Common Errors in Open and Compact Approximation

Three Checks Before Using an Approximation

Regularity statements are easy to misapply when the set difference is written in the wrong direction, when a finite-measure hypothesis is silently dropped, or when one-sided control is mistaken for control of the symmetric difference. These are not merely notational slips: each can turn a valid estimate into a false conclusion. The reliable checks are to draw the containments, name the set whose measure is being bounded, and verify that any subtraction involves finite quantities.

Throughout, \(E,F,G\) denote Lebesgue measurable subsets of \(\mathbb{R}\), and \(m\) is Lebesgue measure on \((\mathbb{R},\mathcal{L},m)\). Open sets and compact sets are Borel, hence Lebesgue measurable. A set has finite measure when its measure is less than infinity; this does not mean that the whole measure space is finite. We use the regularity facts established earlier: if \(m(E)<\infty\), then for every \(\varepsilon>0\) there is an open \(G\supseteq E\) with \(m(G\setminus E)<\varepsilon\), and there is a compact \(K\subseteq E\) with \(m(E\setminus K)<\varepsilon\).

Containment Determines the Error Term

For an outer approximation, the open set contains the set being approximated: \(E\subseteq G\). The points added by the approximation are \(G\setminus E\). In particular, \(E\setminus G=\varnothing\), so bounding \(m(E\setminus G)\) gives no information about the outer error.

For an inner approximation, the compact set lies inside the set being approximated: \(K\subseteq E\). The omitted points are \(E\setminus K\). Here \(K\setminus E=\varnothing\), so bounding that difference says nothing about how much of \(E\) was left out.

Proof. If \(x\in B\), then either \(x\in A\), or \(x\notin A\), in which case \(x\in B\setminus A\). Also, no point belongs to both \(A\) and \(B\setminus A\). This proves the disjoint union identity. Countable additivity of Lebesgue measure, applied to these two disjoint measurable sets, gives \(m(B)=m(A)+m(B\setminus A)\), with values in \([0,\infty]\). When both terms on the right are finite, ordinary subtraction is valid and gives the stated formula. If the terms are infinite, the identity remains valid as an extended-real equality, but subtracting one infinite quantity from another is undefined. \(\square\)

Finite Measure and Infinite Measure Are Not Interchangeable

The formula \(m(G)-m(E)=m(G\setminus E)\) is useful only when the subtraction is defined. For \(E\subseteq G\) with finite measures, it follows from the proposition. If both measures are infinite, the difference \(m(G)-m(E)\) is not a real number and cannot be used as an error estimate. The set difference \(G\setminus E\), however, may still have finite measure. Regularity can control that difference directly, including for some sets of infinite measure; it does not make infinite-minus-infinite meaningful.

There is a corresponding issue for compact inner approximations. Every compact subset of \(\mathbb{R}\) is bounded and has finite Lebesgue measure. Thus, if \(m(E)=\infty\), no compact \(K\subseteq E\) can satisfy \(m(E\setminus K)<\varepsilon\) for finite \(\varepsilon\): otherwise \(E=K\cup(E\setminus K)\) would have finite measure. For an infinite-measure set, inner approximation must instead be understood locally or through a different stated target, not as a globally small omitted part.

One-Sided Error Is Not Symmetric Error

For any measurable \(A,B\), the two pieces \(A\setminus B\) and \(B\setminus A\) are disjoint and their union is \(A\,\triangle\,B\). Therefore \(m(A\,\triangle\,B)=m(A\setminus B)+m(B\setminus A)\), with the value \(+\infty\) allowed. A bound on just one term does not bound their sum. If one set is contained in the other, the other term is empty, and then the one-sided error is exactly the symmetric-difference error. Without that containment, both directions must be controlled.

Putting Open and Compact Approximation Together

Proof. Apply outer regularity with tolerance \(\varepsilon/2\) to obtain an open \(G\supseteq E\) with \(m(G\setminus E)<\varepsilon/2\). Apply compact inner regularity with the same tolerance to obtain a compact \(K\subseteq E\) with \(m(E\setminus K)<\varepsilon/2\). These choices give \(K\subseteq E\subseteq G\), so \(K\subseteq G\) and \(G\,\triangle\,K=G\setminus K\). Moreover, \(G\setminus K=(G\setminus E)\cup(E\setminus K)\), a disjoint union: the first piece lies outside \(E\), and the second lies inside \(E\). Countable additivity therefore gives \(m(G\,\triangle\,K)=m(G\setminus E)+m(E\setminus K)<\varepsilon/2+\varepsilon/2=\varepsilon\). The strict final inequality follows because each of the two chosen bounds is strict. \(\square\)

This result also shows why the containment chain is valuable: it reduces the symmetric difference to a single set difference and divides the error budget between the outer and inner approximations. The finite-measure assumption is part of the inner approximation statement used here. It cannot be dropped while retaining a globally small \(m(E\setminus K)\) for compact \(K\) when \(m(E)=\infty\).

Common Error Checklist

Before accepting a regularity calculation, check the following points:

• For an open outer approximation, verify \(E\subseteq G\) and estimate \(m(G\setminus E)\).
• For a compact inner approximation, verify \(K\subseteq E\) and estimate \(m(E\setminus K)\).
• Do not replace a set-difference estimate by subtraction of measures unless the relevant measures are finite.
• For a symmetric-difference estimate, account for both \(A\setminus B\) and \(B\setminus A\), unless containment makes one empty.
• When applying a finite-measure regularity statement, identify the set with finite measure; do not assume the entire space has finite measure.