Continuity of Measures From Above

Measures of Decreasing Sequences

For an increasing sequence, a point belongs to the union as soon as it appears at one finite stage. For a decreasing sequence, the relevant limiting set is instead the intersection: its points remain in every stage. The measures of the stages are nonincreasing, but their limit agrees with the measure of the intersection only under a finiteness condition. This condition prevents infinite measure from obscuring what is removed at each stage.

Throughout, let $(X,\mathcal{F},\mu)$ be a measure space. A sequence of measurable sets $A_1,A_2,\ldots$ is decreasing if $A_{n+1}\subseteq A_n$ for every $n$. Write $A_n\downarrow A$ when the sequence is decreasing and $A=\bigcap_{n=1}^{\infty}A_n$. The intersection is measurable because a sigma-algebra is closed under countable intersections.

Theorem (Continuity from Above for Finite Measure): If $A_1\supseteq A_2\supseteq\cdots$ are measurable and $\mu(A_1)<\infty$, then $$ \mu\left(\bigcap_{n=1}^{\infty}A_n\right) = \lim_{n\to\infty}\mu(A_n). $$

This theorem was established in “Definition of a Measure.” Its hypothesis is attached to the first set, not merely to each later stage: since $A_n\subseteq A_1$, it guarantees that every stage and the intersection have finite measure. By monotonicity, the sequence $\mu(A_n)$ is nonincreasing, so its limit exists. Continuity from above identifies that limit with the measure of the points that remain at every stage.

Worked Example: Closed Intervals Shrinking to a Limit

Use Lebesgue measure on the Borel subsets of $\mathbb{R}$, and set $$ A_n=[-1/n,\,2+1/n],\qquad n\geq 1. $$ The left endpoints increase toward $0$, while the right endpoints decrease toward $2$, so $A_{n+1}\subseteq A_n$. A number belongs to every interval exactly when it is at least $0$ and at most $2$; hence $$ \bigcap_{n=1}^{\infty}A_n=[0,2]. $$ For each $n$, the interval length is $$ (2+1/n)-(-1/n)=2+2/n. $$ Thus $\mu(A_n)=2+2/n$, and $\mu(A_1)=4<\infty$. The measures converge to $2$, in agreement with $$ \mu\left(\bigcap_{n=1}^{\infty}A_n\right)=\mu([0,2])=2. $$ The endpoints of the intervals move at every stage, but the intersection retains precisely the interval between their limiting positions.

Controlling What Lies Outside the Intersection

Continuity from above also gives a useful quantitative conclusion. If $A_n\downarrow A$, then $A_n\setminus A$ is the part of stage $n$ that will eventually be removed. Under the finite-measure hypothesis, the measure of this remainder tends to zero. This is not the same as saying the sets themselves eventually equal the intersection: the remainder can be nonempty at every finite stage.

Theorem (Approximation from Above in Measure): Suppose $A_n\downarrow A=\bigcap_{n=1}^{\infty}A_n$ and $\mu(A_1)<\infty$. Then $$ \lim_{n\to\infty}\mu(A_n\setminus A)=0. $$ Consequently, for every $\varepsilon>0$, there is an $N$ such that $\mu(A_n\setminus A)<\varepsilon$ for all $n\geq N$.

Proof. Since $A\subseteq A_n\subseteq A_1$, both $A$ and $A_n$ have finite measure. The increment formula for nested sets gives $$ \mu(A_n)=\mu(A)+\mu(A_n\setminus A). $$ By Continuity from Above for Finite Measure, $\mu(A_n)\to\mu(A)$. Subtracting the finite number $\mu(A)$ in the displayed identity yields $$ \mu(A_n\setminus A)=\mu(A_n)-\mu(A)\longrightarrow 0. $$ The definition of convergence then gives, for each $\varepsilon>0$, an $N$ such that the remainder has measure less than $\varepsilon$ whenever $n\geq N$. $\square$

The requirement that the remainder have small measure is weaker than requiring it to be empty. For example, intervals can keep narrowing toward a point while retaining a positive-length interval at every finite stage. Their remaining lengths can tend to zero even though no stage has yet become the intersection.

Worked Example: Open Intervals Shrinking to the Empty Set

On $\mathbb{R}$ with Lebesgue measure, let $A_n=(0,1/n)$. These sets decrease, and no positive real number belongs to every one of them: for any $x>0$, choose $n>1/x$, giving $1/n<x$. Therefore $$ \bigcap_{n=1}^{\infty}A_n=\varnothing. $$ The measure of $A_n$ is $1/n$, so $\mu(A_n)\to 0=\mu(\varnothing)$. Here $A_n\setminus\bigcap_k A_k=A_n$, and the measure of the part outside the intersection is exactly $1/n$. The sets are nonempty at every stage, even though their measures tend to zero.

One Finite Stage Is Enough

The finiteness condition can be checked at any stage, not only at the first one. A decreasing sequence may start with an infinite-measure set and later enter a set of finite measure. Once that happens, the tail of the sequence satisfies the usual theorem, and removing the finitely many earlier stages does not change the intersection.

Corollary (Continuity from Above After a Finite Stage): Suppose $A_n\downarrow A$ and $\mu(A_N)<\infty$ for some $N$. Then $$ \mu(A)=\lim_{n\to\infty}\mu(A_n). $$

Proof. Consider the tail $B_k=A_{N+k-1}$ for $k\geq1$. It is decreasing and $\mu(B_1)=\mu(A_N)<\infty$. Continuity from Above for Finite Measure gives $$ \mu\left(\bigcap_{k=1}^{\infty}B_k\right)=\lim_{k\to\infty}\mu(B_k). $$ The intersection of this tail equals $\bigcap_{n=1}^{\infty}A_n$: every set before $A_N$ contains $A_N$, so adding those earlier sets does not further restrict the intersection. Also, the tail measures are the original measures with only finitely many initial terms omitted, and therefore have the same limit. This proves the claim. $\square$

Worked Example: An Infinite-Measure First Stage

On $\mathbb{R}$ with Lebesgue measure, define $A_1=\mathbb{R}$ and $A_n=[0,1]$ for every $n\geq2$. This is a decreasing sequence, and $\mu(A_1)=\infty$, so the finite-measure hypothesis at the first stage does not apply. But $A_2$ has finite measure, and $$ \bigcap_{n=1}^{\infty}A_n=[0,1]. $$ For every $n\geq2$, $\mu(A_n)=1$. Hence $$ \lim_{n\to\infty}\mu(A_n)=1=\mu\left(\bigcap_{n=1}^{\infty}A_n\right). $$ This example illustrates the finite-stage corollary: an infinite initial measure does not prevent continuity from above when a later stage has finite measure.

Localizing to a Finite-Measure Set

Sometimes the sets $A_n$ do not have finite measure, and no stage is known to have finite measure. A useful alternative is to restrict attention to a fixed measurable set of finite measure. The intersections with that set form a decreasing sequence whose first term has finite measure, so continuity from above applies there. This localization lets us draw conclusions about finite portions of a larger measure space.

Theorem (Localized Continuity from Above): Let $A_n$ be a decreasing sequence of measurable sets, let $A=\bigcap_{n=1}^{\infty}A_n$, and let $B\in\mathcal{F}$ satisfy $\mu(B)<\infty$. Then $$ \lim_{n\to\infty}\mu(B\cap A_n)=\mu(B\cap A). $$

Proof. Each $B\cap A_n$ is measurable, and the sequence $B\cap A_n$ is decreasing. Its first set satisfies $\mu(B\cap A_1)\leq\mu(B)<\infty$, by monotonicity. Its intersection is $$ \bigcap_{n=1}^{\infty}(B\cap A_n) = B\cap\left(\bigcap_{n=1}^{\infty}A_n\right) = B\cap A. $$ Continuity from Above for Finite Measure, applied to the sequence $B\cap A_n$, now gives the asserted limit. $\square$

Worked Example: Restricting an Unbounded Sequence to a Bounded Window

For Lebesgue measure on $\mathbb{R}$, take $A_n=[n,\infty)$. The sets decrease and their intersection is empty, but each $A_n$ has infinite measure. Now choose the finite-measure window $B=[0,5]$. For $1\leq n\leq5$, $$ B\cap A_n=[n,5], $$ whose measure is $5-n$; for $n\geq5$, the intersection is either the singleton $\{5\}$ when $n=5$, or empty when $n>5$, and in either case has measure zero at and after $n=5$. In particular, the measures $\mu(B\cap A_n)$ tend to zero. The localized theorem gives the same conclusion from $$ B\cap\bigcap_{n=1}^{\infty}A_n=B\cap\varnothing=\varnothing. $$ Although the original sets all have infinite measure, their portions inside a fixed finite window obey continuity from above.

Why the Finiteness Condition Matters

Without a finite-measure condition, continuity from above can fail completely. Consider $A_n=[n,\infty)$ in $\mathbb{R}$ with Lebesgue measure. Their intersection is empty: for any real $x$, an integer $n>x$ gives $x\notin A_n$. Yet every stage has infinite measure. Thus $$ \mu\left(\bigcap_{n=1}^{\infty}A_n\right)=0, \qquad \lim_{n\to\infty}\mu(A_n)=+\infty. $$ The measures do not converge to the measure of the intersection.

The contrast with continuity from below is important. For increasing sequences, the measure of the union equals the limit of the measures without any finiteness assumption. For decreasing sequences, an infinite amount of measure can persist at every stage even as the intersection becomes small or empty. The finite-measure hypothesis makes subtraction meaningful: it allows the measure removed between a stage and the intersection to be expressed as the difference of two finite quantities.

Common pitfall: Decreasing sets have nonincreasing measures, but that fact alone does not identify the limit with the measure of the intersection. Check that the first set, or at least some later set, has finite measure; otherwise continuity from above may fail.

When applying the theorem, first identify the intersection and check the finiteness condition. If a finite-measure stage occurs later, apply the theorem to that tail. If no such stage is available but the question concerns a finite-measure region $B$, intersect every set with $B$ and use localized continuity from above. These steps distinguish a valid limiting argument from one that silently subtracts infinite measures.

Check Your Understanding

Use continuity from above and its consequences to answer the following questions.

What is the limiting set associated with a decreasing sequence of measurable sets?
Which finiteness condition allows the Continuity from Above Theorem to apply directly?
Under that condition, why does the measure of $A_n\setminus A$ tend to zero?
If $\mu(A_1)=\infty$ but $\mu(A_N)<\infty$ for some later stage, how can continuity from above still be applied?
How does restricting a decreasing sequence to a finite-measure set $B$ give a continuity result?
Give an example showing why decreasing measures alone do not guarantee continuity from above.