From Open Approximation to a G_delta Set
Outer regularity approximates a measurable set from outside by open sets, with their measures as close as desired to the measure of the set. A single open approximation may have positive excess measure. Choosing a sequence of increasingly accurate open approximations and intersecting them gives a stronger conclusion: the resulting G_delta set contains the original set, and the difference has measure zero.
Throughout, the measure space is \((\mathbb{R},\mathcal{L},m)\), where \(\mathcal{L}\) is the Lebesgue sigma-algebra and \(m\) is Lebesgue measure. This measure space is complete, but it is not finite. We assume the set \(E\in\mathcal{L}\) has finite measure, \(m(E)<\infty\). A G_delta set is a countable intersection of open sets. We use outer regularity for finite-measure Lebesgue measurable sets: for every \(\varepsilon>0\), there is an open set \(O\supseteq E\) with \(m(O)<m(E)+\varepsilon\).
Proof. For each integer \(n\ge1\), apply outer regularity with \(\varepsilon=2^{-n}\). This gives an open set \(O_n\supseteq E\) satisfying \(m(O_n)<m(E)+2^{-n}\). In particular, \(m(O_n)<\infty\). Define \(G=\bigcap_{n=1}^{\infty}O_n\). This is a G_delta set, and \(E\subseteq G\) because every \(O_n\) contains \(E\). Every open subset of \(\mathbb{R}\) is Borel, so \(G\) is Borel and hence Lebesgue measurable. The difference \(O_n\setminus E\) is measurable. Since \(O_n\) is the disjoint union of \(E\) and \(O_n\setminus E\), and both \(m(O_n)\) and \(m(E)\) are finite, additivity gives $$ m(O_n\setminus E)=m(O_n)-m(E)<2^{-n}. $$ Also \(G\setminus E\subseteq O_n\setminus E\) for every \(n\), so \(0\le m(G\setminus E)<2^{-n}\) for every \(n\). Given any \(\varepsilon>0\), choose \(n\) so large that \(2^{-n}<\varepsilon\). Then \(m(G\setminus E)<\varepsilon\). A nonnegative number less than every positive \(\varepsilon\) must be zero; therefore \(m(G\setminus E)=0\). \(\square\)
The error bound is applied at the same index as the open set: \(G\setminus E\) lies inside every individual excess \(O_n\setminus E\). The intersection is what forces its measure below every \(2^{-n}\). The construction does not assert that the open sets \(O_n\) are nested; nesting is not needed.
Worked Constructions
Worked Example: Removing the Rationals from an Interval
Let \(E=[0,1]\setminus\mathbb{Q}\), the irrational points in \([0,1]\). Define \(O_n=(-1/n,1+1/n)\) for \(n\ge1\), and set \(G=\bigcap_{n=1}^{\infty}O_n\). Each \(O_n\) is open and contains \(E\). Its measure, computed from its endpoints, is $$ m(O_n)=(1+1/n)-(-1/n)=1+2/n. $$
A point \(x\) belongs to every \(O_n\) exactly when \(-1/n<x<1+1/n\) for every \(n\). If \(0\le x\le1\), these inequalities hold for every \(n\). If \(x<0\), choose \(n\) with \(1/n\le -x\); then \(x\notin O_n\). If \(x>1\), choose \(n\) with \(1/n\le x-1\); then \(x\notin O_n\). Thus \(G=[0,1]\). The rationals in \([0,1]\) form a countable set and have measure zero, so \(m(E)=1\) and \(m(G\setminus E)=m(\mathbb{Q}\cap[0,1])=0\). For each \(n\), the excess \(O_n\setminus E\) consists of the two outside intervals, with total measure \(1/n+1/n=2/n\), together with rationals inside \(O_n\), which have measure zero. This agrees with \(m(O_n)-m(E)=(1+2/n)-1=2/n\). Check: at \(n=2\), \(m(O_2)=1+2/2=2\), and the excess measure is \(2/2=1\); the limiting excess in \(G\) is zero, not one.
Worked Example: A Set Missing Its Endpoints and a Point
Take \(E=\{0\}\cup(2,3)\), which is measurable and has \(m(E)=0+ (3-2)=1\). For \(n\ge2\), let $$ O_n=(-1/n,1/n)\ \cup\ (2-1/n,3+1/n). $$
The two intervals are disjoint: the first ends at \(1/n\le1/2\), while the second begins at \(2-1/n\ge3/2\). Their lengths are \(2/n\) and \((3+1/n)-(2-1/n)=1+2/n\), respectively. Hence \(m(O_n)=2/n+1+2/n=1+4/n\), and each open set contains \(E\). The intersection of the first intervals is \(\{0\}\): zero belongs to all of them, while any nonzero \(x\) fails \(|x|<1/n\) once \(n\ge1/|x|\). The intersection of the second intervals is \([2,3]\): every point strictly between 2 and 3 belongs to all of them, both endpoints belong to all because the intervals are open extensions beyond the endpoints, and a point outside \([2,3]\) eventually fails one of the two endpoint inequalities. Since the two components stay separated for every \(n\ge2\), it follows that \(G=\bigcap_{n\ge2}O_n=\{0\}\cup[2,3]\). Therefore \(G\setminus E=\{2,3\}\) and \(m(G\setminus E)=0\). Check: \(m(G)=0+(3-2)=1=m(E)\); the two added endpoints change the set but not its measure.
Worked Example: A G_delta Superset of a Countable Dense Set
Enumerate \(\mathbb{Q}\cap[0,1]\) as \(q_1,q_2,\ldots\), and let \(E=\mathbb{Q}\cap[0,1]\). For \(n,k\ge1\), define $$ I_{n,k}=(q_k-2^{-(n+k+2)},q_k+2^{-(n+k+2)}),\qquad O_n=\bigcup_{k=1}^{\infty}I_{n,k}. $$
Each \(O_n\) is open and contains every \(q_k\), so \(E\subseteq G=\bigcap_{n=1}^{\infty}O_n\). The length of \(I_{n,k}\) is \(2\cdot2^{-(n+k+2)}=2^{-(n+k+1)}\). Countable subadditivity gives $$ m(O_n)\le\sum_{k=1}^{\infty}2^{-(n+k+1)} =2^{-(n+1)}\sum_{k=1}^{\infty}2^{-k} =2^{-(n+1)}. $$
Because \(G\subseteq O_n\) for every \(n\), \(0\le m(G)\le2^{-(n+1)}\) for every \(n\), and hence \(m(G)=0\). Also \(E\) is countable, so \(m(E)=0\); consequently \(G\setminus E\subseteq G\) has measure zero. Check: at \(n=1\), the displayed bound is \(m(O_1)\le2^{-2}=1/4\); as \(n\) increases, the bound tends to zero, forcing the intersection to be null. This example shows that the G_delta superset need not be the original dense set.
What Is Unique, and What Is Not
The theorem guarantees a G_delta superset with null excess, not equality with the original set. The first example makes this distinction explicit: the irrational points in \([0,1]\) are contained in the G_delta set \([0,1]\), and the additional rational points form a null set. G_delta sets are Borel, whereas a Lebesgue measurable set need not itself be presented as a G_delta set. The construction concerns containment and measure of the excess, not pointwise equality.
Proof. Since \(E\subseteq G\) and \(E\subseteq H\), a point in \(G\setminus H\) cannot belong to \(E\), so \(G\setminus H\subseteq G\setminus E\). Likewise, \(H\setminus G\subseteq H\setminus E\). The symmetric difference is the disjoint union \((G\setminus H)\cup(H\setminus G)\), and therefore $$ G\triangle H\subseteq(G\setminus E)\cup(H\setminus E). $$ The right-hand side is a union of two null sets, so it has measure zero by countable subadditivity. Monotonicity of measure now gives \(m(G\triangle H)=0\). \(\square\)
Common mistake: assuming the intersection equals the set. In the first example, it would be wrong to conclude \(\bigcap_n O_n=E\) just because each \(O_n\) contains \(E\) and their measures approach \(m(E)\). The actual intersection is \([0,1]\), which also contains \(\mathbb{Q}\cap[0,1]\). The correct conclusion is \(E\subseteq G\) and \(m(G\setminus E)=0\). Small excess measure in each open approximation does not by itself exclude every point outside \(E\); the proof controls the measure of the excess in the intersection instead.
The finite-measure hypothesis is used in a precise place: it lets us subtract \(m(E)\) from the finite number \(m(O_n)\) to obtain \(m(O_n\setminus E)\). Without a finite-measure setting for \(E\), this subtraction need not be valid; an inequality of the form \(m(O_n)<m(E)+2^{-n}\) would not provide a finite error bound when \(m(E)=\infty\). Here the hypotheses ensure each open approximation has finite measure and each excess can be estimated.
Check Your Understanding
For each question, distinguish the set containment from the assertion about measure.
- Why does \(G=\bigcap_n O_n\) contain \(E\) when every \(O_n\) contains \(E\)?
- In the first worked example, identify \(G\setminus E\) and explain why it is null although it is not empty.
- If \(m(O_n)<m(E)+3^{-n}\), what bound follows for \(m(G\setminus E)\) when \(G=\bigcap_nO_n\)?
- In the second example, why do the two endpoints 2 and 3 belong to every second component of \(O_n\), even though \(O_n\) is open?
- Why does the proof of the theorem use \(m(E)<\infty\) when calculating \(m(O_n\setminus E)\)?