From Closed Inner Approximation to Compactness
A closed subset of \(\mathbb{R}\) need not be compact: it may be unbounded. For a bounded measurable set, however, a closed inner approximation can be made compact without any additional loss of measure. This observation turns a measure approximation into a topological one.
Throughout, the ambient measure space is \((\mathbb{R},\mathcal{L},m)\), where \(m\) is Lebesgue measure. This measure space is complete and \(\sigma\)-finite, but not finite. A set \(E\in\mathcal{L}\) is bounded if it is contained in some bounded interval; such an \(E\) has finite measure. All set containments below hold everywhere, not merely almost everywhere.
The route to the theorem is the closed inner approximation. We use outer regularity of Lebesgue measure: if \(A\in\mathcal{L}\) has finite measure and \(\delta>0\), there is an open set \(G\supseteq A\) with \(m(G)<m(A)+\delta\). This standard result was established in the earlier material on outer regularity.
Proof. If \(E=\varnothing\), take \(F=\varnothing\). Otherwise choose \(R>0\) such that \(E\subseteq I=[-R,R]\). The set \(A=I\setminus E\) is Lebesgue measurable and has finite measure because \(A\subseteq I\) and \(m(I)=2R\). By outer regularity, choose an open \(G\supseteq A\) with \(m(G)<m(A)+\delta\). Set \(F=I\setminus G\). Since \(I\) is closed and \(G\) is open, \(F\) is closed. If \(x\in F\), then \(x\in I\) and \(x\notin G\); since \(A\subseteq G\), this implies \(x\notin A=I\setminus E\), so \(x\in E\). Also \(E\setminus F=E\cap G\subseteq G\setminus A\). As \(A\subseteq G\) and both measures are finite, \(m(G\setminus A)=m(G)-m(A)<\delta\). Therefore \(m(E\setminus F)<\delta\). \(\square\)
Proof of the theorem. If \(E=\varnothing\), choose \(K=\varnothing\). Otherwise apply the lemma with \(\delta=\varepsilon\) to obtain a closed \(F\subseteq E\) with \(m(E\setminus F)<\varepsilon\). Because \(E\) is bounded, \(F\) is bounded as well. The Heine–Borel theorem says that a subset of \(\mathbb{R}\) is compact if and only if it is closed and bounded. Thus \(K=F\) is compact, lies in \(E\), and has the required measure loss. Since \(F\subseteq E\) and \(m(E)<\infty\), additivity gives \(m(E)=m(F)+m(E\setminus F)\), so \(m(F)>m(E)-\varepsilon\). \(\square\)
The boundedness assumption is used for compactness, not for the closed approximation itself. The lemma produces a closed subset with small measure loss; the theorem then uses boundedness to invoke Heine–Borel. This is why the result concerns compact approximation of bounded sets, rather than simply repeating a closed inner approximation.
Worked Constructions
Worked Example: A Compact Core of an Open Interval
Let \(E=(0,1)\subset\mathbb{R}\) and take \(\varepsilon=\frac15\). Both are interpreted in \((\mathbb{R},\mathcal{L},m)\). Choose \(K=[\frac1{20},\frac{19}{20}]\). Every point of this closed interval lies strictly between \(0\) and \(1\), so \(K\subseteq E\); it is closed and bounded, hence compact.
The interval \(E\) has measure \(m(E)=1\), while the length of \(K\) is \(\frac{19}{20}-\frac1{20}=\frac{18}{20}=\frac9{10}\). Thus \(m(K)=\frac9{10}\), and \(m(E\setminus K)=1-\frac9{10}=\frac1{10}<\frac15\). Check: the two omitted end intervals have lengths \(\frac1{20}\) each, so their total measure is \(\frac1{20}+\frac1{20}=\frac1{10}\), agreeing with the subtraction calculation.
Worked Example: Removing Countably Many Points
In \((\mathbb{R},\mathcal{L},m)\), let \(E=[0,1]\setminus\{1/n:n\ge2\}\). This is Lebesgue measurable because the removed set is countable and hence null; therefore \(m(E)=m([0,1])=1\). We construct an explicit compact subset. For each integer \(n\ge2\), define \(U_n=(\frac1n-2^{-(n+5)},\frac1n+2^{-(n+5)})\), let \(U=\bigcup_{n\ge2}U_n\), and set \(K=[0,1]\setminus U\).
Each \(U_n\) lies inside \([0,1]\): its center is at least \(\frac12\), and its radius is at most \(2^{-7}=\frac1{128}<\frac12\); its upper endpoint is also less than \(1\), since for \(n\ge2\), \(\frac1n\le\frac12\) and \(2^{-(n+5)}\le\frac1{128}\). Every removed point \(1/n\) belongs to \(U_n\), so \(K\subseteq E\). The set \(U\) is relatively open in \([0,1]\), making \(K\) closed in \([0,1]\); hence \(K\) is compact.
We verify that the intervals are disjoint, so their lengths can be added. Consecutive centers are separated by \(\frac1n-\frac1{n+1}=\frac1{n(n+1)}\), whereas the sum of their radii is \(2^{-(n+5)}+2^{-(n+6)}=3\cdot2^{-(n+6)}\). For \(n\ge2\), \(3n(n+1)<2^{n+6}\): at \(n=2\), \(18<256\), and from \(n\) to \(n+1\) the left side grows by a factor \(\frac{n+2}{n}\le2\), while the right side doubles. Thus the radius sum is strictly less than the center separation. The centers are ordered, so separation of each consecutive pair implies all the intervals are disjoint.
Each interval has length \(2\cdot2^{-(n+5)}=2^{-(n+4)}\). Therefore, by countable additivity on these disjoint measurable intervals, \(m(U)=\sum_{n=2}^{\infty}2^{-(n+4)}=\frac1{64}\cdot\frac1{1-\frac12}=\frac1{32}\). The set \(U\) is contained in \([0,1]\), so \(m(K)=1-\frac1{32}=\frac{31}{32}\). Since \(E\setminus K=E\cap U\) differs from \(U\) only by a subset of the countable removed set, \(m(E\setminus K)=m(U)=\frac1{32}\). Check: \(m(E)=m(K)+m(E\setminus K)=\frac{31}{32}+\frac1{32}=1\), as required.
Worked Example: A Set with Infinitely Many Components
In \((\mathbb{R},\mathcal{L},m)\), define \(E=\bigcup_{n=1}^{\infty}(2^{-n},2^{-n}+2^{-(n+2)})\). These intervals are disjoint: the upper endpoint of the interval indexed by \(n+1\) is \((\frac12+\frac18)2^{-n}=\frac58\,2^{-n}\), strictly below the lower endpoint \(2^{-n}\) of the interval indexed by \(n\). All intervals lie in \([0,1]\), so \(E\) is bounded and measurable. Their lengths are \(2^{-(n+2)}\), and hence \(m(E)=\sum_{n=1}^{\infty}2^{-(n+2)}=\frac18\cdot\frac1{1-\frac12}=\frac14\).
To make a compact inner approximation, retain the first four intervals and trim each end by \(2^{-(n+5)}\). Put \(K=\bigcup_{n=1}^{4}[2^{-n}+2^{-(n+5)},\,2^{-n}+2^{-(n+2)}-2^{-(n+5)}]\). Each retained closed interval has positive length: the original length is \(2^{-(n+2)}\), and the total trim is \(2^{-(n+4)}\), which is smaller. Each is contained in its corresponding open component of \(E\); thus \(K\subseteq E\). As a finite union of closed bounded intervals, \(K\) is compact.
The omitted tail has measure \(\sum_{n=5}^{\infty}2^{-(n+2)}=\frac1{128}\cdot\frac1{1-\frac12}=\frac1{64}\). The trimming loss in the first four intervals is \(\sum_{n=1}^{4}2\cdot2^{-(n+5)}=2(2^{-6}+2^{-7}+2^{-8}+2^{-9})=2\cdot\frac{15}{512}=\frac{15}{256}\). The retained intervals and all omitted pieces are disjoint, so \(m(E\setminus K)=\frac1{64}+\frac{15}{256}=\frac{19}{256}\), and \(m(K)=\frac14-\frac{19}{256}=\frac{45}{256}\). Check: \(\frac{45}{256}+\frac{19}{256}=\frac{64}{256}=\frac14=m(E)\). In particular, the loss \(\frac{19}{256}\) is less than \(\frac1{10}\), since \(190<256\).
A Common Pitfall: The Ambient Interval Matters
It is tempting to take an open set \(G\) covering the complement of \(E\) and declare \(E\setminus G\) closed. That conclusion is false when \(E\) is not itself closed. For a concrete example, take \(E=(0,1)\) and \(G=(-\frac1{10},\frac1{10})\cup(\frac9{10},\frac{11}{10})\). This open set covers the complement of \(E\) inside \(I=[0,1]\), namely \(\{0,1\}\). But \(E\setminus G=(\frac1{10},\frac9{10})\), which is not closed in \(\mathbb{R}\): it does not contain its limit points \(\frac1{10}\) and \(\frac9{10}\).
The correction is to take the complement inside the closed ambient interval: \(F=I\setminus G=[\frac1{10},\frac9{10}]\). This set is closed and bounded, hence compact, and it lies inside \(E\). Its measure is \(\frac9{10}-\frac1{10}=\frac8{10}=\frac45\), so the loss from \(E\) is \(1-\frac45=\frac15\). The construction in the proof uses exactly this relative-complement step: \(F=I\setminus G\), not \(E\setminus G\).
The compact approximation theorem is useful because compact sets are closed and bounded, making them amenable to finite subcover arguments and topological constructions. Here the measure control comes from outer regularity applied to the part of a containing interval outside \(E\); boundedness then supplies compactness through Heine–Borel. The two roles should be kept distinct: regularity controls the measure lost, and boundedness converts closedness into compactness.
Check Your Understanding
In each question, work in \((\mathbb{R},\mathcal{L},m)\) and state whether the set inclusion is pointwise.
- For bounded measurable \(E\) and \(\varepsilon>0\), why does a closed \(F\subseteq E\) with \(m(E\setminus F)<\varepsilon\) become compact?
- In the closed-inner-approximation proof, why is \(I\setminus G\subseteq E\) when \(G\supseteq I\setminus E\)?
- For \(E=(0,1)\), compute the measure loss for \(K=[\frac1{10},\frac9{10}]\), and determine whether \(K\subseteq E\).
- Why does \(E\setminus G\) fail to be closed in the pitfall example, while \(I\setminus G\) is closed?
- For the infinite-component example, identify separately the tail loss and the trimming loss, and verify that their sum is \(\frac{19}{256}\).