Lebesgue Measure

Approximating Measurable Sets

The construction of Lebesgue measure starts with interval covers, so open sets are naturally suited to approximating measurable sets from the outside. This connection is useful even when a set is not itself open or Borel: if it has finite measure, it can be covered by an open set whose excess measure is as small as we choose. For bounded measurable sets, a complementary argument also gives compact approximations from the inside.

Throughout, \(\mathcal{L}\) denotes the Lebesgue sigma-algebra and \(\lambda\) denotes Lebesgue measure, as in the previous tutorial. We use the established facts that Borel sets belong to \(\mathcal{L}\), and that Lebesgue measure is complete. Approximation here concerns the measure of the difference between sets; it does not generally make the sets equal.

Proof. Because \(E\) is Lebesgue measurable, the definition of \(\lambda\) gives \(m^*(E)=\lambda(E)\). The definition of Lebesgue outer measure as an infimum over countable open-interval covers therefore provides open intervals \(I_1,I_2,\ldots\) covering \(E\) such that

where \(|I_j|\) denotes the length of \(I_j\). Let \(G=\bigcup_{j=1}^{\infty} I_j\). This set is open and contains \(E\). It is Borel, hence Lebesgue measurable, and countable subadditivity gives

The sets \(E\) and \(G\setminus E\) are disjoint and measurable, and their union is \(G\). Since \(\lambda(G)<\infty\), countable additivity (or finite additivity) gives \(\lambda(G)=\lambda(E)+\lambda(G\setminus E)\). Subtracting the finite number \(\lambda(E)\) yields \(\lambda(G\setminus E)<\varepsilon\), as required. \(\square\)

The finite-measure hypothesis matters in this formulation: if \(\lambda(E)=\infty\), the bound \(\lambda(G)<\lambda(E)+\varepsilon\) gives no useful control, and subtraction of infinite quantities is not valid. Infinite-measure sets can still be approximated locally by applying the theorem to their intersections with bounded intervals.

Compact Approximation from the Inside

For bounded measurable sets, the outside approximation theorem also gives an inside approximation. The idea is to approximate the complement of the set within a containing closed interval, and then remove that open approximation. The resulting set is closed and bounded, hence compact by the Heine–Borel Theorem.

Proof. Choose real numbers \(a\leq b\) such that \(E\subseteq[a,b]\), and let \(C=[a,b]\setminus E\). The set \(C\) is measurable because \([a,b]\) is Borel and \(\mathcal{L}\) is a sigma-algebra. Also \(\lambda(C)\leq b-a<\infty\). Apply the Open Approximation Theorem to \(C\): there is an open set \(G\supseteq C\) with \(\lambda(G\setminus C)<\varepsilon\). Set \(K=[a,b]\setminus G\). This is closed and bounded, so it is compact. Since \(G\) contains \(C\), every point of \(K\) lies in \([a,b]\setminus C=E\), and therefore \(K\subseteq E\). Finally, if \(x\in E\setminus K\), then \(x\in E\) and \(x\notin [a,b]\setminus G\). As \(x\in[a,b]\), this implies \(x\in G\), so \(x\in G\setminus C\). Hence \(E\setminus K\subseteq G\setminus C\), and monotonicity gives

This proves the theorem. \(\square\)

Every Lebesgue Measurable Set Has a Borel Representative

Open approximation has a further consequence, including for sets of infinite measure. Every Lebesgue measurable set differs from a Borel set by a null set. The Borel set need not equal the original set: the point is that their symmetric difference has measure zero. This explains one sense in which the Lebesgue sigma-algebra extends the Borel sigma-algebra—by including arbitrary subsets of Borel null sets, as guaranteed by completeness.

Proof. For each positive integer \(k\), set \(E_k=E\cap[-k,k]\). These sets are measurable, have finite measure, and satisfy \(E=\bigcup_{k=1}^{\infty}E_k\). For every positive integer \(j\), the Open Approximation Theorem gives an open set \(U_{k,j}\supseteq E_k\) such that

Define \(H_k=\bigcap_{j=1}^{\infty}U_{k,j}\). This is a Borel set because it is a countable intersection of open sets, and it contains \(E_k\). The measurable set \(H_k\setminus E_k\) is contained in \(U_{k,j}\setminus E_k\) for every \(j\). Consequently,

A nonnegative real number bounded above by \(2^{-j}\) for every positive integer \(j\) must be zero, so \(\lambda(H_k\setminus E_k)=0\). Now let \(B=\bigcup_{k=1}^{\infty}H_k\), which is Borel. Since \(E_k\subseteq H_k\) for every \(k\), we have \(E\subseteq B\). Moreover, if \(x\in B\setminus E\), then \(x\in H_k\) for some \(k\), and \(x\notin E_k\) because \(E_k\subseteq E\). Thus

Each set on the right is null, so countable subadditivity gives \(\lambda(B\setminus E)=0\). Because \(E\subseteq B\), the symmetric difference \(E\mathbin{\triangle}B\) equals \(B\setminus E\), and is null. \(\square\)

What Approximation Does—and Does Not—Say

Open and compact approximations are useful because they let us replace a complicated measurable set by a topologically simple set while controlling the measure of the error. For a finite-measure set, an open superset can add less than any prescribed positive amount of measure. For a bounded measurable set, a compact subset can omit less than any prescribed positive amount. The Borel representative theorem makes a related statement for sets that may have infinite measure: a Borel set can represent the same Lebesgue-measure information up to a null difference.

These statements concern measure, not pointwise equality. An open approximation may contain many points outside \(E\), and a compact approximation may leave out points of \(E\); the theorems control only the measure of those differences. Also, the compact approximation theorem is stated for bounded sets. Its proof relies on enclosing \(E\) in a finite-length interval; one should not apply that proof unchanged to an unbounded set.