Approximating from Inside by Closed Sets
A Lebesgue measurable set can be irregular in shape and unbounded, so it need not itself be closed. Nevertheless, when its measure is finite, we can find a set made from countably many closed pieces that lies inside it and misses as little measure as we prescribe. The key is to approximate bounded pieces first, then make the errors across all pieces summable.
Throughout, sets are subsets of \(\mathbb{R}\), and \((\mathbb{R},\mathcal{L},m)\) is the complete measure space of the real line with Lebesgue measure. A set is closed in the usual topology of \(\mathbb{R}\). A set is \(F_\sigma\) if it is a countable union of closed sets. An approximation from within of \(E\) is a set \(F\subseteq E\); the quantity \(m(E\setminus F)\) measures the part of \(E\) omitted.
This earlier result supplies the individual closed pieces. It does not by itself give a suitable approximation of an unbounded set: the pieces must be chosen so their combined omissions remain small. The construction below assigns an error budget to each bounded piece and uses countable subadditivity of Lebesgue measure to control their union.
The \(F_\sigma\) Approximation Theorem
Proof. For each integer \(n\ge1\), let \(E_n=E\cap[-n,n]\). Each \(E_n\) is measurable and bounded. Apply the compact approximation result with \(\eta_n=\varepsilon 2^{-(n+1)}\). It gives a compact set \(K_n\subseteq E_n\) satisfying \(m(E_n\setminus K_n)<\varepsilon 2^{-(n+1)}\). Define \(F=\bigcup_{n=1}^{\infty}K_n\). Each \(K_n\) is closed, so \(F\) is \(F_\sigma\); also \(K_n\subseteq E_n\subseteq E\) for every \(n\), hence \(F\subseteq E\).
We check the omitted part carefully. Since \(\bigcup_{n=1}^{\infty}E_n=E\), any \(x\in E\setminus F\) belongs to some \(E_n\), but belongs to no \(K_n\). In particular, for that index \(n\), \(x\in E_n\setminus K_n\). Thus \(E\setminus F\subseteq\bigcup_{n=1}^{\infty}(E_n\setminus K_n)\), and countable subadditivity gives
The geometric series used here starts with \(2^{-2}\), and \(\sum_{n=1}^{\infty}2^{-(n+1)}=1/4+1/8+\cdots=1/2\). This proves the claimed estimate. If \(E\) is empty, the same construction works with every \(K_n=\varnothing\), and \(F=\varnothing\). The finiteness assumption is consistent with the stated theorem and ensures that \(E\) has finite measure; the local construction itself uses bounded pieces. \(\square\)
Notice the order of the argument: first choose each \(K_n\) with a specified error, then form their union, and finally bound the omitted set by a union of the individual errors. No limit is exchanged with an integral, and no subtraction of infinite measures is involved.
Worked Example: Two Bounded Intervals
Let \(E=(0,1)\cup(2,4)\) and choose \(\varepsilon=1/10\). We construct a closed subset and compute exactly how much measure it omits.
Set \(\delta=1/100\) and let \(F=[\delta,1-\delta]\cup[2+\delta,4-\delta]\). Both intervals are closed and lie inside \(E\), so \(F\) is closed and \(F\subseteq E\). The first interval has length \(1-2\delta=1-2/100=98/100\). The second has length \(2-2\delta=2-2/100=198/100\). They are disjoint, so \(m(F)=98/100+198/100=296/100=74/25\).
The measure of \(E\) is \(1+2=3\), because its two intervals are disjoint. The omitted pieces at the four ends each have length \(\delta=1/100\); endpoints do not affect Lebesgue measure. Therefore \(m(E\setminus F)=4/100=1/25\), which is less than \(1/10\). Check: \(m(F)+m(E\setminus F)=74/25+1/25=75/25=3=m(E)\).
Summable Errors for Countably Many Pieces
The theorem uses the bounded sets \(E_n=E\cap[-n,n]\), which overlap. Overlap causes no difficulty: the proof only needs the inclusion \(E\setminus F\subseteq\bigcup_n(E_n\setminus K_n)\), followed by countable subadditivity. The errors need not describe disjoint pieces, and they do not need to add up to the exact omitted measure.
Summability is essential to this estimate. If each of infinitely many approximations is allowed an error merely less than the same fixed number, adding those bounds gives an infinite upper bound. By contrast, the budgets \(\varepsilon 2^{-(n+1)}\) have finite sum. The theorem in fact leaves room: their sum is \(\varepsilon/2\), so the resulting omitted measure is strictly less than \(\varepsilon/2\), and therefore strictly less than the requested \(\varepsilon\).
Worked Example: A Countable Set with No Omission
Let \(E=\mathbb{Q}\cap[0,1]\), viewed as a subset of \(\mathbb{R}\), and take \(\varepsilon=1/100\). Enumerate its elements as \(q_1,q_2,\ldots\), with each rational in \([0,1]\) listed once. For each \(N\ge1\), set \(K_N=\{q_1,\ldots,q_N\}\). A finite set is closed, and \(K_N\subseteq E\). Hence \(F=\bigcup_{N=1}^{\infty}K_N=E\) is \(F_\sigma\).
The set \(E\) is countable, so \(m(E)=0\). Also \(E\setminus F=\varnothing\), giving \(m(E\setminus F)=0<1/100\). Check: the countable union of singleton sets has measure zero because each singleton has Lebesgue measure zero and Lebesgue measure is countably subadditive. Thus the computed omitted measure is indeed zero.
Worked Example: An Unbounded Set of Finite Measure
Consider \(E=\bigcup_{n=1}^{\infty}(n,n+2^{-n})\). These open intervals are disjoint: the right endpoint \(n+2^{-n}\) is less than \(n+1\), the left endpoint of the next interval. Their lengths are \(2^{-n}\), so countable additivity gives \(m(E)=\sum_{n=1}^{\infty}2^{-n}=1\).
For each \(n\ge1\), define \(\delta_n=2^{-(n+4)}\) and \(K_n=[n+\delta_n,n+2^{-n}-\delta_n]\). To check that this is a nonempty closed interval inside \((n,n+2^{-n})\), note that \(\delta_n=2^{-n}/16\), which is less than half the interval length \(2^{-n}/2\). Let \(F=\bigcup_{n=1}^{\infty}K_n\). This is an \(F_\sigma\) subset of \(E\).
In the \(n\)-th interval, the omitted length is \(2\delta_n=2^{-(n+3)}\). Consequently, \(m(E\setminus F)=\sum_{n=1}^{\infty}2^{-(n+3)}=1/8\). For \(\varepsilon=1/4\), this is less than \(\varepsilon\). The retained length in the \(n\)-th interval is \(2^{-n}-2^{-(n+3)}=(7/8)2^{-n}\), so \(m(F)=\sum_{n=1}^{\infty}(7/8)2^{-n}=7/8\). Check: \(m(F)+m(E\setminus F)=7/8+1/8=1=m(E)\).
A Consequence and a Common Pitfall
Proof. Apply the theorem to obtain \(F\subseteq E\) with \(m(E\setminus F)<\varepsilon\). Since \(F\subseteq E\), we have \(F\setminus E=\varnothing\), and therefore \(E\triangle F=E\setminus F\). It follows that \(m(E\triangle F)=m(E\setminus F)<\varepsilon\). \(\square\)
Common mistake: using the same error allowance for every piece. Suppose one chooses compact \(K_n\subseteq E_n\) with \(m(E_n\setminus K_n)<\varepsilon/2\) for every \(n\). The resulting estimate is only \(\sum_{n=1}^{\infty}m(E_n\setminus K_n)\le\sum_{n=1}^{\infty}\varepsilon/2=\infty\); it does not establish a finite bound for \(m(E\setminus\bigcup_nK_n)\), let alone a bound below \(\varepsilon\). The correction is to choose a summable sequence of allowances, such as \(\varepsilon 2^{-(n+1)}\). Then the sum of the allowances is \(\varepsilon/2\), and the proof controls the entire omitted part.
The distinction between a finite union and a countable union matters here. A finite number of errors, each made sufficiently small, can be added directly; for infinitely many pieces, the allowances must be summable. This is why the construction explicitly records its error budget before taking the union.
Check Your Understanding
Use the construction and the error estimate above to answer the following questions.
- For a measurable \(E\subseteq\mathbb{R}\) with \(m(E)<\infty\), explain why \(E_n=E\cap[-n,n]\) is eligible for compact approximation.
- If the error for the \(n\)-th compact approximation is less than \(\varepsilon 3^{-n}\), compute the sum of these error allowances for \(n\ge1\). Does this sum suffice to guarantee an omitted measure less than \(\varepsilon\)?
- In the unbounded-interval example, verify that \(K_n\subset(n,n+2^{-n})\) by comparing \(\delta_n\) with half the interval length.
- Why does \(F\subseteq E\) imply \(E\triangle F=E\setminus F\)?
- Give a concrete sequence of positive error allowances whose sum is less than \(\varepsilon\), and state how countable subadditivity uses those allowances.