Constructing a Measurable Set That Is Not Borel

Why Cardinality Produces a Non-Borel Measurable Set

The middle-third Cantor set \(C\) has Lebesgue measure zero, but it is uncountable. Completeness of Lebesgue measure therefore makes every subset of \(C\) measurable. The question is whether all those measurable subsets could also be Borel. Comparing cardinalities shows that they cannot: the family of subsets of \(C\) is strictly larger than the family of Borel subsets of the real line.

One point needs care in stating this comparison. The Borel family is not countable. There are continuum many Borel subsets of \(\mathbb{R}\). What matters is that the power set of \(C\) has cardinality strictly greater than the continuum. Write \(\mathfrak c=|\mathbb{R}|=2^{\aleph_0}\), where \(\aleph_0\) is the cardinality of the natural numbers.

Counting Points and Subsets of the Cantor Set

Every point of \(C\) has a ternary expansion using only the digits \(0\) and \(2\). Thus infinite binary sequences provide codes for points of \(C\). The key distinction is between counting the codes and counting all collections of codes: there are \(\mathfrak c\) binary sequences, but \(2^{\mathfrak c}\) subsets of the set of binary sequences.

Since \(|C|=\mathfrak c\), its power set \(\mathcal{P}(C)\), the family of all subsets of \(C\), has cardinality \(2^{\mathfrak c}\). Cantor’s theorem says that \(2^{\mathfrak c}>\mathfrak c\). Its diagonal argument applies to any set \(S\): no proposed map from \(S\) onto \(\mathcal{P}(S)\) can be onto, because the subset \(\{s\in S:s\notin f(s)\}\) differs from \(f(s)\) at \(s\) for every \(s\in S\). In particular, there are more subsets of \(C\) than there are real numbers.

Bounding the Number of Borel Sets

Open sets are only the starting point: Borel sets are formed by repeatedly taking complements and countable unions. To bound their number, it is important to account for every countable stage of this process, rather than assume that a Borel set is obtained after a fixed finite number of operations.

Proof. For each countable ordinal \(\alpha\), define families \(\mathcal{B}_\alpha\) recursively. Let \(\mathcal{B}_0\) be the open subsets of \(\mathbb{R}\). At a successor stage, let \(\mathcal{B}_{\alpha+1}\) contain \(\mathcal{B}_\alpha\), complements of members of \(\mathcal{B}_\alpha\), and countable unions of members of \(\mathcal{B}_\alpha\). At a countable limit ordinal \(\lambda\), let \(\mathcal{B}_\lambda=\bigcup_{\alpha<\lambda}\mathcal{B}_\alpha\). Every set in every stage is Borel, because the Borel family contains open sets and is closed under complements and countable unions.

Conversely, put \(\mathcal{B}_*=\bigcup_{\alpha<\omega_1}\mathcal{B}_\alpha\), where \(\omega_1\) is the first uncountable ordinal. This family contains all open sets. If \(A\in\mathcal{B}_\alpha\), then \(A^c\in\mathcal{B}_{\alpha+1}\), so \(\mathcal{B}_*\) is closed under complements. If \(A_n\in\mathcal{B}_*\) for every \(n\), choose countable ordinals \(\alpha_n\) with \(A_n\in\mathcal{B}_{\alpha_n}\). Their supremum \(\gamma=\sup_n\alpha_n\) is countable: a countable union of countable ordinals is countable. The families are increasing, so every \(A_n\in\mathcal{B}_\gamma\), and \(\bigcup_n A_n\in\mathcal{B}_{\gamma+1}\). Thus \(\mathcal{B}_*\) is a sigma-algebra containing the open sets, so it contains \(\mathcal{B}_{\mathbb{R}}\). Together with the first inclusion, \(\mathcal{B}_*=\mathcal{B}_{\mathbb{R}}\).

It remains to count. The open sets have cardinality at most \(\mathfrak c\), as shown above. If \(|\mathcal{B}_\alpha|\le\mathfrak c\), then the number of sequences of its members is at most \(\mathfrak c^{\aleph_0}=\mathfrak c\); complements and countable unions therefore leave the successor stage of cardinality at most \(\mathfrak c\). Here \(\mathfrak c^{\aleph_0}=\mathfrak c\) follows by coding a countable sequence of binary sequences as one binary sequence indexed by \(\mathbb{N}\times\mathbb{N}\), and using a bijection between \(\mathbb{N}\times\mathbb{N}\) and \(\mathbb{N}\). At a countable limit stage, a countable union of families of size at most \(\mathfrak c\) still has size at most \(\mathfrak c\). Transfinite induction gives this bound at every countable stage. There are at most \(\mathfrak c\) countable ordinals: each can be represented by a well-ordering on a subset of \(\mathbb{N}\), and there are at most \(2^{\aleph_0}=\mathfrak c\) such relations. Hence \(|\mathcal{B}_{\mathbb{R}}|\le\mathfrak c\).

For the reverse inequality, every singleton \(\{x\}\) is closed and hence Borel, and distinct real numbers give distinct singletons. Therefore \(|\mathcal{B}_{\mathbb{R}}|\ge|\mathbb{R}|=\mathfrak c\). The two inequalities prove \(|\mathcal{B}_{\mathbb{R}}|=\mathfrak c\). \(\square\)

The Measurable Set That Is Not Borel

Proof. Consider the family of subsets of \(C\) that are Borel as subsets of \(\mathbb{R}\): \(\mathcal{D}=\{B\cap C:B\in\mathcal{B}_{\mathbb{R}}\}\). The map \(B\mapsto B\cap C\) shows \(|\mathcal{D}|\le|\mathcal{B}_{\mathbb{R}}|=\mathfrak c\). But \(|\mathcal{P}(C)|=2^{\mathfrak c}>\mathfrak c\). Therefore \(\mathcal{D}\ne\mathcal{P}(C)\); choose \(A\in\mathcal{P}(C)\setminus\mathcal{D}\). This means \(A\subseteq C\) and \(A\) is not Borel in \(\mathbb{R}\).

The Cantor set is Lebesgue measurable and \(m(C)=0\). Lebesgue measure is complete: every subset of a measurable null set is Lebesgue measurable. Applying completeness to \(A\subseteq C\) gives \(A\in\mathcal{L}\). Thus \(A\) is Lebesgue measurable and not Borel. \(\square\)

What the Argument Does—and Does Not—Construct

The construction is an existence argument: it defines a nonempty difference of families and chooses a member of that difference. It does not give a simple formula for a particular set \(A\). The proof does not identify which subset of \(C\) is outside the Borel family; it establishes that at least one must be.

Common mistake: “There are only countably many Borel sets, since the Borel sigma-algebra is generated from a countable base.” This confuses a countable collection of generators with the sigma-algebra they generate. For instance, even the possible subfamilies of a countable base are counted by \(2^{\aleph_0}=\mathfrak c\), not by \(\aleph_0\). The correct comparison is \(|\mathcal{B}_{\mathbb{R}}|=\mathfrak c\) versus \(|\mathcal{P}(C)|=2^{\mathfrak c}>\mathfrak c\). The strict inequality, not countability of the Borel family, forces the conclusion.

The distinction matters because “measurable” and “Borel” are not interchangeable descriptions. Every Borel set is Lebesgue measurable, but completeness adds all subsets of null Borel sets to \(\mathcal{L}\), whether or not those subsets are Borel. Here the null set \(C\) supplies an entire power set of measurable sets, while the Borel cardinality bound shows that not all of them belong to \(\mathcal{B}_{\mathbb{R}}\).