Regularity Calculations for Intervals and the Cantor Set

Approximation Errors as Exact Measure Calculations

Regularity becomes concrete when an approximating set is written down and its measure is calculated. For a bounded measurable set $E\subset\mathbb{R}$, an open set $U\supset E$ has outer approximation error $m(U\setminus E)$, while a compact set $K\subset E$ has inner approximation error $m(E\setminus K)$. In the examples below, these errors can be computed directly, rather than inferred only from a general regularity theorem.

All sets and measures in this tutorial are in $(\mathbb{R},\mathcal{L},m)$, where $\mathcal{L}$ is the Lebesgue sigma-algebra and $m$ is Lebesgue measure. This measure space is complete and sigma-finite, but not finite. The sets considered here are bounded and have finite measure. If $E\subset U$ and $m(U)<\infty$, then finite additivity gives $m(U\setminus E)=m(U)-m(E)$; likewise, if $K\subset E$ and $m(E)<\infty$, then $m(E\setminus K)=m(E)-m(K)$. The finiteness conditions matter: subtracting two infinite measures would not determine an error.

Definition: An open outer approximation to a measurable set $E$ is an open set $U$ with $E\subset U$. Its error is $m(U\setminus E)$, when this difference is measurable. A compact inner approximation is a compact set $K\subset E$. Its error is $m(E\setminus K)$, when this difference is measurable. The errors measure the excess outside $E$ and the omitted part of $E$, respectively.

Intervals: Open and Compact Approximations

Let $E=[a,b]$ with $a<b$, and let $\delta>0$. Expanding each endpoint by $\delta$ gives an open outer approximation; moving each endpoint inward by $\delta$ gives a compact inner approximation when $2\delta<b-a$. The formulas below give both errors exactly.

Proposition: In $(\mathbb{R},\mathcal{L},m)$, let $E=[a,b]$, where $a<b$. For every $\delta>0$, the open set $U_\delta=(a-\delta,b+\delta)$ contains $E$ and has $m(U_\delta\setminus E)=2\delta$. If $0<\delta<(b-a)/2$, the compact set $K_\delta=[a+\delta,b-\delta]$ is contained in $E$ and has $m(E\setminus K_\delta)=2\delta$.

Proof. The interval $U_\delta$ is open and contains every point of $E$. Its length is $(b+\delta)-(a-\delta)=(b-a)+2\delta$, whereas $m(E)=b-a$. Both measures are finite, so $m(U_\delta\setminus E)=m(U_\delta)-m(E)=2\delta$. If $0<\delta<(b-a)/2$, then $a+\delta<b-\delta$, and $K_\delta$ is a closed bounded interval, hence compact, contained in $E$. Its measure is $(b-\delta)-(a+\delta)=(b-a)-2\delta$; therefore $m(E\setminus K_\delta)=m(E)-m(K_\delta)=2\delta$. $\square$

Worked Example: Approximating [2, 5]

Take $E=[2,5]$ and $\delta=1/4$. The open outer approximation is $U=(7/4,21/4)$, since $2-1/4=7/4$ and $5+1/4=21/4$. Its measure is $21/4-7/4=14/4=7/2$. Since $m(E)=5-2=3$, the outer error is $m(U\setminus E)=7/2-3=1/2$.

The compact inner approximation is $K=[9/4,19/4]$, since $2+1/4=9/4$ and $5-1/4=19/4$. The condition $1/4<(5-2)/2=3/2$ holds, so this is a nonempty compact interval. Its measure is $19/4-9/4=10/4=5/2$, and the inner error is $m(E\setminus K)=3-5/2=1/2$. Check: the two outer pieces have lengths $2-7/4=1/4$ and $21/4-5=1/4$, totaling $1/2$. The two omitted inner pieces also have lengths $9/4-2=1/4$ and $5-19/4=1/4$, totaling $1/2$.

For any desired error bound $\varepsilon>0$, choosing $0<\delta<\varepsilon/2$ makes both errors smaller than $\varepsilon$, provided also that $\delta<(b-a)/2$ for the inner approximation. The calculations show why an interval can be approximated from both sides with errors tending to zero as the endpoint displacement tends to zero.

Open Outer Approximations to the Cantor Set

Write $C\subset[0,1]$ for the middle-third Cantor set, and $C_n$ for the set remaining after $n$ stages of its construction, with $C_0=[0,1]$. The finite-stage set $C_n$ is a union of $2^n$ closed intervals, each of length $3^{-n}$, and $C=\bigcap_{n=0}^{\infty}C_n$. We use the established facts that $C$ is compact and $m(C)=0$. The finite stages are not the Cantor set: $m(C_n)=(2/3)^n$, whereas the limiting set has measure zero.

For each $n\geq1$, expand every component interval of $C_n$ by $\delta_n=3^{-n}/4$ at both endpoints, and let $O_n$ be the union of the resulting open intervals. The expansions remain disjoint. Indeed, each gap between consecutive components of $C_n$ was removed at some stage $k\leq n$ and has length $3^{-k}\geq3^{-n}$. Expanding both adjacent intervals reduces that gap by $2\delta_n=3^{-n}/2$, leaving a gap of at least $3^{-n}/2>0$.

Proposition: For every integer $n\geq1$, $O_n$ is an open set containing $C$, and $$ m(O_n\setminus C)=m(O_n)=\frac{3}{2}\left(\frac{2}{3}\right)^n. $$ In particular, these outer errors tend to zero as $n\to\infty$.

Proof. Each component of $C_n$ is contained in its expanded open interval, so $C\subset C_n\subset O_n$. By the gap calculation above, the $2^n$ expanded intervals are pairwise disjoint. Each has length $3^{-n}+2\delta_n=3^{-n}+3^{-n}/2=(3/2)3^{-n}$. Additivity over this finite disjoint union gives $m(O_n)=2^n(3/2)3^{-n}=(3/2)(2/3)^n$. Because $C\subset O_n$, $m(C)=0$, and $m(O_n)<\infty$, finite additivity gives $m(O_n\setminus C)=m(O_n)-m(C)=m(O_n)$. Finally, $0<2/3<1$, so $(2/3)^n\to0$, and the displayed errors tend to zero. $\square$

Worked Example: The Expanded Stage-Two Cantor Set

At stage two, $$ C_2=[0,1/9]\cup[2/9,1/3]\cup[2/3,7/9]\cup[8/9,1]. $$ Here $\delta_2=3^{-2}/4=1/36$. Expanding each interval gives $$ O_2=(-1/36,5/36)\cup(7/36,13/36)\cup(23/36,29/36)\cup(31/36,37/36). $$ For instance, the second interval starts at $2/9-1/36=8/36-1/36=7/36$ and ends at $1/3+1/36=12/36+1/36=13/36$. Each expanded interval has length $6/36=1/6$, so $m(O_2)=4(1/6)=2/3$. Since $m(C)=0$, the outer error is also $2/3$. Check: the general formula gives $(3/2)(2/3)^2=(3/2)(4/9)=2/3$.

The gaps between these expanded intervals have lengths $7/36-5/36=2/36=1/18$, $23/36-13/36=10/36=5/18$, and $31/36-29/36=2/36=1/18$. They are all positive, verifying directly that the four intervals do not overlap.

The error formula also gives a concrete choice for a prescribed tolerance. For example, to make the outer error less than $1/100$, choose $n=13$. The exact error is $$ \frac{3}{2}\left(\frac{2}{3}\right)^{13} =\frac{3}{2}\cdot\frac{8192}{1594323} =\frac{4096}{531441}. $$ This is less than $1/100$, since $100\cdot4096=409600<531441$. Thus $O_{13}$ is an explicit open set containing $C$ with error below the stated tolerance.

Compact Inner Approximation of the Cantor Set

For a null set such as $C$, inner measure approximation behaves differently from outer approximation. The set $C$ is itself compact and is therefore already a compact inner approximation to $C$: its inner error is exactly zero. More generally, every compact subset $K\subset C$ has $m(C\setminus K)=0$, because $C\setminus K\subset C$ and $m(C)=0$. This is a statement about measure error; it does not assert that every such $K$ is geometrically close to all points of $C$.

Worked Example: A Finite Compact Subset of the Cantor Set

Let $K=\{0,1\}$. Both endpoints remain in every finite stage of the construction, so $K\subset C$. The set $K$ is finite and hence compact. Each singleton has Lebesgue measure zero, so $m(K)=0$; also $C\setminus K\subset C$, giving $m(C\setminus K)=0$. Thus the compact inner error is exactly zero, even though $K$ omits most of the points of $C$. Check: finite additivity and disjointness give $m(K)=m(\{0\})+m(\{1\})=0+0=0$, consistent with $m(C)=0$ and $m(C\setminus K)=0$.

One can also use the entire set $C$ as the compact inner approximation; then $K=C$ and $m(C\setminus K)=m(\varnothing)=0$. The finite example emphasizes that for null sets the measure-based inner error alone cannot express how many points are omitted. For a positive-measure interval, by contrast, moving the endpoints inward omits pieces whose total length is exactly the inner error.

A Pitfall: Confusing the Stage-Two Gaps

A tempting but incorrect calculation is to say that all gaps between the four stage-two components have length $1/9$. The two outer gaps do have length $1/9$, but the central gap is larger. The three original gaps, in order, have lengths $$ \frac{2}{9}-\frac{1}{9}=\frac{1}{9},\qquad \frac{2}{3}-\frac{1}{3}=\frac{1}{3},\qquad \frac{8}{9}-\frac{7}{9}=\frac{1}{9}. $$ The middle gap has length $1/3$, not $1/9$. For the expansion by $1/36$ at each side, the smallest gap $1/9=4/36$ loses $2/36$ to the two expansions and retains $2/36=1/18>0$. The middle gap retains $12/36-2/36=10/36=5/18>0$. Therefore none of the expanded intervals overlap. Using the smallest gap is a valid uniform check; claiming all gaps have that length is not.

A second error is to identify the measure of a finite stage with the measure of the limiting Cantor set. At stage two, for example, $m(C_2)=4(1/9)=4/9$, while $m(C)=0$. The open outer set $O_2$ has still another measure, $2/3$, because its component intervals have been expanded. Keeping $C$, $C_n$, and $O_n$ distinct is essential to each error calculation.

Check Your Understanding

Use containment and finite measure when turning differences of measures into approximation errors.

For $E=[-2,4]$ and $\delta=1/3$, write down $U_\delta$ and $K_\delta$, then calculate both errors.
Why are the expanded stage-$n$ intervals disjoint when each component is expanded by $3^{-n}/4$ at both ends?
Calculate $m(O_3)$ from the formula and verify it by multiplying the number of expanded intervals by the length of one.
Explain why every compact $K\subset C$ has $m(C\setminus K)=0$, even if $K\ne C$.
For the stage-two gaps, calculate separately the two outer gap lengths and the central gap length before expansion.