The Construction and Its Measure Budget
The rationals in \([0,1]\) are dense, so removing the rationals themselves would leave a set that is not closed and therefore not compact. Instead, surround every rational by a small open interval. If the sum of the interval lengths is controlled, countable subadditivity bounds the measure removed. The complement of this open union inside \([0,1]\) will be compact, contain only irrationals, and retain nearly the full measure of the interval.
Throughout, \(m\) is Lebesgue measure on \(\mathbb{R}\), and all measures below are Lebesgue measures of measurable subsets of \(\mathbb{R}\). Fix an enumeration without repetitions, \((q_n)_{n\ge1}\), of \(\mathbb{Q}\cap[0,1]\). Such an enumeration exists because this set is countable; it is dense in \([0,1]\). Given a target error \(\eta>0\), define the radius \(r_n=\eta/2^{n+1}\), the open interval \(I_n=(q_n-r_n,q_n+r_n)\), and the open set \(U=\bigcup_{n=1}^{\infty}I_n\). Each \(I_n\) has length \(2r_n=\eta/2^n\).
The radii are chosen to make the total length budget a convergent geometric series. Since \(U\) is a countable union of open intervals, it is open and measurable. Countable subadditivity gives \(m(U)\le\sum_{n=1}^{\infty}m(I_n)=\sum_{n=1}^{\infty}\eta/2^n=\eta\). The intervals may overlap, so this inequality need not be an equality; overlap only makes the measure removed smaller.
Why the Remaining Set Is Compact and Irrational
Proof. Because \(U\) is open in \(\mathbb{R}\), its complement \(U^c\) is closed in \(\mathbb{R}\). The interval \([0,1]\) is closed, so \(K_\eta=[0,1]\cap U^c\) is closed in \(\mathbb{R}\). It is also bounded, since \(K_\eta\subseteq[0,1]\). By the Heine–Borel theorem, a closed bounded subset of \(\mathbb{R}\) is compact.
For each \(n\), \(q_n\in I_n\), since \(|q_n-q_n|=0<r_n\) and \(r_n>0\). Thus \(q_n\notin K_\eta\). The enumeration includes every rational in \([0,1]\), so \(K_\eta\cap\mathbb{Q}=\varnothing\). Finally, \(K_\eta\) is closed and hence Lebesgue measurable. Its complement relative to \([0,1]\) is \([0,1]\cap U\), and therefore \(m([0,1]\setminus K_\eta)=m([0,1]\cap U)\le m(U)\le\eta\). The interval \([0,1]\) has measure \(1\), and \(K_\eta\) and \([0,1]\setminus K_\eta\) are disjoint measurable sets whose union is \([0,1]\). Finite additivity now gives \(m(K_\eta)=1-m([0,1]\setminus K_\eta)\ge1-\eta\). This proves all three claims. \(\square\)
Worked Example: A Compact Irrational Set of Measure at Least Four-Fifths
Take \(\eta=1/5\), and use the enumeration beginning \(q_1=0\), \(q_2=1\), \(q_3=1/2\), \(q_4=1/3\), \(q_5=2/3\). The radius formula is \(r_n=(1/5)/2^{n+1}\). For the first three rationals, the radii and interval lengths are \(r_1=1/20\), \(m(I_1)=2(1/20)=1/10\); \(r_2=1/40\), \(m(I_2)=2(1/40)=1/20\); and \(r_3=1/80\), \(m(I_3)=2(1/80)=1/40\). For instance, the first interval is \((-1/20,1/20)\), centered at \(0\), and the second is \((39/40,41/40)\), centered at \(1\).
The first three lengths sum to \(1/10+1/20+1/40=4/40+2/40+1/40=7/40\). For \(n\ge4\), the tail of the length budget is \(\sum_{n=4}^{\infty}(1/5)/2^n=(1/5)(1/8)=1/40\). Thus the full sum of interval lengths is \(7/40+1/40=8/40=1/5\). By the theorem, the complement \(K_{1/5}\) is compact, contains no rational, and satisfies \(m(K_{1/5})\ge1-1/5=4/5\). Check: the first three lengths together with the tail account for \(1/5\), and the resulting lower bound is \(1-1/5=4/5\), as required.
Worked Example: A One-Percent Error Budget
Now set \(\eta=1/100\) and use the same radii \(r_n=\eta/2^{n+1}\). With the stated enumeration, \(q_5=2/3\), so its neighborhood has radius \(r_5=(1/100)/2^6=1/6400\). Explicitly, this interval is \((2/3-1/6400,\,2/3+1/6400)\); it contains \(2/3\), since the distance from its center to \(2/3\) is \(0<1/6400\).
The total length removed is bounded by \(\sum_{n=1}^{\infty}2r_n=\sum_{n=1}^{\infty}(1/100)/2^n=(1/100)\cdot1=1/100\). Consequently the resulting compact set has measure at least \(1-1/100=99/100\), contains no rational in \([0,1]\), and is contained in \([0,1]\). The displayed interval illustrates how an individual rational is excluded; the sum over every index, not just the first five, is what guarantees exclusion of all rationals.
Check: the geometric series \(\sum_{n=1}^{\infty}2^{-n}\) equals \(1\), so the length budget is \(1/100\), and subtracting it from the interval measure \(1\) gives \(99/100\).
Other Length Budgets and a Geometric Consequence
The particular powers of \(2\) are convenient rather than essential. The argument only needs positive radii whose doubled sum is at most the desired error. For example, the alternative choice \(r_n=\eta/3^n\) gives \(\sum_{n=1}^{\infty}2r_n=2\eta\sum_{n=1}^{\infty}3^{-n}=2\eta(1/2)=\eta\). The same proof then gives a compact set of irrationals with measure at least \(1-\eta\). What matters is that each rational receives a positive radius and the total interval length is controlled.
Worked Example: Using a Different Geometric Series
Choose \(\eta=1/8\) and \(r_n=\eta/3^n\). The first two radii are \(r_1=(1/8)/3=1/24\) and \(r_2=(1/8)/9=1/72\), so the first two interval lengths are \(2r_1=1/12\) and \(2r_2=1/36\). Their sum is \(1/12+1/36=3/36+1/36=4/36=1/9\).
The remaining lengths, for \(n\ge3\), sum to \(\sum_{n=3}^{\infty}(1/4)3^{-n}=(1/4)(1/27)/(1-1/3)=(1/4)(1/18)=1/72\). The total is \(1/9+1/72=8/72+1/72=9/72=1/8\). Thus the compact complement has measure at least \(1-1/8=7/8\). Check: the first two lengths and the tail total \(8/72+1/72=9/72=1/8\), matching the prescribed error.
Proof. Suppose \(K_\eta\) contained a nonempty open interval \(J\) in \(\mathbb{R}\). Since \(K_\eta\subseteq[0,1]\), any such interval contained in \(K_\eta\) would include a nonempty open subinterval of \((0,1)\). The rationals are dense in \(\mathbb{R}\), so that subinterval contains a rational in \([0,1]\). But \(K_\eta\) contains no such rational, a contradiction. Hence \(K_\eta\) has empty interior. Since \(K_\eta\) is closed, its closure is itself, so its closure also has empty interior; this is exactly the stated nowhere-dense property. \(\square\)
This consequence distinguishes large measure from containing an interval: the construction gives measure as close to \(1\) as desired, but no nonempty interval can lie inside the set. Indeed, every interval of positive length contains rationals, while the constructed set excludes every rational in \([0,1]\).
A Common Length-Budget Mistake
A tempting but invalid choice is to give every rational the same positive radius. For example, with target error \(1/10\), assigning radius \(1/20\) to every rational makes each interval have length \(1/10\). The sum over the first \(N\) intervals is then \(N/10\); at \(N=2\), this is \(2/10=1/5\), already greater than the target \(1/10\). As \(N\) increases, this sum has no finite upper bound. Countable subadditivity therefore supplies no useful small upper bound for the measure of the union.
The correction is to make the radii decrease with \(n\) and verify the entire infinite sum. With radius \((1/10)/2^{n+1}\), the \(n\)-th interval has length \((1/10)/2^n\), and the sum over all \(n\ge1\) is \((1/10)\sum_{n=1}^{\infty}2^{-n}=1/10\). A finite list of rational neighborhoods is not enough either: it leaves all rationals not on the list in the complement. The countable union is what excludes every rational, while the convergent length budget controls the measure of that full union.
Check Your Understanding
Use the construction and its hypotheses to answer the following questions.
- For the radii \(r_n=\eta/2^{n+1}\), compute the length of the \(n\)-th interval and the sum of all interval lengths.
- Explain why \(K_\eta\) is closed in \(\mathbb{R}\), and identify the theorem that then gives compactness.
- Why does every rational in \([0,1]\) fail to belong to \(K_\eta\)? Where is positivity of each radius used?
- If \(\eta=1/50\), what lower bound does the construction give for \(m(K_\eta)\)?
- What fails in the measure estimate if every rational is assigned the same radius \(1/20\)?