Why Unbounded Sets Need a Different Construction
For a bounded Lebesgue-measurable set, compact approximation gives a compact subset that leaves out as little measure as desired. For an unbounded set, applying that result just once is not enough: one must first divide the set into bounded pieces. There is a second issue. A countable union of compact sets need not be closed, so the pieces must be assembled in a way that controls their accumulation points.
Throughout, \(m\) denotes Lebesgue measure on the Lebesgue sigma-algebra \(\mathcal{L}\) of \(\mathbb{R}\). All sets \(E\) under discussion are Lebesgue measurable, and “closed” refers to the usual topology on \(\mathbb{R}\). We use the compact approximation result for bounded measurable sets: if \(A\in\mathcal{L}\) is bounded and \(\delta>0\), there is a compact set \(K\subseteq A\) such that \(m(A\setminus K)<\delta\). The construction below extends the approximation to unbounded sets, without requiring \(m(E)<\infty\).
The quantity controlled here is \(m(E\setminus F)\), the measure of the part omitted. When \(m(E)=m(F)=\infty\), the expression \(m(E)-m(F)\) is not defined as a difference of extended real numbers, so it cannot serve as the error.
Why Local Finiteness Preserves Closedness
A family of sets in \(\mathbb{R}\) is locally finite if every point has an open neighborhood that meets only finitely many members of the family. For a locally finite family of closed sets, the union is closed. The condition matters: it prevents infinitely many pieces from accumulating at a point that is missing from the union.
Proof. Let \(x\notin\bigcup_{k\in\mathbb{Z}}C_k\). By local finiteness, there is an open neighborhood \(U\) of \(x\) that meets only finitely many of the sets, say \(C_{k_1},\ldots,C_{k_r}\). The finite union \(C_{k_1}\cup\cdots\cup C_{k_r}\) is closed, so its complement is open and contains \(x\). Intersect that complement with \(U\); the resulting open neighborhood of \(x\) meets none of the \(C_k\). Thus the complement of \(\bigcup_k C_k\) is open, and the union is closed. \(\square\)
The lemma addresses the topological part of the construction. The measure part is handled by dividing \(\mathbb{R}\) into the disjoint bounded intervals \([k,k+1)\), indexed by \(k\in\mathbb{Z}\). Each point belongs to exactly one of these intervals, and every bounded region meets only finitely many of them.
Proof of the Closed Approximation Theorem
Proof. Fix \(E\in\mathcal{L}\) and \(\varepsilon>0\). For each integer \(k\), set \(E_k=E\cap[k,k+1)\). These sets are measurable, bounded, pairwise disjoint, and their union is \(E\). Assign to each piece the positive error allowance \(\delta_k=\varepsilon 2^{-|k|-2}\). The total allowance is finite: \(\sum_{k\in\mathbb{Z}}\delta_k=\varepsilon(1/4+2\sum_{n=1}^{\infty}2^{-n-2})=3\varepsilon/4\).
By compact approximation for bounded measurable sets, for each \(k\) choose a compact set \(K_k\subseteq E_k\) such that \(m(E_k\setminus K_k)<\delta_k\). This also applies when \(E_k\) is empty, by taking \(K_k=\varnothing\). Define \(F=\bigcup_{k\in\mathbb{Z}}K_k\). Since \(K_k\subseteq E_k\subseteq E\), we have \(F\subseteq E\).
Each \(K_k\) is closed and lies in \([k,k+1]\). The family \((K_k)_{k\in\mathbb{Z}}\) is locally finite: a bounded neighborhood can meet \([k,k+1]\) for only finitely many integers \(k\). The lemma therefore shows that \(F\) is closed.
Finally, the sets \(E_k\setminus K_k\) are pairwise disjoint and their union is \(E\setminus F\). Countable additivity of Lebesgue measure and the choices of \(K_k\) give \[ m(E\setminus F)=\sum_{k\in\mathbb{Z}}m(E_k\setminus K_k) \leq\sum_{k\in\mathbb{Z}}\delta_k =\frac{3\varepsilon}{4}<\varepsilon. \] This estimate also holds if \(m(E)=\infty\), since it concerns the disjoint omitted pieces, not a subtraction of infinite measures. The required closed set has been constructed. \(\square\)
Worked Constructions
Worked Example: A Closed Approximation to the Positive Half-Line
Let \(E=[0,\infty)\). For each integer \(n\geq0\), put \(\delta_n=2^{-n-3}\) and \(K_n=[n+\delta_n,n+1-\delta_n]\). These are nonempty closed intervals: \(\delta_n\leq1/8<1/2\), so \(n+\delta_n<n+1-\delta_n\). Each lies in \(E\), and the family is locally finite. Hence \(F=\bigcup_{n=0}^{\infty}K_n\) is closed and \(F\subseteq E\).
Within each interval \([n,n+1)\), the omitted part consists of an initial interval of length \(\delta_n\) and a final interval of length \(\delta_n\), up to endpoints that do not affect Lebesgue measure. Thus \[ m(E\setminus F)=\sum_{n=0}^{\infty}2\delta_n =2\sum_{n=0}^{\infty}2^{-n-3} =2\left(\frac{1/8}{1-1/2}\right) =\frac12. \] The set \(E\) has infinite measure, yet this particular closed inner approximation omits only finite measure. Check: the length of \(K_n\) is \(1-2\delta_n\), so the omitted length in each unit interval is \(1-(1-2\delta_n)=2\delta_n\), as used in the sum.
Worked Example: A Finite-Measure Unbounded Set
Consider the measurable set \(E=\bigcup_{n=1}^{\infty}(n,n+2^{-n})\). Its intervals are disjoint, since each has length \(2^{-n}\leq1/2\) and lies between \(n\) and \(n+1\). Countable additivity gives \[ m(E)=\sum_{n=1}^{\infty}2^{-n}=\frac{1/2}{1-1/2}=1. \]
For each \(n\geq1\), let \(K_n=[n+2^{-n-2},\,n+2^{-n}-2^{-n-2}]\). The interval is contained in \((n,n+2^{-n})\), and its length is \(2^{-n}-2\cdot2^{-n-2}=2^{-n}-2^{-n-1}=2^{-n-1}>0\). The family is locally finite, so \(F=\bigcup_{n=1}^{\infty}K_n\) is closed and contained in \(E\). The omitted measure is \[ m(E\setminus F)=\sum_{n=1}^{\infty}2^{-n-1} =\frac{1/4}{1-1/2}=\frac12. \] As an independent check, \(m(F)=\sum_{n=1}^{\infty}2^{-n-1}=1/2\); since \(F\subseteq E\) and \(m(E)=1\), finite additivity gives \(m(E\setminus F)=m(E)-m(F)=1/2\).
Worked Example: Why a Compact Truncation Does Not Work for the Whole Line
Take \(E=\mathbb{R}\). The compact truncation \(F_N=[-N,N]\), for any finite \(N>0\), leaves out \((-\infty,-N)\cup(N,\infty)\). This remainder has infinite Lebesgue measure: for every \(R>N\), it contains \((N,R)\), of measure \(R-N\), and these lengths have no finite upper bound as \(R\) increases. Thus no such truncation gives \(m(E\setminus F_N)<\varepsilon\) for finite \(\varepsilon\).
The theorem still applies: choose \(F=\mathbb{R}\), which is closed, is contained in \(E\), and satisfies \(m(E\setminus F)=m(\varnothing)=0\). This example shows why the conclusion asks for a closed set, not necessarily a compact one. Check: the omitted measure for \(F=\mathbb{R}\) is zero, while for every finite \(N\) it is infinite; the two constructions have different outcomes because one retains the entire unbounded set.
Common Pitfalls and the Role of the Construction
Pitfall: assuming a countable union of closed pieces is closed. For example, every singleton \(\{1/n\}\) is closed, but their union \(\{1/n:n\geq1\}\) is not closed: it omits its limit point \(0\). In the approximation theorem, the sets \(K_k\) are not assembled arbitrarily. They lie in bounded intervals indexed by integers, so their family is locally finite; the lemma is what guarantees that their union is closed.
Pitfall: subtracting infinite measures. If \(m(E)=\infty\), the expression \(m(E)-m(F)\) may be undefined, even when \(F\subseteq E\). The theorem makes the precise claim \(m(E\setminus F)<\varepsilon\), and its proof establishes that claim by summing the measures of disjoint omitted pieces. It does not require the measure of \(E\) itself to be finite.
The bounded partition has two jobs. It makes the earlier compact approximation result available on every piece, and it ensures that the chosen compact sets cannot accumulate in a bounded region from infinitely many different pieces. The error allowances are summable, so even infinitely many local choices leave a globally controlled remainder. Together, these points explain how closed inner approximation extends from bounded measurable sets to arbitrary measurable subsets of \(\mathbb{R}\).
Check Your Understanding
Use the bounded-piece construction and the local finiteness argument in your answers.
- Why does the proof use the half-open intervals \([k,k+1)\) to partition \(\mathbb{R}\), and why does each point belong to exactly one piece?
- For \(\delta_k=\varepsilon2^{-|k|-2}\), compute \(\sum_{k\in\mathbb{Z}}\delta_k\).
- Explain why the union of the compact sets in the theorem is closed, even though an arbitrary countable union of closed sets need not be closed.
- For \(E=\mathbb{R}\), why does \([-N,N]\) fail to give a finite omitted measure, while \(F=\mathbb{R}\) succeeds?
- Why is \(m(E\setminus F)\), rather than \(m(E)-m(F)\), the appropriate error when \(m(E)=\infty\)?