Lebesgue Sets as Borel Sets Up to Null Sets

A Borel Representative Modulo a Null Set

Lebesgue measurable sets need not themselves be Borel. However, the additional sets allowed by Lebesgue measurability have a precise form: each differs from a Borel set only on a null set. For finite-measure sets, outer regularity gives a direct construction of the Borel representative. This observation also explains how Lebesgue measure is obtained by completing Borel measure.

Throughout, \(m\) is Lebesgue measure on the measure space \((\mathbb{R},\mathcal{L},m)\), where \(\mathcal{L}\) is the Lebesgue sigma-algebra. The restriction of \(m\) to the measurable space \((\mathbb{R},\mathcal{B}_{\mathbb{R}})\) will be denoted by \(m_{\mathcal{B}}\); it is Borel measure. A set is null if it has measure zero. We use the established outer regularity property: if \(E\in\mathcal{L}\) and \(m(E)<\infty\), then for every \(\varepsilon>0\) there is an open set \(U\supseteq E\) such that \(m(U\setminus E)<\varepsilon\).

Proof. For each positive integer \(n\), apply outer regularity with \(\varepsilon=1/n\) to choose an open set \(U_n\supseteq E\) such that \(m(U_n\setminus E)<1/n\). Define \(B=\bigcap_{n=1}^{\infty}U_n\). Since every \(U_n\) is open, \(B\) is a \(G_\delta\) set and therefore Borel. Also \(E\subseteq B\), so \(E\triangle B=B\setminus E\). For every \(n\), \(B\setminus E\subseteq U_n\setminus E\), and hence \(0\le m(B\setminus E)\le m(U_n\setminus E)<1/n\). A nonnegative number less than \(1/n\) for every positive integer \(n\) must be zero. Thus \(m(E\triangle B)=0\), as required. \(\square\)

The estimates do not assert that \(B\) has finite measure by an independent argument; rather, it follows here that \(m(B)=m(E)+m(B\setminus E)=m(E)<\infty\). The constructed set \(B\) is a Borel superset, and its excess over \(E\) is null. This direction of containment is useful: the exceptional set is exactly \(B\setminus E\).

From Finite Measure to Arbitrary Measure

The finite-measure result extends to every Lebesgue measurable set on \(\mathbb{R}\), even when its measure is infinite. The key is to first restrict the set to bounded intervals, where every piece has finite measure, and then combine the Borel representatives. The conclusion in the infinite-measure case is about a null symmetric difference; it does not require subtracting one infinite measure from another.

Proof. For each positive integer \(n\), let \(E_n=E\cap[-n,n]\). This set is Lebesgue measurable and has finite measure because \(E_n\subseteq[-n,n]\) and \(m([-n,n])=2n\). By the theorem, choose a Borel set \(B_n\) with \(m(E_n\triangle B_n)=0\). Put \(B=\bigcup_{n=1}^{\infty}B_n\), which is Borel. Since \(E=\bigcup_{n=1}^{\infty}E_n\), any point in \(E\setminus B\) belongs to some \(E_n\setminus B_n\), and any point in \(B\setminus E\) belongs to some \(B_n\setminus E_n\). Therefore \(E\triangle B\subseteq\bigcup_{n=1}^{\infty}(E_n\triangle B_n)\). Each set on the right is null, so countable subadditivity gives \(m(E\triangle B)=0\). Both \(E\) and \(B\) are Lebesgue measurable; since their symmetric difference is null, their measures agree. More explicitly, \(E\setminus B\) and \(B\setminus E\) are null, so \(m(E)=m(E\cap B)=m(B)\), where the equalities hold also when the common value is infinite. \(\square\)

Completion of Borel Measure

For a measure space, its completion adjoins every subset of every measurable null set. Applied to Borel measure on \((\mathbb{R},\mathcal{B}_{\mathbb{R}},m_{\mathcal{B}})\), the completion therefore includes sets that need not be Borel. Precisely, define \(\mathcal{C}\) to be the family of sets \(A\subseteq\mathbb{R}\) for which there are Borel sets \(B,Z\) with \(m_{\mathcal{B}}(Z)=0\) and \(A\triangle B\subseteq Z\). The set \(Z\) is a Borel null set containing all points where \(A\) and \(B\) may differ.

Proof. First verify that \(\mathcal{C}\) is a sigma-algebra. If \(A\triangle B\subseteq Z\) with \(B,Z\) Borel and \(Z\) null, then \(A^c\triangle B^c=A\triangle B\subseteq Z\), so \(A^c\in\mathcal{C}\). For a sequence \(A_k\in\mathcal{C}\), choose Borel \(B_k,Z_k\) with \(A_k\triangle B_k\subseteq Z_k\) and \(m_{\mathcal{B}}(Z_k)=0\). The set \(B=\bigcup_k B_k\) is Borel, as is \(Z=\bigcup_k Z_k\), and \(Z\) is null by countable subadditivity. If a point lies in exactly one of \(\bigcup_k A_k\) and \(\bigcup_k B_k\), it lies in some \(A_k\triangle B_k\); hence \((\bigcup_k A_k)\triangle B\subseteq Z\). Thus \(\bigcup_k A_k\in\mathcal{C}\). Every Borel set belongs to \(\mathcal{C}\) by taking \(Z=\varnothing\), and every subset of a Borel null set belongs to \(\mathcal{C}\) by taking \(B=\varnothing\). This is the completion of Borel measure.

Next, if \(A\in\mathcal{C}\), choose Borel \(B,Z\) as in its definition. Borel sets are Lebesgue measurable, and Lebesgue measure is complete: every subset of a Lebesgue null set is Lebesgue measurable. Since \(A\triangle B\subseteq Z\), this completeness and closure of \(\mathcal{L}\) under set operations imply \(A\in\mathcal{L}\).

Conversely, suppose \(A\in\mathcal{L}\). The corollary gives a Borel \(B\) with \(m(A\triangle B)=0\). The set \(A\triangle B\) is Lebesgue measurable and null. By outer regularity, for each positive integer \(n\) there is an open set \(V_n\supseteq A\triangle B\) with \(m(V_n)<1/n\). Set \(Z=\bigcap_{n=1}^{\infty}V_n\). This is Borel, contains \(A\triangle B\), and satisfies \(0\le m(Z)\le m(V_n)<1/n\) for every \(n\), so \(m_{\mathcal{B}}(Z)=0\). Thus \(A\in\mathcal{C}\). We have proved \(\mathcal{C}=\mathcal{L}\).

Finally, the completed measure of \(A\in\mathcal{C}\) is defined as \(m_{\mathcal{B}}(B)\) for a Borel representative \(B\). This is well-defined: if \(A\triangle B_1\subseteq Z_1\) and \(A\triangle B_2\subseteq Z_2\) for Borel null sets \(Z_1,Z_2\), then \(B_1\triangle B_2\subseteq Z_1\cup Z_2\). Hence \(m_{\mathcal{B}}(B_1\triangle B_2)=0\), which implies \(m_{\mathcal{B}}(B_1)=m_{\mathcal{B}}(B_2)\). The corollary shows this value equals \(m(A)\). Therefore the completed measure is exactly Lebesgue measure. \(\square\)

A Common Mistake and Its Correction

A common mistake is to replace “Lebesgue measurable” with “Borel.” The example of a non-Borel subset \(N\) of the Cantor set shows why this is false: \(N\) is Lebesgue measurable by completeness, but not Borel. The correct statement is that there is a Borel \(B\) with \(m(N\triangle B)=0\); here \(B=\varnothing\). Conversely, it would also be wrong to infer that every set differing from a Borel set by an arbitrary set of measure zero is Lebesgue measurable without specifying measurability of that difference. The completion formulation avoids this issue by requiring the difference to lie inside a Borel null set, whose every subset is measurable in \(\mathcal{L}\).

The distinction matters whenever a proof uses Borel structure, such as a construction from open sets, but the set under study is only known to be Lebesgue measurable. One may pass to a Borel representative for measure calculations, while keeping track that the original set and representative agree only outside a null set. This is not a claim that the sets are equal pointwise.