From Two Approximations to One Sandwich
For a measurable set, an open superset and a compact subset give complementary ways to approximate it: the open set controls what has been added, while the compact set controls what has been left out. The useful conclusion is not merely that each approximation is close to the original set. They can be chosen together so the entire region between them has arbitrarily small measure.
Throughout the main result, work in \((\mathbb{R}^n,\mathcal{L},m)\), where \(n\ge 1\), \(\mathcal{L}\) is the Lebesgue sigma-algebra, and \(m\) is Lebesgue measure. Sets in the sandwich are subsets of \(\mathbb{R}^n\); inclusions are literal set inclusions, not inclusions merely almost everywhere. An open set is open in \(\mathbb{R}^n\), and a compact set is compact in \(\mathbb{R}^n\).
We use the following two regularity results for Lebesgue measure. Outer regularity: if \(E\in\mathcal{L}\) and \(m(E)<\infty\), then for every \(\eta>0\) there is an open \(G\supseteq E\) such that \(m(G\setminus E)<\eta\). Compact approximation for bounded measurable sets: if \(E\in\mathcal{L}\) is bounded, then for every \(\eta>0\) there is a compact \(K\subseteq E\) such that \(m(E\setminus K)<\eta\). These are the outer- and inner-approximation results for Lebesgue measure. Since a bounded measurable set has finite measure, both apply to the sets considered here.
Proof. Apply compact approximation to \(E\) with tolerance \(\varepsilon/2\). It gives a compact set \(K\subseteq E\) with \(m(E\setminus K)<\varepsilon/2\). Apply outer regularity to \(E\), also with tolerance \(\varepsilon/2\). It gives an open set \(G\supseteq E\) with \(m(G\setminus E)<\varepsilon/2\). The inclusions give \(K\subseteq E\subseteq G\). Moreover, $$ G\setminus K=(G\setminus E)\,\cup\,(E\setminus K). $$ The two sets on the right are disjoint and measurable. By additivity of \(m\) on disjoint measurable sets, $$ m(G\setminus K)=m(G\setminus E)+m(E\setminus K) <\frac{\varepsilon}{2}+\frac{\varepsilon}{2} =\varepsilon. $$ Thus the compact and open approximations together have the required gap. \(\square\)
How the Error Budget Works
The proof uses a simple identity that makes the error accounting exact. In particular, it does not require subtracting the measures of two possibly large sets.
Proof. Suppose first that \(x\in G\setminus K\). If \(x\notin E\), then \(x\in G\setminus E\). If \(x\in E\), then \(x\notin K\), so \(x\in E\setminus K\). This proves that \(G\setminus K\) is contained in the displayed union. Conversely, \(G\setminus E\subseteq G\setminus K\) because \(K\subseteq E\), and \(E\setminus K\subseteq G\setminus K\) because \(E\subseteq G\). The union is disjoint: a point in \(G\setminus E\) is not in \(E\), while a point in \(E\setminus K\) is in \(E\). Countable additivity, applied to these two disjoint measurable sets, gives the measure identity, including when one of the measures is infinite. \(\square\)
For the sandwich theorem, the budget \(\varepsilon/2\) is assigned once to the outer error \(m(G\setminus E)\) and once to the inner error \(m(E\setminus K)\). More generally, any positive numbers \(\eta_{\mathrm{out}}\) and \(\eta_{\mathrm{in}}\) with \(\eta_{\mathrm{out}}+\eta_{\mathrm{in}}\le\varepsilon\) can be used as the two tolerances. The equal split is convenient, not necessary.
Worked Example: An Open Interval
Take \(E=(0,1)\subset\mathbb{R}\), with Lebesgue measure \(m\), and choose \(\varepsilon=1/5\). Let \(\delta=1/40\), and define $$ K=[\delta,1-\delta]=[1/40,39/40], \qquad G=(-\delta,1+\delta)=(-1/40,41/40). $$ The set \(K\) is compact, \(G\) is open, and \(K\subseteq(0,1)\subseteq G\). The two parts of the gap are $$ G\setminus E=(-1/40,0]\cup[1,41/40), \qquad E\setminus K=(0,1/40)\cup(39/40,1). $$ Their measures are \(m(G\setminus E)=1/40+1/40=1/20\) and \(m(E\setminus K)=1/40+1/40=1/20\). The gap decomposition therefore gives $$ m(G\setminus K)=\frac{1}{20}+\frac{1}{20}=\frac{1}{10}<\frac{1}{5}. $$ As a direct check, \(G\setminus K=(-1/40,1/40)\cup(39/40,41/40)\), whose two intervals each have length \(2/40=1/20\). Their total length is \(1/10\), as above.
Worked Example: A Set Missing Every Rational
Let \(E=[0,1]\setminus\mathbb{Q}\subset\mathbb{R}\), and take \(\varepsilon=1/10\). The set is Lebesgue measurable: \(\mathbb{Q}\cap[0,1]\) is countable and hence has measure zero. Enumerate its points without repetition as \(q_1,q_2,\ldots\). For each \(j\ge1\), let $$ I_j=(q_j-r_j,q_j+r_j), \qquad r_j=\frac{1}{10\cdot 2^{j+3}}, \qquad U=\bigcup_{j=1}^{\infty}I_j. $$ Then \(U\) is open and contains every rational in \([0,1]\). By countable subadditivity and the lengths of the intervals, $$ m(U)\le\sum_{j=1}^{\infty}2r_j =\sum_{j=1}^{\infty}\frac{1}{10\cdot2^{j+2}} =\frac{1}{40}\sum_{j=1}^{\infty}\frac{1}{2^j} =\frac{1}{40}. $$ Define \(K=[0,1]\setminus U\). This is compact, since it is the intersection of the closed sets \([0,1]\) and \(U^c\). No rational in \([0,1]\) lies in \(K\), so \(K\subseteq E\). Also, $$ m(E\setminus K)\le m(U)\le\frac{1}{40}. $$ Choose the open set \(G=(-1/80,81/80)\). It contains \(E\), and the portion outside \([0,1]\) has measure \(1/80+1/80=1/40\). Since \(K\subseteq[0,1]\), the gap splits into that outside portion and \([0,1]\setminus K\). Thus $$ m(G\setminus K) =\frac{1}{40}+m([0,1]\setminus K) \le\frac{1}{40}+\frac{1}{40} =\frac{1}{20}<\frac{1}{10}. $$ This verifies the sandwich by explicit estimates. Independently, \(m(E)=1\), since \([0,1]\) has measure \(1\) and its rational points have measure zero; the bound \(m(E\setminus K)\le1/40\) therefore ensures \(m(K)\ge39/40\).
Worked Example: Two Separated Pieces
In \(\mathbb{R}\), let $$ E=(0,1)\setminus(1/3,2/3)=(0,1/3]\cup[2/3,1), $$ and set \(\varepsilon=1/10\). This set is measurable and has measure \(1/3+1/3=2/3\). Choose \(\delta=1/120\), and define $$ K=[\delta,1/3]\cup[2/3,1-\delta], \qquad G=(-\delta,1/3+\delta)\cup(2/3-\delta,1+\delta). $$ Here \(K\) is a finite union of closed bounded intervals, hence compact. Since \(\delta<1/3\), both intervals in \(K\) lie in \(E\); and the two open intervals in \(G\) cover the two pieces of \(E\). The open intervals are disjoint because $$ 1/3+\delta<2/3-\delta \quad\Longleftrightarrow\quad 2\delta<1/3, $$ which holds for \(\delta=1/120\). In each piece, the portion of \(G\) not in \(K\) has length \(3\delta\): for the first piece it consists of lengths \(\delta\), \( \delta\), and \(\delta\) at the left end, the right extension, and the portion between \((0,1/3]\) and \([\delta,1/3]\), respectively; equivalently, the full open interval has length \(1/3+2\delta\) and its compact subinterval has length \(1/3-\delta\), a difference of \(3\delta\). The second piece has the same lengths. Therefore $$ m(G\setminus K)=2(3\delta)=6\cdot\frac{1}{120}=\frac{1}{20}<\frac{1}{10}. $$ As a check using the separate errors, \(m(G\setminus E)=4\delta\) (two extensions of length \(\delta\) at each piece), and \(m(E\setminus K)=2\delta\) (one omitted end of length \(\delta\) in each piece). Their sum is \(6\delta=1/20\), agreeing with the direct gap calculation.
A Common Error in Choosing the Tolerances
It is not enough to make each approximation error less than \(\varepsilon\) and then conclude that the gap is less than \(\varepsilon\). The gap decomposition adds the two errors. For example, with \(E=(0,1)\) and \(\varepsilon=1/10\), choose \(\delta=1/30\), \(K=[\delta,1-\delta]\), and \(G=(-\delta,1+\delta)\). Each error is $$ m(G\setminus E)=\frac{1}{15}<\frac{1}{10}, \qquad m(E\setminus K)=\frac{1}{15}<\frac{1}{10}, $$ but the total gap is $$ m(G\setminus K)=\frac{1}{15}+\frac{1}{15}=\frac{2}{15}>\frac{1}{10}. $$ The correction is to assign each approximation a smaller tolerance, such as \(\varepsilon/2\). For the interval example, using \(\delta=1/40\) makes each error \(1/20\), and the total is \(1/10\), which is still not strictly below the requested \(1/10\). To guarantee a strict bound, use a smaller choice, such as \(\delta=1/80\); then each error is \(1/40\) and the total is \(1/20<1/10\). In the theorem proof, the strict inequalities in the two regularity estimates ensure the final strict inequality.
The sandwich result is useful because it gives a single small region \(G\setminus K\) that contains every point added by the open approximation and every point omitted by the compact approximation. It does not say that \(E\) itself is open or compact, nor that the set difference is empty. Rather, it makes the measure of the difference as small as desired while preserving the exact inclusions \(K\subseteq E\subseteq G\).
Check Your Understanding
Use the gap decomposition and track strict inclusions and measure bounds in each case.
- If \(m(G\setminus E)<\varepsilon/3\) and \(m(E\setminus K)<2\varepsilon/3\), what bound follows for \(m(G\setminus K)\)? Is it necessarily strictly less than \(\varepsilon\)?
- For \(E=(0,1)\), take \(K=[1/10,9/10]\) and \(G=(-1/10,11/10)\). Compute \(m(G\setminus E)\), \(m(E\setminus K)\), and \(m(G\setminus K)\).
- Why does \(K=[0,1]\setminus U\) remain compact when \(U\) is open in \(\mathbb{R}\)?
- Which two regularity statements are used to obtain the compact set and the open set in the sandwich theorem, respectively?
- Give a tolerance split other than \(\varepsilon/2\) and \(\varepsilon/2\) that guarantees a total gap strictly below \(\varepsilon\).