Open and Compact Approximations
Lebesgue measure allows a measurable set to be replaced by geometrically simpler sets while controlling the measure of what changes. For a set of finite measure in \(\mathbb{R}^n\), the outer approximation is an open superset and the inner approximation is a compact subset. The containment directions matter: a set that differs from the target only by a null set need not satisfy the required containment.
Throughout, the measure space is \((\mathbb{R}^n,\mathcal{L},m)\), where \(\mathcal{L}\) is the Lebesgue sigma-algebra and \(m\) is Lebesgue measure. This measure space is complete and sigma-finite, but not finite. We consider Lebesgue measurable sets \(E\in\mathcal{L}\) with \(m(E)<\infty\). An open set is open in the Euclidean topology; a compact set is a set that is compact in that topology. In \(\mathbb{R}^n\), compact sets are closed and bounded.
For the outer construction, we use the defining cover property of Lebesgue outer measure \(m^*\): if \(E\) has finite measure, then for every \(\eta>0\) there is a countable cover by axis-parallel boxes whose total volume is less than \(m(E)+\eta\). Each box in such a cover can be enlarged to an open box with an arbitrarily small increase in volume. Indeed, its side lengths can be increased by sufficiently small positive amounts; the product of the new side lengths tends to the original volume as those amounts tend to zero. We also use continuity from below: if measurable sets \(A_j\) increase to \(A\), then \(m(A_j)\to m(A)\). This follows by writing \(A\) as the disjoint union of \(A_1\) and the successive differences \(A_j\setminus A_{j-1}\), and applying countable additivity.
The Regular Approximation Theorem
Proof. First construct \(U\). By the box-cover property, choose countably many boxes \(R_j\) covering \(E\) with \(\sum_j m(R_j)<m(E)+\varepsilon/2\). Enlarge each \(R_j\) to an open box \(G_j\supseteq R_j\) so that \(m(G_j)<m(R_j)+\varepsilon 2^{-j-1}\). The sum of the added allowances is \(\sum_{j=1}^{\infty}\varepsilon 2^{-j-1}=\varepsilon/2\). Thus \(U=\bigcup_jG_j\) is open, contains \(E\), and satisfies \(m(U)\le\sum_jm(G_j)<m(E)+\varepsilon\). Since \(E\) is measurable, \(U\setminus E\) is measurable and \(m(U)=m(E)+m(U\setminus E)\); consequently \(m(U\setminus E)<\varepsilon\).
For the compact approximation, let \(Q_R=[-R,R]^n\) for positive integers \(R\). The sets \(E\cap Q_R\) increase to \(E\), so continuity from below and \(m(E)<\infty\) give an \(R\) for which \(m(E\setminus Q_R)<\varepsilon/2\). Set \(A=Q_R\setminus E\). This is a measurable set of finite measure. Apply the outer construction, with error less than \(\varepsilon/2\), to obtain an open set \(G\supseteq A\) with \(m(G\setminus A)<\varepsilon/2\). Put \(K=Q_R\setminus G\). The set \(K\) is closed and bounded, hence compact. Also \(K\subseteq E\), because every point of \(Q_R\setminus E=A\) lies in \(G\). Finally, $$ E\setminus K\subseteq (E\setminus Q_R)\cup(E\cap Q_R\cap G) \subseteq (E\setminus Q_R)\cup(G\setminus A). $$ The two sets on the right have measures less than \(\varepsilon/2\), so \(m(E\setminus K)<\varepsilon\). This proves both approximations. \(\square\)
The finite-measure hypothesis is used in the outer argument to choose a cover with finite total volume close to \(m(E)\), and in the inner argument to make the part outside a large cube small. The theorem gives strict error bounds, not merely errors less than or equal to the chosen tolerance.
Worked Approximations
Worked Example: A Closed Interval
In \((\mathbb{R},\mathcal{L},m)\), take \(E=[1,4]\) and \(\delta=1/20\). Its measure is \(m(E)=4-1=3\). Define the open superset \(U=(1-\delta,4+\delta)=(19/20,81/20)\). Its length is \(81/20-19/20=62/20=31/10\), so $$ m(U\setminus E)=\frac{31}{10}-3=\frac{1}{10}. $$ For an inner approximation, take \(K=[1+\delta,4-\delta]=[21/20,79/20]\). This is compact and contained in \(E\). Its length is \(79/20-21/20=58/20=29/10\), giving $$ m(E\setminus K)=3-\frac{29}{10}=\frac{1}{10}. $$ Thus both errors are exactly \(1/10\), which is less than \(\varepsilon=1/5\). As a check, the interval \(U\setminus E\) consists of two pieces of length \(1/20\), and \(E\setminus K\) also consists of two pieces of length \(1/20\); each total is \(2/20=1/10\).
Worked Example: A Rectangle in the Plane
In \((\mathbb{R}^2,\mathcal{L},m)\), let \(E=[0,2]\times[0,1]\). Its area is \(2\cdot1=2\). With \(\delta=1/20\), set $$ U=(-\delta,2+\delta)\times(-\delta,1+\delta),\qquad K=[\delta,2-\delta]\times[\delta,1-\delta]. $$ The first set is open and contains \(E\); the second is compact and contained in \(E\). The side lengths of \(U\) are \(2+2\delta=21/10\) and \(1+2\delta=11/10\). Therefore \(m(U)=231/100\), and \(m(U\setminus E)=231/100-2=31/100\). The side lengths of \(K\) are \(2-2\delta=19/10\) and \(1-2\delta=9/10\). Hence \(m(K)=171/100\), and \(m(E\setminus K)=2-171/100=29/100\). Both errors are less than \(32/100\). As an independent check, the four corner rectangles removed from \(E\) have total area \(4(\delta^2)=4/400=1/100\), and the four strips along the edges have total area \(28/100\), giving \(29/100\) altogether.
The same calculation applies to a box \(E=\prod_{i=1}^n[a_i,b_i]\) with positive side lengths \(L_i=b_i-a_i\). Expanding each side by \(\delta\) gives outer excess \(\prod_{i=1}^n(L_i+2\delta)-\prod_{i=1}^nL_i\). If \(2\delta\le\min_iL_i\), shrinking each side by \(\delta\) gives inner loss \(\prod_{i=1}^nL_i-\prod_{i=1}^n(L_i-2\delta)\). These formulas follow from the volume formula for boxes and the inclusions \(K\subseteq E\subseteq U\); they let one choose \(\delta\) to meet any prescribed positive error.
Worked Example: An Unbounded Set of Finite Measure
In \((\mathbb{R},\mathcal{L},m)\), consider $$ E=\bigcup_{k=1}^{\infty}(3k,3k+2^{-k}). $$ The intervals are disjoint, since each has length at most \(1/2\) and consecutive starting points differ by \(3\). Countable additivity gives \(m(E)=\sum_{k=1}^{\infty}2^{-k}=1\). This set is unbounded, so compact inner approximations must omit some of its distant intervals.
For the outer approximation define $$ U=\bigcup_{k=1}^{\infty}(3k-2^{-(k+4)},\,3k+2^{-k}+2^{-(k+4)}). $$ These enlarged intervals remain disjoint: the right endpoint for index \(k\) is at most \(3k+1/2+1/32\), while the next left endpoint is at least \(3k+3-1/32\). The extra length for interval \(k\) is \(2\cdot2^{-(k+4)}=2^{-(k+3)}\). Thus $$ m(U\setminus E)=\sum_{k=1}^{\infty}2^{-(k+3)} =\frac{1/16}{1-1/2}=\frac18. $$ Endpoints do not affect these measures, because a countable set of endpoints has Lebesgue measure zero. Hence \(U\) is an open approximation with error \(1/8<1/4\).
For an inner approximation, take the compact finite union $$ K=\bigcup_{k=1}^{4}[3k+2^{-(k+4)},\,3k+2^{-k}-2^{-(k+4)}]. $$ Each closed interval lies inside its corresponding open interval of \(E\). Its length is \(2^{-k}-2\cdot2^{-(k+4)}=(7/8)2^{-k}\). The first four original interval lengths sum to \(1/2+1/4+1/8+1/16=15/16\), so \(m(K)=(7/8)(15/16)=105/128\). Therefore \(m(E\setminus K)=1-105/128=23/128<1/4\). The omitted tail has measure \(2^{-4}=1/16=8/128\), and the trimming removes \(15/128\); together these give \(23/128\), confirming the calculation.
Why the Containments Matter
A common mistake is to treat a small measure error as a substitute for the required inclusion. For \(E=[0,1]\subset\mathbb{R}\), the set \(V=(0,1)\) satisfies \(m(V\setminus E)=0\), but it is not an outer approximation as defined: \(E\not\subseteq V\), since \(0,1\in E\) and neither endpoint belongs to \(V\). The correction is to use \(U=(-\delta,1+\delta)\) for \(\delta>0\); then \(E\subseteq U\) and \(m(U\setminus E)=2\delta\). Null boundary differences can make measures equal without making set containments true.
There is a parallel issue for inner approximation. If a finite-measure set is unbounded, choosing a compact subset inside a large cube captures only part of it. The error must include both what was trimmed inside the cube and what lies outside it. The theorem’s proof controls the latter by first choosing the cube so that \(m(E\setminus Q_R)\) is small. In the unbounded example, omitting the intervals after \(k=4\) costs exactly \(1/16\), which is part of the total inner error.
These approximations also give a geometric interpretation of indicator functions. For measurable sets \(E,F\subseteq\mathbb{R}^n\), their real-valued indicators satisfy \(|\chi_E-\chi_F|=\chi_{E\triangle F}\) pointwise. Consequently, when \(m(E\triangle F)<\infty\) and \(1\le p<\infty\), the definition of the \(L^p\) norm gives \(\|\chi_E-\chi_F\|_p=m(E\triangle F)^{1/p}\). Thus controlling the measure of the set error controls the \(L^p\) distance of the indicators. Regular approximation supplies such control while retaining the open or compact geometry of the approximating set.
Check Your Understanding
For each question, keep the set containment and the measure error as separate requirements.
- For \(E=[2,5]\subset\mathbb{R}\), construct an open outer approximation with error less than \(1/10\), and calculate its exact error.
- For \(E=[0,1]\times[0,3]\subset\mathbb{R}^2\), write the outer and inner errors obtained by expanding and shrinking every side by \(\delta\).
- In the inner-approximation proof, explain why \(K=Q_R\setminus G\) is compact and why it is contained in \(E\).
- Why does \(m(V\setminus E)=0\) alone not imply that \(V\) is an outer approximation of \(E\)? Give a specific example.
- For the unbounded set in the third worked example, calculate the measure of the intervals omitted after the first \(N\) intervals.