Measurable Sets

Measurability Along a Sequence of Sets

In the previous tutorial, a measurable set was defined as a member of the sigma-algebra in a measurable space $(X,\mathcal{F})$. That definition makes measurability relative to the chosen structure: the same subset of $X$ may be measurable for one sigma-algebra and not for another. Here we examine what happens when we have a sequence of measurable sets and ask how membership in those sets changes.

A sigma-algebra is closed under countable unions and intersections, as established in the earlier tutorials on sigma-algebras. This countable closure lets us form two useful sets from a sequence: the points that belong to infinitely many sets, and the points that belong to every set from some point onward. These constructions are called the set limsup and set liminf. They are set-theoretic limits; they do not require a measure or a numerical notion of distance.

Definition: Let $(E_n)_{n=1}^{\infty}$ be a sequence of subsets of $X$. Its set limsup and set liminf are $$ \limsup_{n\to\infty} E_n =\bigcap_{n=1}^{\infty}\bigcup_{k=n}^{\infty}E_k, \qquad \liminf_{n\to\infty} E_n =\bigcup_{n=1}^{\infty}\bigcap_{k=n}^{\infty}E_k. $$

The inner union in the limsup collects all points that occur at least once in a given tail of the sequence. Intersecting these tail unions means a point must occur in every tail, or equivalently, in infinitely many of the sets. The inner intersection in the liminf collects points that remain in every set in a given tail. Taking the union over all starting indices means membership must hold eventually, though not necessarily from the first set.

Takeaway: A point is in $\limsup E_n$ exactly when it belongs to infinitely many $E_n$. It is in $\liminf E_n$ exactly when it belongs to every $E_n$ from some index onward.

Measurability of Set Limits

Suppose now that every $E_n$ is measurable in $(X,\mathcal{F})$. The tail unions and tail intersections appearing in the definitions are countable operations on members of $\mathcal{F}$. The resulting set limits are therefore measurable as well.

Theorem (Measurability of Set Limsup and Liminf): If $(X,\mathcal{F})$ is a measurable space and $E_n\in\mathcal{F}$ for every positive integer $n$, then $$ \limsup_{n\to\infty}E_n\in\mathcal{F} \qquad\text{and}\qquad \liminf_{n\to\infty}E_n\in\mathcal{F}. $$

Proof. Fix $n$. Since every $E_k$ is in $\mathcal{F}$, the countable union $\bigcup_{k=n}^{\infty}E_k$ belongs to $\mathcal{F}$. The limsup is the countable intersection of these tail unions, so it also belongs to $\mathcal{F}$, by closure under countable intersections.

For the liminf, fix $n$ again. The tail intersection $\bigcap_{k=n}^{\infty}E_k$ belongs to $\mathcal{F}$ by the same closure property. The liminf is the countable union of these tail intersections, and hence belongs to $\mathcal{F}$ by closure under countable unions. Both set limits are measurable. $\square$

This theorem applies in any measurable space; it does not depend on a topology on $X$. In a Borel measurable space, for example, any sequence of Borel sets has Borel set limsup and liminf. The important ingredients are that each $E_n$ is measurable and that only countably many unions and intersections are used.

Worked Example: A Decreasing Sequence of Intervals

Work in $(\mathbb{R},\mathcal{B}(\mathbb{R}))$, and let $E_n=[0,1/n]$. Each $E_n$ is a closed interval, so it is Borel measurable. The sets are decreasing: if $k\geq n$, then $1/k\leq 1/n$, and consequently $[0,1/k]\subseteq[0,1/n]$.

A point belongs to $\liminf E_n$ if it belongs to all sets from some index onward. Since the sets are decreasing, this requires membership in every $E_n$ from the start of that tail. The point $0$ belongs to every $E_n$. If $x>0$, choose a positive integer $n>1/x$. Then $1/n<x$, so $x\notin[0,1/n]$. Negative points are in none of the intervals. Thus no point other than $0$ belongs to every set in a tail.

For the limsup, a positive $x$ is in $[0,1/n]$ only when $n\leq 1/x$, so it can belong to only finitely many of the sets. The point $0$ belongs to all of them, and negative points belong to none. Therefore $$ \limsup_{n\to\infty}E_n =\liminf_{n\to\infty}E_n =\{0\}. $$

When the Two Set Limits Agree

The liminf is always contained in the limsup: a point that belongs to every set from some index onward must belong to infinitely many sets. Agreement of the two limits has a more specific meaning. It says that every point’s membership in the sequence eventually settles down: from some index onward, the point either always belongs or never belongs.

Theorem (Equality and Eventual Membership): For any sequence of subsets $(E_n)$ of $X$, the equality $$ \limsup_{n\to\infty}E_n=\liminf_{n\to\infty}E_n $$ holds if and only if, for every $x\in X$, there is an index $N$ such that either $x\in E_n$ for every $n\geq N$, or $x\notin E_n$ for every $n\geq N$.

Proof. First, $\liminf E_n\subseteq\limsup E_n$. Indeed, if $x\in\liminf E_n$, there is an $N$ such that $x\in E_k$ for every $k\geq N$. For any $n$, there is a $k\geq n$ with $k\geq N$, so $x\in E_k$ and hence $x\in\bigcup_{j=n}^{\infty}E_j$. This is true for every $n$, so $x\in\limsup E_n$.

Now suppose the two limits are equal. Fix $x\in X$. If $x$ belongs to the common set, then it belongs to $\liminf E_n$, so it belongs to every $E_n$ from some index onward. If $x$ does not belong to the common set, it is not in $\limsup E_n$. By the definition of limsup, there is some $N$ for which $x\notin\bigcup_{k=N}^{\infty}E_k$. Thus $x\notin E_k$ for every $k\geq N$. In either case, membership eventually settles down.

Conversely, suppose membership settles down for every $x$. If $x$ is eventually always in $E_n$, then $x$ is in both the liminf and the limsup. If $x$ is eventually always outside $E_n$, then $x$ is in neither: it does not belong to every set in any tail, and it cannot belong to infinitely many sets. Thus the two limits contain exactly the same points, so they are equal. $\square$

Worked Example: Alternating Intervals

In the Borel measurable space on $\mathbb{R}$, define $E_n=[0,1]$ when $n$ is even and $E_n=[1,2]$ when $n$ is odd. Each of these two closed intervals occurs infinitely often. Every point of $[0,1]\cup[1,2]=[0,2]$ therefore belongs to infinitely many $E_n$, and a point outside $[0,2]$ belongs to none. Hence $$ \limsup_{n\to\infty}E_n=[0,2]. $$

Every tail of the sequence contains at least one even-indexed set and at least one odd-indexed set. Its intersection is therefore $$ [0,1]\cap[1,2]=\{1\}. $$ The intersection of every tail is $\{1\}$, so their union is also $\{1\}$, giving $\liminf E_n=\{1\}$. The two limits differ. In fact, points in $[0,1)$ belong at every even index but not at odd indices, while points in $(1,2]$ belong at every odd index but not at even indices. Their membership does not eventually settle down.

Membership in Infinitely Many Sets

A set limsup can also describe repeated occurrence. For a sequence of measurable sets, the points that occur infinitely often form a measurable set by the Measurability of Set Limsup and Liminf Theorem. The complementary description is often useful: points that occur only finitely often are eventually absent. De Morgan’s laws give the identity $$ X\setminus\left(\limsup_{n\to\infty}E_n\right) =\liminf_{n\to\infty}(X\setminus E_n). $$ The left side consists of points belonging to only finitely many $E_n$; the right side consists of points eventually belonging to every complement.

Likewise, a point fails to be in $\liminf E_n$ exactly when it fails to belong to infinitely many of the sets. In set notation, $$ X\setminus\left(\liminf_{n\to\infty}E_n\right) =\limsup_{n\to\infty}(X\setminus E_n). $$ These identities follow by applying De Morgan’s laws to the countable unions and intersections in the definitions. They also help prevent a common confusion: “infinitely often” and “eventually always” are different conditions, and their complements have different descriptions.

Worked Example: A Set Appearing on Even Indices

Let $$ E_n= \begin{cases} [0,1/n]\cup\{2\},&\text{if }n\text{ is even},\\ [0,1/n],&\text{if }n\text{ is odd}. \end{cases} $$

Every $E_n$ is Borel measurable because it is a union of a closed interval and, when applicable, a singleton. The point $0$ belongs to every $E_n$. The point $2$ belongs to every even-indexed set and no odd-indexed set, so it occurs infinitely often but not eventually always. If $x>0$ and $x\neq 2$, choose $N>1/x$. For $n\geq N$, $x>1/n$, so $x\notin[0,1/n]$; since $x\neq2$, it is not in the other part either. Negative points are never in any $E_n$. Therefore $$ \limsup_{n\to\infty}E_n=\{0,2\}, \qquad \liminf_{n\to\infty}E_n=\{0\}. $$ The example separates infinite recurrence from eventual membership.

Why Countability Matters

The countable operations in the definitions are not incidental. A sigma-algebra guarantees closure under countable unions and intersections, but it need not be closed under unions of arbitrarily many measurable sets. For a concrete illustration, let $X=\mathbb{R}$ and use the countable-cocountable sigma-algebra: a set is measurable when it or its complement is countable. Every singleton $\{x\}$ is measurable because it is countable. However, the uncountable union $$ \bigcup_{x\in(0,1)}\{x\}=(0,1) $$ is not in that sigma-algebra: both $(0,1)$ and its complement in $\mathbb{R}$ are uncountable.

Thus, even when every set in a family is measurable, its uncountable union need not be measurable. The limsup and liminf avoid this problem because they are constructed from countably many sets using countable unions and intersections. This is precisely the level of closure supplied by a sigma-algebra.

Measurable set limits provide a way to describe recurring and persistent membership without assigning sizes to sets. The next step in measure theory is to define a measure on a measurable space: a function that assigns sizes to its measurable sets. Keeping the set operations and their measurability clear now will make those later definitions easier to use correctly.

Check Your Understanding

Use the definitions and results in this tutorial to answer the following questions.

What membership condition characterizes a point in the limsup of a sequence of sets?
What membership condition characterizes a point in the liminf?
Why are the limsup and liminf measurable when every set in the sequence is measurable?
What does equality of the limsup and liminf imply about each point’s eventual membership?
In the alternating-interval example, why is the limsup larger than the liminf?
Why does a sigma-algebra not guarantee that every uncountable union of measurable sets is measurable?

What You'll Learn