Why Infinite Measure Requires a Different Argument
For a measurable set of finite measure, regularity gives open supersets and closed subsets whose excess or loss in measure is as small as desired. If a measurable set has infinite measure, the same goal still makes sense: the excess outside an open superset, or the part missed by a closed subset, can have finite measure smaller than any prescribed positive number. What fails is not regularity, but a shortcut based on subtracting the measures of two infinite sets.
Throughout, the measure space is \((\mathbb{R},\mathcal{L},m)\), where \(\mathcal{L}\) is the Lebesgue sigma-algebra and \(m\) is Lebesgue measure. A set \(E\in\mathcal{L}\) has infinite measure when \(m(E)=\infty\). We use the finite-measure regularity results established earlier: if \(A\in\mathcal{L}\) and \(m(A)<\infty\), then for every \(\eta>0\) there is an open \(U\supseteq A\) with \(m(U\setminus A)<\eta\); if \(A\) is also bounded, there is a compact \(K\subseteq A\) with \(m(A\setminus K)<\eta\). The key is to apply these results to bounded pieces of \(E\), with error allowances whose sum is finite.
Regularity for Infinite-Measure Sets
- there is an open set \(O\supseteq E\) such that \(m(O\setminus E)<\varepsilon\);
- there is a closed set \(F\subseteq E\) such that \(m(E\setminus F)<\varepsilon\).
Proof. Partition \(\mathbb{R}\) into the disjoint half-open intervals \(I_k=[k,k+1)\), indexed by \(k\in\mathbb{Z}\), and put \(A_k=E\cap I_k\). Each \(A_k\) is Lebesgue measurable and has measure at most \(1\), so the finite-measure regularity results apply. Define positive weights \(w_k=\frac13\,2^{-|k|}\). Their sum is \(\frac13(1+2\sum_{j=1}^{\infty}2^{-j})=1\). Set \(\eta_k=(\varepsilon/4)w_k\), so \(\sum_{k\in\mathbb Z}\eta_k=\varepsilon/4\).
For the open approximation, choose for each \(k\) an open set \(U_k\supseteq A_k\) with \(m(U_k\setminus A_k)<\eta_k\), and let \(O=\bigcup_{k\in\mathbb Z}U_k\). This set is open and contains \(E=\bigcup_k A_k\). If \(x\in O\setminus E\), then \(x\in U_k\) for some \(k\), while \(x\notin A_k\); hence \(O\setminus E\subseteq\bigcup_k(U_k\setminus A_k)\). Countable subadditivity therefore gives \(m(O\setminus E)\le\sum_k m(U_k\setminus A_k)\le\sum_k\eta_k=\varepsilon/4<\varepsilon\). The error set is measurable because \(O\) is open and \(E\) is Lebesgue measurable.
For the inner approximation, each \(A_k\) is bounded. Choose a compact set \(K_k\subseteq A_k\) with \(m(A_k\setminus K_k)<\eta_k\), and let \(F=\bigcup_{k\in\mathbb Z}K_k\). Then \(F\subseteq E\). The family \((K_k)\) is locally finite: every bounded interval can meet \(K_k\subseteq[k,k+1]\) for only finitely many integers \(k\). A locally finite union of closed sets is closed. In this case, to verify that fact, take a convergent sequence in \(F\). Its terms lie in a bounded interval, so they belong to a finite union of the compact sets \(K_k\). That finite union is closed, and the limit belongs to it, hence to \(F\). Thus \(F\) is closed. Since the \(A_k\) partition \(E\) and \(K_k\subseteq A_k\), we have \(E\setminus F=\bigcup_k(A_k\setminus K_k)\), a disjoint union. Countable additivity gives \(m(E\setminus F)=\sum_k m(A_k\setminus K_k)\le\sum_k\eta_k=\varepsilon/4<\varepsilon\). This proves both assertions. \(\square\)
Proof. Apply the theorem with error tolerance \(\varepsilon/4\) to obtain \(O\supseteq E\) and \(F\subseteq E\). The inclusions give the disjoint decomposition \(O\setminus F=(O\setminus E)\cup(E\setminus F)\). Each part has measure less than \(\varepsilon/4\), so \(m(O\setminus F)=m(O\setminus E)+m(E\setminus F)<\varepsilon/2<\varepsilon\). In particular, the difference has finite measure even though \(m(E)=\infty\). \(\square\)
Worked Approximations
Worked Example: A Half-Line
Let \(E=[0,\infty)\), a Lebesgue measurable set in \((\mathbb{R},\mathcal{L},m)\) with infinite measure. For \(\varepsilon=1/10\), choose \(\delta=1/40\) and set \(O=(-\delta,\infty)\) and \(F=[\delta,\infty)\). The set \(O\) is open, \(F\) is closed, and \(F\subseteq E\subseteq O\).
The excess \(O\setminus E=(-\delta,0)\) has measure \(\delta=1/40\). The missed part \(E\setminus F=[0,\delta)\) also has measure \(1/40\). These sets are disjoint, so \(m(O\setminus F)=1/40+1/40=1/20\), which is less than \(1/10\). As a check, \(O\setminus F=(-1/40,1/40)\), an interval of length \(2/40=1/20\). The open and closed approximations work even though both \(E\) and \(F\) have infinite measure.
Worked Example: The Irrationals
Let \(E=\mathbb{R}\setminus\mathbb{Q}\), the set of irrationals. It is Lebesgue measurable, and it has infinite measure: for every positive integer \(N\), the set \([0,N]\setminus\mathbb{Q}\) has measure \(N\), since \(\mathbb{Q}\) is countable and has measure zero. Enumerate the rationals without repetition as \((q_j)_{j\ge1}\), and define \(U=\bigcup_{j=1}^{\infty}(q_j-1/(100\cdot2^j),q_j+1/(100\cdot2^j))\). This is an open set containing every rational. Its measure is at most the sum of the interval lengths: \(m(U)\le\sum_{j=1}^{\infty}2/(100\cdot2^j)=\frac1{50}\sum_{j=1}^{\infty}2^{-j}=1/50\).
Take \(O=\mathbb{R}\) and \(F=\mathbb{R}\setminus U\). The set \(O\) is open, \(F\) is closed, and \(F\subseteq E\), because \(U\) contains all rationals. The outer error is \(O\setminus E=\mathbb{Q}\), so \(m(O\setminus E)=0\). Also \(E\setminus F=E\cap U\), and therefore \(m(E\setminus F)\le m(U)\le1/50\). Thus \(m(O\setminus F)=m(U)\le1/50\). The numerical bound follows from the geometric series just computed; the outer error is zero independently, by countability of \(\mathbb{Q}\).
Worked Example: Infinitely Many Separated Intervals
In \((\mathbb{R},\mathcal{L},m)\), let \(E=\bigcup_{n=0}^{\infty}[2n,2n+1]\). The intervals are disjoint and each has measure \(1\), so countable additivity gives \(m(E)=\sum_{n=0}^{\infty}1=\infty\). For \(n\ge0\), put \(\delta_n=1/(100\cdot2^{n+1})\), and define \(O=\bigcup_{n=0}^{\infty}(2n-\delta_n,2n+1+\delta_n)\) and \(F=\bigcup_{n=0}^{\infty}[2n+\delta_n,2n+1-\delta_n]\). The set \(O\) is open and contains \(E\), while \(F\subseteq E\). Since the intervals making up \(F\) form a locally finite family of closed sets, \(F\) is closed.
We have \(\sum_{n=0}^{\infty}\delta_n=\frac1{100}\sum_{n=0}^{\infty}2^{-(n+1)}=1/100\). Each expanded interval contributes two excess pieces, each of length \(\delta_n\). These pieces do not overlap: the gap between successive expanded intervals is \(1-\delta_n-\delta_{n+1}>0\), since \(\delta_n+\delta_{n+1}\le 1/100+1/200<1\). Consequently \(m(O\setminus E)=2\sum_n\delta_n=1/50\). In each original interval, the part removed to form \(F\) has two pieces of length \(\delta_n\), so \(m(E\setminus F)=2\sum_n\delta_n=1/50\). Hence \(m(O\setminus F)=1/50+1/50=1/25\). This exact calculation checks the two errors separately and their sum.
A Common Pitfall
A tempting but invalid argument is: “Since \(m(E)=\infty\), choose a closed \(F\subseteq E\) with \(m(F)>m(E)-\varepsilon\); then \(m(E\setminus F)<\varepsilon\).” The expression \(\infty-\varepsilon\) is not a real number, and subtracting \(m(F)\) from \(m(E)\) does not compute the measure of \(E\setminus F\) when both measures are infinite. For a concrete failure, take \(E=\mathbb{R}\) and \(F=[-1,1]\). Both are measurable and \(F\) is closed, but \(m(E)=\infty\), \(m(F)=2\), and \(m(E\setminus F)=\infty\), not a small error.
The valid argument controls the set difference directly. Partition \(E\) into bounded pieces, approximate each piece, and choose the individual error allowances to have a finite sum smaller than the desired tolerance. The proof also explains why the inner construction is a union of compact approximations rather than an arbitrary union of closed sets: local finiteness ensures that the resulting union remains closed.
Check Your Understanding
For each question, distinguish the measure of the entire set from the measure of the approximation error.
- Why does applying finite-measure regularity directly to \(E\) fail when \(m(E)=\infty\), and which bounded pieces are used in the proof?
- In the half-line example, calculate \(m(O\setminus E)\), \(m(E\setminus F)\), and \(m(O\setminus F)\) for the stated \(\delta\).
- For the irrational-set example, explain why \(F=\mathbb{R}\setminus U\) is closed and why it contains no rational numbers.
- Verify that the weights \(w_k=\frac13\,2^{-|k|}\), for \(k\in\mathbb{Z}\), sum to \(1\).
- Why does local finiteness matter when forming the closed inner approximation from the compact sets \(K_k\)?