Set-Limit and Sigma-Generation Problem Workshop

A Reliable Procedure for Set Limits

Set limits record how membership in a sequence of sets behaves far out in the sequence. The distinction is between belonging to infinitely many sets and belonging to every set from some point onward. Confusing these two conditions can reverse a calculation, so it is useful to begin with the tail definitions and check membership point by point.

In this tutorial, $X$ is an arbitrary set and $A_1,A_2,\ldots$ are subsets of $X$. These are set-theoretic statements: no measure or measure space is involved. For $N\ge1$, the $N$-th tail union is $\bigcup_{n\ge N}A_n$, and the $N$-th tail intersection is $\bigcap_{n\ge N}A_n$.

Definition: The limit superior and limit inferior of $(A_n)$ are the sets $$ \limsup_{n\to\infty}A_n=\bigcap_{N=1}^{\infty}\bigcup_{n\ge N}A_n, \qquad \liminf_{n\to\infty}A_n=\bigcup_{N=1}^{\infty}\bigcap_{n\ge N}A_n. $$ A point belongs to the limit superior exactly when it belongs to infinitely many $A_n$; it belongs to the limit inferior exactly when it belongs to every $A_n$ from some index onward.

Proof. Fix $x\in X$. By the definition of union and intersection, $x\in\limsup A_n$ means that for every $N\ge1$, there is an $n\ge N$ for which $x\in A_n$. This is precisely the condition that membership occurs at arbitrarily large indices, or equivalently in infinitely many sets. Also, $x\in\liminf A_n$ means that there is an $N\ge1$ such that $x\in A_n$ for every $n\ge N$. This is exactly eventual membership. $\square$

Proposition: For every sequence of subsets of $X$, $\liminf A_n\subseteq\limsup A_n$. Moreover, $\liminf A_n=\limsup A_n$ if and only if each point of $X$ is either in all sufficiently late sets or in only finitely many sets.

Proof. If $x\in\liminf A_n$, there is an $N$ such that $x\in A_n$ for every $n\ge N$. Thus, for every tail beginning at any index $K$, some $n\ge K$ has $x\in A_n$: one may take $n=\max\{N,K\}$. Hence $x\in\limsup A_n$, proving the inclusion. By the membership characterizations just proved, a point outside $\liminf A_n$ fails to belong eventually always, and a point outside $\limsup A_n$ belongs only finitely often. Equality of the two sets therefore says there is no point that belongs infinitely often but not eventually always. This is equivalent to the stated alternative for every point. $\square$

Worked Set-Limit Problems

Worked Example: An Alternating Sequence of Intervals

Let $X=\mathbb{R}$, and define $A_n=[0,1]$ for even $n$ and $A_n=[1,2]$ for odd $n$. Compute both set limits.

Every point in $[0,1]$ belongs to each even-indexed set, and every point in $[1,2]$ belongs to each odd-indexed set. Both even and odd indices occur infinitely often. Therefore every point in $[0,2]$ belongs to infinitely many $A_n$, including the shared endpoint $1$. No point outside $[0,2]$ belongs to any $A_n$. Thus $$ \limsup_{n\to\infty}A_n=[0,2]. $$ For eventual membership, points in $[0,1)$ are missing from every odd-indexed set, and points in $(1,2]$ are missing from every even-indexed set. The point $1$, however, belongs to both intervals and hence to every $A_n$. No other point is eventually always present, so $$ \liminf_{n\to\infty}A_n=\{1\}. $$ Check: $\{1\}\subseteq[0,2]$, as required by the general inclusion.

Worked Example: Intervals That Move to the Right

On $X=\mathbb{R}$, let $A_n=(n,n+1)$ for $n\ge1$. For a fixed $x\in\mathbb{R}$, membership $x\in A_n$ requires $n<x<n+1$. There is at most one integer $n$ satisfying these strict inequalities. In particular, $x$ cannot belong to infinitely many of the intervals, so no $x$ belongs to the limsup. Since eventual membership would imply membership in infinitely many intervals, no $x$ belongs to the liminf either: $$ \limsup_{n\to\infty}A_n=\liminf_{n\to\infty}A_n=\varnothing. $$ Check: for any fixed $x$, every integer $n\ge x$ fails the strict condition $n<x$; thus all sufficiently late intervals exclude $x$.

These examples show why a picture of a few initial sets is not enough: both limits depend on every tail of the sequence. The membership test reduces the calculation to a fixed point $x$ and the question of how often the condition $x\in A_n$ holds.

Generated Sigma-Algebras from Partitions

A sigma-algebra on $X$ is a family $\mathcal{S}$ of subsets of $X$ that contains $X$, is closed under complements in $X$, and is closed under countable unions. For a family $\mathcal{E}$ of subsets of $X$, $\sigma(\mathcal{E})$ denotes the smallest sigma-algebra containing every member of $\mathcal{E}$. Equivalently, it is the intersection of all sigma-algebras on $X$ that contain $\mathcal{E}$; such sigma-algebras exist because the power set of $X$ is one.

Proposition: If $X$ is partitioned into finitely many nonempty, pairwise disjoint blocks $P_1,\ldots,P_k$, then $\sigma(\{P_1,\ldots,P_k\})$ consists of all unions of these blocks, including the empty union $\varnothing$.

Proof. Let $\mathcal{U}$ be the family of all unions of subfamilies of $\{P_1,\ldots,P_k\}$. The empty union is $\varnothing$, and the union of all blocks is $X$, so both belong to $\mathcal{U}$. The complement of a union of blocks is the union of the remaining blocks, because the blocks are disjoint and cover $X$. A countable union of members of $\mathcal{U}$ is also a union of blocks: it contains precisely those blocks that occur in at least one of the countably many unions. Thus $\mathcal{U}$ is a sigma-algebra containing every $P_i$, which gives $\sigma(\{P_1,\ldots,P_k\})\subseteq\mathcal{U}$. Conversely, every sigma-algebra containing the blocks contains each finite union of them, so $\mathcal{U}\subseteq\sigma(\{P_1,\ldots,P_k\})$. The two families are equal. $\square$

Worked Example: The Sigma-Algebra of a Three-Block Partition

Let $X=\{a,b,c,d\}$ and let the partition blocks be $P_1=\{a,b\}$, $P_2=\{c\}$, and $P_3=\{d\}$. A member of the generated sigma-algebra is a union of selected blocks. There are $2^3=8$ choices of selected blocks, giving $$ \sigma(\{P_1,P_2,P_3\})= \{\varnothing,\{a,b\},\{c\},\{d\},\{a,b,c\},\{a,b,d\},\{c,d\},X\}. $$ Check: the complement of $\{a,b,c\}$ is $\{d\}$, and both are in the list. Every listed set is a union of whole blocks, so no set that splits $\{a,b\}$, such as $\{a\}$, belongs to the generated sigma-algebra.

Rational Generators for Borel Sets

The Borel sigma-algebra $\mathcal{B}_{\mathbb{R}}$ is the sigma-algebra generated by the open subsets of $\mathbb{R}$. A useful countable generating family is given by rational half-lines. Countability matters because open sets can then be assembled using only countable unions of basic sets.

Theorem: On $\mathbb{R}$, $$ \mathcal{B}_{\mathbb{R}}=\sigma\bigl(\{(-\infty,q):q\in\mathbb{Q}\}\bigr). $$

Proof. Each half-line $(-\infty,q)$ is open, hence Borel. Therefore the sigma-algebra generated by these half-lines is contained in $\mathcal{B}_{\mathbb{R}}$. For the reverse inclusion, first fix $a\in\mathbb{Q}$. The closed half-line $(-\infty,a]$ can be written as $$ (-\infty,a]=\bigcap_{\substack{q\in\mathbb{Q}\\q>a}}(-\infty,q). $$ Indeed, every $x\le a$ lies below every such $q$, while if $x>a$, density of the rationals gives a rational $q$ with $a<q<x$, so $x\notin(-\infty,q)$. Consequently $(a,b)=(-\infty,b)\setminus(-\infty,a]$ belongs to the generated sigma-algebra whenever $a,b\in\mathbb{Q}$ and $a<b$. Every open set $U\subseteq\mathbb{R}$ is the union of the rational-endpoint intervals $(a,b)$ contained in $U$: for each $x\in U$, openness gives an interval around $x$ contained in $U$, and density of $\mathbb{Q}$ supplies rational endpoints on either side of $x$ inside that interval. There are only countably many rational pairs, so this is a countable union. Thus every open set belongs to the generated sigma-algebra, proving the reverse inclusion. $\square$

Worked Example: Generating the Borel Sigma-Algebra of the Plane

For $i\in\{1,2\}$ and $q\in\mathbb{Q}$, set $H_{i,q}=\{x\in\mathbb{R}^2:x_i<q\}$, where $x=(x_1,x_2)$. We identify a generating family for $\mathcal{B}_{\mathbb{R}^2}$.

Each $H_{i,q}$ is open, so $\sigma(\{H_{i,q}:i=1,2,\ q\in\mathbb{Q}\})\subseteq\mathcal{B}_{\mathbb{R}^2}$. For rational $a<b$, the coordinate strip condition $a<x_i<b$ is generated by these half-spaces: $\{x_i<b\}=H_{i,b}$, while $$ \{x_i\le a\}=\bigcap_{\substack{q\in\mathbb{Q}\\q>a}}H_{i,q}, \qquad \{a<x_i<b\}=H_{i,b}\setminus\{x_i\le a\}. $$ It follows by intersecting the conditions for $i=1$ and $i=2$ that every rational open rectangle $(a_1,b_1)\times(a_2,b_2)$ belongs to the generated sigma-algebra. These rectangles form a countable base: for any open $U\subseteq\mathbb{R}^2$ and any $x\in U$, an open ball about $x$ lies in $U$, and density of $\mathbb{Q}$ gives a rational rectangle containing $x$ inside that ball. Hence $U$ is a union of rational rectangles contained in $U$; because there are countably many such rectangles, this union is countable. Every open set is therefore in the generated sigma-algebra, proving $$ \mathcal{B}_{\mathbb{R}^2}=\sigma(\{H_{i,q}:i=1,2,\ q\in\mathbb{Q}\}). $$ Check: the proof gives both inclusions: the chosen half-spaces are open, and every open set is a countable union of sets generated from them.

A Common Pitfall and a Practical Checklist

A frequent error is to interchange the tail union and tail intersection when computing the limit inferior. In the alternating-interval example, an incorrect calculation might say that $\liminf A_n=\bigcap_{n\ge1}A_n=[1,1]$, treating the first intersection as the definition. But the limit inferior is a union of tail intersections, not just the intersection of the entire sequence. In this example every tail intersection is $\{1\}$, so the union of the tail intersections is indeed $\{1\}$; the shortcut happens to give the right answer only because the point $1$ lies in every set. In general it does not. For $A_n=(n,n+1)$, the intersection of the entire sequence is empty, but that alone does not justify either set-limit calculation. The correction is to use the definitions: test whether a fixed point belongs to infinitely many sets and whether it belongs to every set after some index.

For a generated sigma-algebra, the parallel error is to list sets that appear plausible without verifying closure or minimality. A reliable method is to propose a family, show it is a sigma-algebra containing the generators, and then show every sigma-algebra containing the generators must contain the proposed family. For Borel generators, prove both inclusions: the proposed generators are Borel, and they generate every open set by a countable construction.

Check Your Understanding

For each question, use the defining tail operations or the definition of a generated sigma-algebra explicitly.

Let $A_n=\{0\}$ for even $n$ and $A_n=\{1\}$ for odd $n$, as subsets of $\{0,1\}$. Compute $\limsup A_n$ and $\liminf A_n$.
For $A_n=[-1/n,1/n]\subseteq\mathbb{R}$, determine the two set limits and justify membership for an arbitrary fixed real $x$.
Describe $\sigma(\{\{a\},\{b\}\})$ on $X=\{a,b,c\}$, where the remaining partition block is $\{c\}$.
Explain why rational open intervals generate $\mathcal{B}_{\mathbb{R}}$, including why only countably many intervals are needed.
In $\mathbb{R}^n$, define the coordinate half-spaces with rational thresholds and explain how they generate rational open boxes.