) =\mu(A). $$ The last equality follows by applying finite additivity to the disjoint union \(A_1=A\cup(A_1\setminus A)\). This proves the stated identity, including the case \(A_1=\varnothing\), in which every set in the sequence is empty. \(\square\)
The theorem needs only finite measure at the first stage. It does not require a finite measure space: later sets are contained in \(A_1\), and the argument takes place entirely inside \(A_1\). This is why the exact hypothesis is \(\mu(A_1)<\infty\), rather than \(\mu(X)<\infty\).
Worked Example: Decreasing Intervals with a Nontrivial Intersection
For \(n\ge1\), take \(A_n=[0,\frac12+\frac1n]\) in \((\mathbb{R},\mathcal{L},m)\), where \(m\) is Lebesgue measure. The right endpoint decreases as \(n\) increases, so \(A_{n+1}\subseteq A_n\). Also \(m(A_1)=m([0,\frac32])=\frac32<\infty\), so the theorem applies.
A point \(x\in[0,\frac12]\) belongs to every \(A_n\). If \(x>\frac12\), choose an integer \(n\) with \(n>1/(x-\frac12)\). Then \(1/n<x-\frac12\), so \(\frac12+\frac1n<x\), and \(x\notin A_n\). Points outside \([0,\frac12]\) are also not in every set. Thus \(\bigcap_n A_n=[0,\frac12]\). Each interval has length \(m(A_n)=\frac12+\frac1n\), and consequently $$ \lim_{n\to\infty}m(A_n)=\lim_{n\to\infty}\left(\frac12+\frac1n\right)=\frac12 =m\left([0,\frac12]\right). $$
Check: At \(n=1\), the interval is \([0,\frac32]\), of length \(\frac32=\frac12+1\). At \(n=2\), it is \([0,1]\), of length \(1=\frac12+\frac12\). These values decrease toward the intersection's length \(\frac12\), as the theorem predicts.
Worked Example: A Decreasing Sequence with a Null Intersection
For \(n\ge1\), let \(C_n=[-\frac1n,\frac1n]\) in \((\mathbb{R},\mathcal{L},m)\). The radii decrease, so \(C_{n+1}\subseteq C_n\), and \(m(C_1)=2<\infty\). The point \(0\) is in every \(C_n\). For any \(x\ne0\), choose \(n>1/|x|\); then \(1/n<|x|\), so \(x\notin C_n\). Therefore \(\bigcap_n C_n=\{0\}\).
The length of \(C_n\) is \(2/n\), which tends to zero. A singleton has Lebesgue measure zero, so the theorem gives $$ \lim_{n\to\infty}m(C_n)=\lim_{n\to\infty}\frac2n=0 =m(\{0\}). $$
Check: The first two lengths are \(2/1=2\) and \(2/2=1\); both are at least the limiting value \(0\). Independently, since \(\{0\}\subseteq C_n\) for every \(n\), monotonicity gives \(m(\{0\})\le m(C_n)\), consistent with these lengths and the limit.
Worked Example: A Finite Discrete Measure
Let \(X=\{a,b,c\}\), \(\mathcal{M}=\mathcal{P}(X)\), and \(\mu=2\delta_a+3\delta_b+5\delta_c\), where \(\delta_x(E)=1\) if \(x\in E\) and \(0\) otherwise. Thus a set's measure is the sum of the weights of the points it contains. Define \(D_1=\{a,b,c\}\), \(D_2=\{a,b\}\), and \(D_n=\{a\}\) for \(n\ge3\). The sets decrease, and \(\mu(D_1)=2+3+5=10<\infty\).
The intersection is \(\{a\}\): \(a\) belongs to every set, while \(b\) and \(c\) are absent from every \(D_n\) with \(n\ge3\). The set measures are \(\mu(D_1)=2+3+5=10\), \(\mu(D_2)=2+3=5\), and \(\mu(D_n)=2\) for \(n\ge3\). Hence $$ \lim_{n\to\infty}\mu(D_n)=2=\mu(\{a\}). $$
Check: The intersection contains only \(a\), whose assigned mass is \(2\). The sequence of measures is \(10,5,2,2,\ldots\), so its limit is also \(2\).
A Useful Zero-Measure Criterion
Proof. The theorem gives \(\mu(\bigcap_n A_n)=\lim_n\mu(A_n)\). Equality of these two nonnegative real numbers shows that one is zero exactly when the other is zero. \(\square\)
This criterion is useful when a decreasing sequence describes increasingly restrictive conditions. For instance, the symmetric intervals in the second example shrink to a point of Lebesgue measure zero; the criterion confirms that their measures must tend to zero. The intersection need not be empty: the relevant conclusion is that it has measure zero.
Common Pitfall: Replacing the Intersection by the Union
The theorem concerns \(\bigcap_n A_n\), not \(\bigcup_n A_n\). In the first worked example, the sets decrease from \(A_1=[0,\frac32]\) toward \([0,\frac12]\). Their union is \([0,\frac32]\), of measure \(\frac32\), while the limit of their measures is \(\frac12\). Thus the calculation “the limit of the measures equals the measure of the union” would give \(\frac12=\frac32\), which is false. The intersection is \([0,\frac12]\), and its measure is indeed \(\frac12\).
A second useful check is to verify the hypotheses before applying the theorem: the sets must be measurable and decreasing, and the first set must have finite measure. Once those conditions hold, monotonicity ensures all later measures are finite and nonincreasing. The theorem then identifies their limit with the measure of the intersection, whether that intersection is empty, nonempty and null, or of positive measure.
Check Your Understanding
For each question, first identify the sequence's direction of inclusion and check the finite-measure hypothesis.
- In \((\mathbb{R},\mathcal{L},m)\), let \(E_n=[0,2+\frac1n]\). Find \(\bigcap_n E_n\) and compute the limit of \(m(E_n)\).
- Let \((A_n)\) be decreasing and measurable with \(\mu(A_1)=7\). What does continuity from above say about \(\mu(\bigcap_n A_n)\), and why are all the \(\mu(A_n)\) finite?
- For \(F_n=[-\frac{3}{n},\frac{3}{n}]\) in \(\mathbb{R}\) with Lebesgue measure, find the intersection and verify the theorem by computing the measures.
- In the finite discrete example, what would the limit of the measures be if \(D_n=\{b,c\}\) for every \(n\ge2\), with \(D_1=\{a,b,c\}\)?
- Explain why the first worked example does not give the measure of the union when one takes the limit of the set measures.