Continuity of Measures From Below

Measures of Increasing Sequences

Countable subadditivity bounds the measure of a union by the sum of the measures of its pieces. For an increasing sequence, however, there is a more precise relationship: the measure of the union is determined by the measures of the successive sets. This lets us compute an infinite union by studying finite stages and taking a limit.

Throughout, let $(X,\mathcal{F},\mu)$ be a measure space. A sequence of measurable sets $A_1,A_2,\ldots$ is increasing if $A_n\subseteq A_{n+1}$ for every $n$. Its union is the set of all points that appear at some finite stage. The Continuity from Below Theorem, established in “Definition of a Measure,” states that

\mu\left(\bigcup_{n=1}^{\infty} A_n\right) = \lim_{n\to\infty}\mu(A_n) = \sup_{n\geq 1}\mu(A_n).

The limit and supremum here may be $+\infty$. Because the sets are nested, their measures are nondecreasing, so the limit exists as an extended nonnegative number. The theorem says that no measure is lost or gained in passing from all finite stages to their union.

Key idea: For increasing measurable sets, the measure of the union is the limit of the measures of the stages. This remains true whether the limit is finite or infinite.

The theorem is particularly useful when the union is easier to describe than to measure directly, or when each finite stage has a simple measure. It also explains how measure accumulates: each new stage adds a layer not present earlier, and these layers are disjoint.

Decomposing the Union into Layers

Given an increasing sequence, define its first layer to be $A_1$, and each later layer to be the part newly added at that stage:

D_1=A_1,\qquad D_n=A_n\setminus A_{n-1}\quad(n\geq 2).

The sets $D_n$ are measurable and pairwise disjoint. Every point in the union has a first stage at which it appears, so it belongs to exactly one layer. Thus the layers partition the union. The following formula records the measure contributed by each stage.

Theorem (Layer Decomposition for an Increasing Union): If $A_1\subseteq A_2\subseteq\cdots$ are measurable and $D_1=A_1$, $D_n=A_n\setminus A_{n-1}$ for $n\geq 2$, then $$ \mu\left(\bigcup_{n=1}^{\infty}A_n\right) = \sum_{n=1}^{\infty}\mu(D_n). $$

Proof. Each $D_n$ is measurable because sigma-algebras are closed under relative differences. If $i<j$, then $D_i\subseteq A_i\subseteq A_{j-1}$, whereas $D_j\subseteq X\setminus A_{j-1}$. Therefore $D_i\cap D_j=\varnothing$. Also, $\bigcup_{n=1}^{N}D_n=A_N$ for every $N$: the layers through stage $N$ consist exactly of the points in $A_N$. Consequently, $$ \bigcup_{n=1}^{\infty}D_n=\bigcup_{n=1}^{\infty}A_n. $$ Countable additivity applied to the pairwise disjoint measurable sets $D_n$ gives the claimed identity. $\square$

This decomposition also shows how the finite-stage measures accumulate. By finite additivity, $$ \mu(A_N)=\sum_{n=1}^{N}\mu(D_n). $$ The partial sums on the right increase with $N$; their limit is the full sum of the layer measures. The Continuity from Below Theorem identifies that limit with the measure of the union. The layer formula is useful when the new portion at each stage is easier to measure than the entire set.

Worked Example: Increasing Intervals Approaching an Open Interval

Use Lebesgue measure on the Borel subsets of $\mathbb{R}$, and for $n\geq 2$ let $$ A_n=[2^{-n},\,1-2^{-n}]. $$ These intervals increase: $2^{-(n+1)}<2^{-n}$, while $1-2^{-(n+1)}>1-2^{-n}$. Their union is $(0,1)$. Indeed, every $x\in(0,1)$ eventually satisfies both $2^{-n}\leq x$ and $2^{-n}\leq 1-x$, while neither endpoint 0 nor 1 belongs to any $A_n$. Their measures are $$ \mu(A_n)=(1-2^{-n})-2^{-n}=1-2^{1-n}. $$ As $n\to\infty$, $2^{1-n}\to 0$, so $$ \mu\left(\bigcup_{n=2}^{\infty}A_n\right) =\lim_{n\to\infty}(1-2^{1-n}) =1. $$ The union is open, whereas every stage is a closed interval. Continuity from below concerns the measures and the set union; it does not require the union to have the same form as the individual stages.

When a Finite Stage Captures Almost All the Measure

If the increasing union has finite measure, continuity from below gives more than a formula for its total measure. A sufficiently large finite stage leaves only a small amount of measure outside it. The finiteness assumption matters: subtracting an approximation from an infinite total does not give a useful finite error.

Theorem (Finite-Stage Approximation): Let $A_1\subseteq A_2\subseteq\cdots$ be measurable, and set $A=\bigcup_{n=1}^{\infty}A_n$. If $\mu(A)<\infty$, then for every $\varepsilon>0$ there is an $N$ such that $$ \mu(A\setminus A_N)<\varepsilon. $$

Proof. Since $A_N\subseteq A$ and $\mu(A)<\infty$, the increment formula for nested sets gives $$ \mu(A)=\mu(A_N)+\mu(A\setminus A_N). $$ By the Continuity from Below Theorem, $\mu(A_N)\to\mu(A)$. Choose $N$ large enough that $$ \mu(A)-\mu(A_N)<\varepsilon. $$ The increment formula identifies this difference with $\mu(A\setminus A_N)$, proving the claim. $\square$

This result says that all but an arbitrarily small amount of the union’s measure is already present at some finite stage. It does not say that $A\setminus A_N$ is empty, or that every point of $A$ appears by stage $N$. A small measure remainder can still contain points.

Worked Example: Finite Initial Segments under a Probability Measure

Let $X=\mathbb{N}$, let $\mathcal{F}=\mathcal{P}(\mathbb{N})$, and assign probability $2^{-k}$ to the point $k$. These weights define a probability measure because $$ \sum_{k=1}^{\infty}2^{-k}=1. $$ Take $A_n=\{1,2,\ldots,n\}$. The sets increase and their union is $\mathbb{N}$. The finite geometric sum gives $$ \mu(A_n)=\sum_{k=1}^{n}2^{-k}=1-2^{-n}. $$ Therefore $\mu(\bigcup_n A_n)=\lim_{n\to\infty}(1-2^{-n})=1$. For a prescribed $\varepsilon>0$, choose $N$ with $2^{-N}<\varepsilon$. Then $$ \mu(\mathbb{N}\setminus A_N)=\sum_{k=N+1}^{\infty}2^{-k}=2^{-N}<\varepsilon. $$ The omitted points are infinitely many, but their total probability can be made arbitrarily small by taking a sufficiently large finite stage.

Finite Stages for Unions That Are Not Nested

A sequence of sets need not itself be increasing for continuity from below to be useful. Given arbitrary measurable sets $E_1,E_2,\ldots$, form their finite cumulative unions $$ B_N=\bigcup_{n=1}^{N}E_n. $$ Then $B_N\subseteq B_{N+1}$, and $\bigcup_{N=1}^{\infty}B_N=\bigcup_{n=1}^{\infty}E_n$. Applying continuity from below to the $B_N$ gives

\mu\left(\bigcup_{n=1}^{\infty}E_n\right) = \lim_{N\to\infty}\mu\left(\bigcup_{n=1}^{N}E_n\right).

This is a finite-stage method for analyzing any countable union: first measure the union of the first $N$ sets, then let $N$ increase. The finite unions can overlap, so their measures are not generally the sums of the individual measures. Countable subadditivity remains available for bounds, but continuity from below supplies the exact limiting identity for the cumulative unions.

Worked Example: Expanding Intervals Exhaust the Real Line

Let $A_n=[-n,n]$ in $\mathbb{R}$ with Lebesgue measure. The sequence is increasing and its union is $\mathbb{R}$: for any real $x$, an integer $n\geq |x|$ satisfies $x\in[-n,n]$. Since $\mu(A_n)=2n$, continuity from below gives $$ \mu(\mathbb{R})=\lim_{n\to\infty}2n=+\infty. $$ In this case, the finite stages do not approximate the union in the sense of leaving a small finite-measure remainder. In fact, for every $N$, the difference $\mathbb{R}\setminus[-N,N]$ has infinite Lebesgue measure. The measures of the stages grow without bound, exactly as the measure of their union requires.

Finite and Infinite Limits: A Necessary Distinction

For an increasing sequence, continuity from below always identifies the measure of the union with the limit of the stage measures. If the union has finite measure, the stage measures approach that finite value and the finite-stage approximation theorem applies. If the union has infinite measure, the stage measures tend to $+\infty$. This follows from the same theorem: their limit is $\mu(A)=+\infty$. In particular, for every finite $M$, some stage satisfies $\mu(A_N)>M$.

Common pitfall: A finite-stage approximation with a small measure remainder requires the union to have finite measure. For an infinite-measure union, the stage measures instead grow without bound; one cannot infer that $\mu(A\setminus A_N)$ is small.

For example, the expanding intervals $[-n,n]$ have union $\mathbb{R}$, whose measure is infinite. Although each stage has finite measure, the part left outside any one stage still has infinite measure. By contrast, for the increasing intervals approaching $(0,1)$, the union has finite measure, so the omitted portion has measure tending to zero. Both situations follow the same continuity theorem; it is the size of the union that determines whether finite-stage approximation in measure is available.

When using continuity from below, check three points: the sets must be measurable, they must be increasing, and the limit may be infinite. If the sets are not nested, replace them with their finite cumulative unions. If the total measure is finite, the theorem also tells you that a sufficiently late stage captures all but an arbitrarily small amount of the union’s measure.

Check Your Understanding

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

What does the Continuity from Below Theorem say about the measure of an increasing union?
How are the layers $D_n$ defined, and why are distinct layers disjoint?
Which hypothesis allows the finite-stage approximation theorem to make $\mu(A\setminus A_N)$ arbitrarily small?
For arbitrary measurable sets $E_n$, what increasing sequence can be formed to apply continuity from below to their union?
If an increasing sequence has a union of infinite measure, what happens to the measures of its stages?