Common Errors with Set Limits and Generated Classes

Start with Membership, Not a Memorized Identity

Set limits record how often a point belongs to a sequence of sets. Many false identities arise from exchanging the order of operations before checking what that exchange means for a single point. The same discipline helps with generated classes: finite closure, countable closure, and closure under monotone limits are different requirements.

Throughout, \(X\) is a set, and \(A_n\subseteq X\) for each positive integer \(n\). No measure or measurability assumption is involved. A point belongs to \(\limsup A_n\) when it belongs to infinitely many \(A_n\); it belongs to \(\liminf A_n\) when it belongs to all but finitely many \(A_n\). Equivalently, the definitions are

\limsup_{n\to\infty} A_n=\bigcap_{n=1}^{\infty}\bigcup_{k\ge n}A_k, \qquad \liminf_{n\to\infty} A_n=\bigcup_{n=1}^{\infty}\bigcap_{k\ge n}A_k.

The first formula says that for every starting index \(n\), a point appears at some index \(k\ge n\). The second says that for some starting index \(n\), the point appears at every index \(k\ge n\). The distinction between “infinitely often” and “eventually always” is decisive when testing an identity.

Definition: An algebra of subsets of \(X\) contains \(X\) and is closed under complements in \(X\) and finite unions. A sigma-algebra has the same requirements but is closed under countable unions. A monotone class is a collection of subsets of \(X\) closed under increasing countable unions and decreasing countable intersections. These are three different closure conditions; one does not get to replace one by another without justification.

Which Finite Exchanges Are Valid?

For a fixed finite number of sequences \(A_n^{(1)},\ldots,A_n^{(r)}\), there are useful identities and inclusions. They follow directly from the membership definitions, not from a general rule that limits commute with set operations.

Proposition: For a finite family of sequences of subsets of \(X\), \(\limsup_n(\bigcup_{j=1}^r A_n^{(j)})=\bigcup_{j=1}^r\limsup_n A_n^{(j)}\), and \(\liminf_n(\bigcap_{j=1}^r A_n^{(j)})=\bigcap_{j=1}^r\liminf_n A_n^{(j)}\). Also, \(\bigcup_{j=1}^r\liminf_n A_n^{(j)}\subseteq\liminf_n(\bigcup_{j=1}^r A_n^{(j)})\) and \(\limsup_n(\bigcap_{j=1}^r A_n^{(j)})\subseteq\bigcap_{j=1}^r\limsup_n A_n^{(j)}\). The inclusions need not be equalities.

Proof. Fix \(x\in X\). For the first equality, if \(x\in\limsup_n(\bigcup_{j=1}^r A_n^{(j)})\), then \(x\) belongs to the union for infinitely many indices \(n\). At each such index it belongs to at least one \(A_n^{(j)}\). Since there are only finitely many choices of \(j\), at least one choice occurs for infinitely many indices; otherwise each choice would occur only finitely often, and their finite union of index sets would be finite. Thus \(x\in\bigcup_{j=1}^r\limsup_n A_n^{(j)}\). Conversely, if \(x\in\limsup_n A_n^{(j)}\) for some \(j\), then \(x\) belongs to the unions for infinitely many indices, proving the reverse inclusion.

For the second equality, \(x\in\liminf_n(\bigcap_{j=1}^r A_n^{(j)})\) means there is an index \(N\) such that, for every \(n\ge N\), \(x\) belongs to every \(A_n^{(j)}\). This implies \(x\in\liminf_n A_n^{(j)}\) for each \(j\). Conversely, if \(x\in\liminf_n A_n^{(j)}\) for every \(j\), choose for each \(j\) an index \(N_j\) after which membership always holds. The maximum \(N=\max_j N_j\) exists because the family is finite. For all \(n\ge N\), \(x\) belongs to every \(A_n^{(j)}\), proving the reverse inclusion.

If \(x\in\liminf_n A_n^{(j)}\) for some \(j\), then \(x\) is eventually in the union, which proves the first inclusion. If \(x\) is in the intersections \(\bigcap_j A_n^{(j)}\) infinitely often, it is in each \(A_n^{(j)}\) infinitely often, proving the second inclusion. \(\square\)

Worked Example: Two Finite Identities That Fail in the Reverse Direction

Let \(X=\{a,b\}\). For odd \(n\), set \(A_n=\{a\}\) and \(B_n=\{b\}\); for even \(n\), set \(A_n=\{b\}\) and \(B_n=\{a\}\). In every \(n\), \(A_n\cup B_n=\{a,b\}\), so \(\liminf_n(A_n\cup B_n)=\{a,b\}\): each point belongs at every index. But \(a\) and \(b\) each belong to \(A_n\) only on one parity of indices, so neither belongs eventually always to \(A_n\); hence \(\liminf_n A_n=\varnothing\). The same reasoning gives \(\liminf_n B_n=\varnothing\). Therefore \(\liminf_n A_n\cup\liminf_n B_n=\varnothing\), strictly smaller than \(\liminf_n(A_n\cup B_n)\).

For the intersection failure, keep the same alternating sets. At every index \(A_n\cap B_n=\varnothing\), so \(\limsup_n(A_n\cap B_n)=\varnothing\). Each of \(a,b\) belongs to \(A_n\) infinitely often and to \(B_n\) infinitely often. Thus \(\limsup_n A_n=\limsup_n B_n=\{a,b\}\), and their intersection is \(\{a,b\}\). The valid inclusion is therefore strict in this example. Check: the claimed inclusion gives \(\varnothing\subseteq\{a,b\}\), as required; neither false reverse equality holds.

Countable Families Need a Further Check

The finiteness of the family in the proposition matters. With infinitely many sets in each union, the index responsible for membership can change every time. A point can belong to the union at every \(n\), while belonging to each fixed column only once.

Worked Example: The Diagonal Error in a Countable Union

Take \(X=\{x\}\), and for positive integers \(n,k\), define \(A_{n,k}=X\) if \(n=k\), and \(A_{n,k}=\varnothing\) if \(n\ne k\). For each fixed \(n\), the union over all \(k\) includes \(A_{n,n}=X\), so \(\bigcup_k A_{n,k}=X\). This row union equals \(X\) for every \(n\), giving \(\limsup_n(\bigcup_k A_{n,k})=X\).

Now fix a column \(k\). Its sets are empty at every \(n\ne k\), and equal to \(X\) only at the single index \(n=k\). Therefore \(\limsup_n A_{n,k}=\varnothing\). Taking the union over \(k\) still gives \(\bigcup_k\limsup_n A_{n,k}=\varnothing\). Thus the two sides differ: \(X\ne\varnothing\). Check: each fixed column has only one nonempty term, whereas each row union is nonempty at every index; this matches the “infinitely often” test exactly.

The finite-union limsup identity cannot therefore be extended to a countable family without extra assumptions. In applications, write out the quantifiers: “for infinitely many \(n\), there exists \(k\)” does not in general imply “there exists \(k\) such that for infinitely many \(n\).” The diagonal example is precisely this failure of exchanging those quantifiers.

Generated Classes Are Not Interchangeable

An algebra permits finite set operations; a sigma-algebra permits countable unions; a monotone class permits limits of nested sequences. A sigma-algebra is a monotone class: an increasing union is a countable union, and a decreasing intersection follows by taking complements and using closure under countable unions. The reverse implications do not hold in general.

For a family \(\mathcal E\) of subsets of \(X\), \(\sigma(\mathcal E)\) denotes the smallest sigma-algebra containing \(\mathcal E\). Similarly, the monotone class generated by \(\mathcal E\) is the smallest monotone class containing it; it exists by taking the intersection of all monotone classes containing \(\mathcal E\). The Monotone Class Theorem for Algebras, established in the earlier material, says that if \(\mathcal A\) is an algebra, the monotone class generated by \(\mathcal A\) is \(\sigma(\mathcal A)\). The hypothesis that the starting family is an algebra is essential: this theorem does not say that every monotone class is a sigma-algebra.

Worked Example: An Algebra That Is Neither a Sigma-Algebra nor a Monotone Class

Let \(X=\mathbb N=\{1,2,\ldots\}\), and let \(\mathcal A\) consist of all finite subsets of \(\mathbb N\) and all complements of finite subsets. This is an algebra: complements exchange finite and cofinite sets, and a finite union is finite if all its members are finite, while a union containing a cofinite set is cofinite.

Let \(E=\{2,4,6,\ldots\}\). It is infinite, and its complement \(\{1,3,5,\ldots\}\) is also infinite, so \(E\) is neither finite nor cofinite and hence \(E\notin\mathcal A\). Since \(E=\bigcup_{j=1}^{\infty}\{2j\}\), it is a countable union of members of \(\mathcal A\) not in \(\mathcal A\). Thus \(\mathcal A\) is not a sigma-algebra.

It is also not a monotone class. The sets \(F_n=\{2,4,\ldots,2n\}\) are finite members of \(\mathcal A\), and \(F_n\subseteq F_{n+1}\) for every \(n\). Their increasing union is \(E\notin\mathcal A\). Check: this one increasing sequence violates the required closure under increasing countable unions, independently of the countable-union test for sigma-algebras.

Worked Example: A Monotone Class That Is Not a Sigma-Algebra

Let \(X=[0,1]\), and let \(\mathcal C\) be the collection of all downward-closed subsets: \(C\in\mathcal C\) means that if \(x\in C\) and \(0\le y\le x\), then \(y\in C\). The empty set and \(X\) belong to \(\mathcal C\). If \(C_1\subseteq C_2\subseteq\cdots\) are in \(\mathcal C\), their union is downward-closed: a point in the union belongs to some \(C_n\), and every smaller point belongs to that same \(C_n\). If \(C_1\supseteq C_2\supseteq\cdots\), their intersection is downward-closed because every member of the intersection belongs to each \(C_n\), as does every smaller point. Thus \(\mathcal C\) is a monotone class.

The set \(C=[0,\tfrac12)\) is downward-closed, so \(C\in\mathcal C\). Its complement in \(X\) is \([\tfrac12,1]\), which is not downward-closed: it contains \(\tfrac12\) but does not contain \(0\). Therefore \(\mathcal C\) is not closed under complements and is not a sigma-algebra. Check: the example satisfies both monotone closure tests but fails the separate complement test required of a sigma-algebra.

A Practical Error Check

A common wrong calculation is \(\limsup_n(A_n\cap B_n)=(\limsup_n A_n)\cap(\limsup_n B_n)\), on the grounds that “limsup commutes with intersection.” The alternating example above disproves it: the left side is empty, while the right side is \(\{a,b\}\). The correction is the inclusion \(\limsup_n(A_n\cap B_n)\subseteq(\limsup_n A_n)\cap(\limsup_n B_n)\); membership in the intersection at infinitely many indices forces membership in each set infinitely often, but the infinitely many indices for the two sets need not coincide.

A second error is to apply the Monotone Class Theorem to an arbitrary collection of sets. Its algebra hypothesis cannot be omitted. Before using that theorem, verify that the starting class contains \(X\), is closed under complements, and is closed under finite unions. Otherwise, use only the monotone-class definition and prove the closure property actually needed.

Takeaway: For set limits, reduce an identity to the membership quantifiers “infinitely often” and “eventually always.” For generated classes, check the exact closure operation required: finite unions, countable unions, or nested monotone limits. Similar notation does not make these conditions equivalent.

Check Your Understanding

For each question, test membership directly before using a set identity.

Prove that \(\liminf_n(A_n\cap B_n)=\liminf_n A_n\cap\liminf_n B_n\).
Give sequences \(A_n,B_n\) for which \(\liminf_n(A_n\cup B_n)\) is strictly larger than \(\liminf_n A_n\cup\liminf_n B_n\).
In the diagonal example, what changes if \(A_{n,k}=X\) when \(k=n+1\), and is the countable-union limsup discrepancy preserved?
Explain why every sigma-algebra is a monotone class, including the decreasing-intersection case.
Which hypothesis of the Monotone Class Theorem for Algebras fails for the downward-closed subsets of \([0,1]\)?