Finite-Interval Approximation in Measure

Approximation by a Finite Union of Intervals

A Lebesgue measurable set can have a complicated shape: it need not be open, closed, or a union of finitely many intervals. Nevertheless, when its measure is finite, its membership pattern can be approximated in measure by a particularly simple set. The relevant error is the measure of the symmetric difference: points that belong to exactly one of the two sets. The approximation controls how much of the line is misclassified, not how close the sets are point by point.

All sets below lie in \(\mathbb{R}\), with Lebesgue sigma-algebra \(\mathcal{L}\) and Lebesgue measure \(m\). A finite union of bounded intervals means a union of finitely many bounded open intervals; the empty union is allowed. The set being approximated may be unbounded, as long as it is Lebesgue measurable and has finite measure.

The key input is outer regularity for finite-measure Lebesgue measurable sets: for each \(\delta>0\), there is an open set \(G\supseteq E\) such that \(m(G)<m(E)+\delta\). Since \(E\) is measurable and has finite measure, this also gives \(m(G\setminus E)=m(G)-m(E)<\delta\). Outer regularity is established in the earlier material on Lebesgue measurable sets; the argument here explains how to turn its open cover into a finite union of bounded intervals.

Why an Open Set Can Be Truncated

Proof. For each \(x\in G\), take the union of all open intervals containing \(x\) and contained in \(G\). This union is an open interval: any two of its points lie in intervals containing \(x\), so the intervals overlap and their union contains every point between those two points. The resulting maximal intervals are either equal or disjoint, and together cover \(G\). Each nonempty such interval contains a rational number. Assigning to each interval the first rational it contains in a fixed enumeration of \(\mathbb{Q}\) shows that there are at most countably many intervals. None can be unbounded, since every unbounded interval has infinite Lebesgue measure, contrary to \(m(G)<\infty\).

Write the components as \(I_1,I_2,\ldots\), with a finite list if there are only finitely many. Countable additivity gives \(m(G)=\sum_j m(I_j)<\infty\). If the list is finite, take \(U=G\). If it is infinite, convergence of this nonnegative series implies that for some \(N\), \(\sum_{j>N}m(I_j)<\eta\). Set \(U=\bigcup_{j=1}^{N}I_j\). The intervals are disjoint, so \(m(G\setminus U)=\sum_{j>N}m(I_j)<\eta\), as required. If \(G\) is empty, choose \(U\) empty. \(\square\)

The Finite-Interval Approximation Theorem

Proof. Fix \(\varepsilon>0\) and put \(\delta=\varepsilon/2\). By outer regularity, choose an open \(G\supseteq E\) with \(m(G)<m(E)+\delta\). Both measures on the right are finite, and \(E\) is measurable, so \(m(G\setminus E)=m(G)-m(E)<\delta\). In particular, \(m(G)<\infty\). Apply the lemma with \(\eta=\delta\) to choose a finite union \(U\) of bounded open intervals contained in \(G\) such that \(m(G\setminus U)<\delta\).

Because \(E\subseteq G\), every point of \(E\setminus U\) lies in \(G\setminus U\). Because \(U\subseteq G\), every point of \(U\setminus E\) lies in \(G\setminus E\). The two parts of the symmetric difference are disjoint, so

This proves the assertion for every positive \(\varepsilon\), including when \(m(E)=0\). In that case the same argument applies; alternatively, the empty union has error \(m(E)=0\). \(\square\)

The proof separates two sources of error. The open set \(G\) may include points not in \(E\), but outer regularity makes this excess small. Then \(U\) discards all but finitely many components of \(G\), with small discarded measure. The choice \(\delta=\varepsilon/2\) allocates half the error to each step. More generally, the two tolerances can be chosen separately as any positive numbers whose sum is less than \(\varepsilon\).

Worked Applications

What the Approximation Does—and Does Not—Say

For finite-measure sets, symmetric-difference measure gives a useful way to compare sets: \(m(E\,\triangle\,U)\) is the total measure of points whose membership changes. The theorem says that finite unions of bounded intervals can achieve arbitrarily small error in this sense. It does not say that \(E\) itself is a finite union of intervals, that \(U\subseteq E\), or that every point of \(E\) is close to some point of \(U\). The approximating intervals may include some points outside \(E\), while omitting other points of \(E\).

A common mistake is to infer boundedness from finite measure and then assume an open cover is already a finite union. Here is a concrete correction to both steps. In the first example, \(m(E)=1\), but \(E\) is unbounded because it contains points in intervals starting at \(2n\) for arbitrarily large \(n\). Also, taking the entire open set \(G=E\) does not give a finite union: it has infinitely many components. Taking only the first four components gives a finite union, but its error is not zero; the omitted lengths sum to \(\frac1{16}\). The correct argument controls this tail rather than pretending it vanishes.

It is also important that the theorem assumes \(E\) is Lebesgue measurable and has finite measure. Outer regularity in the form used above provides an open cover with small excess for such a set; the proof does not establish the same assertion for arbitrary nonmeasurable sets. The finiteness hypothesis ensures the open cover can be chosen with finite measure, so its component lengths form a convergent series that can be truncated.