Measurable Subsets of the Cantor Set

Why Completeness Matters for the Cantor Set

A set of measure zero can contain many points: the middle-third Cantor set is uncountable, even though its Lebesgue measure is zero. The key consequence of completeness is stronger than the assertion that the Cantor set itself is measurable: every subset of it is Lebesgue measurable, including subsets with no simple description. This is a statement about Lebesgue measurability; it does not say that every such subset is Borel.

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 and sigma-finite, but not finite: $m([-k,k])=2k$ for each positive integer $k$. Functions below are real-valued unless specified otherwise. The middle-third Cantor set is denoted by $C$.

Definition: A measure space $(X,\mathcal{M},\mu)$ is complete if, whenever $N\in\mathcal{M}$ satisfies $\mu(N)=0$, every subset $A\subseteq N$ belongs to $\mathcal{M}$. In that case every such $A$ also has $\mu(A)=0$, since $A$ is measurable and $0\leq\mu(A)\leq\mu(N)=0$.

The Cantor set is closed, hence Borel and Lebesgue measurable. Recall the standard computation of its measure. At stage $n$, the remaining set $C_n$ consists of $2^n$ closed intervals, each of length $3^{-n}$, so $m(C_n)=2^n3^{-n}=(2/3)^n$. Since $C=\bigcap_{n=0}^{\infty}C_n$, the sets decrease and $m(C_0)=1<\infty$. Continuity from above for a measure therefore gives $m(C)=\lim_{n\to\infty}(2/3)^n=0$.

The Completeness Argument

For clarity, the completeness property can be seen directly from the outer-measure criterion for Lebesgue measurability. Write $m^*$ for Lebesgue outer measure. A set $E\subseteq\mathbb{R}$ is Lebesgue measurable precisely when, for every $T\subseteq\mathbb{R}$, it satisfies $m^*(T)=m^*(T\cap E)+m^*(T\setminus E)$; on such sets $m(E)=m^*(E)$. We use this criterion to verify why subsets of null sets are measurable.

Proposition: In $(\mathbb{R},\mathcal{L},m)$, if $N\in\mathcal{L}$ and $m(N)=0$, then every subset $A\subseteq N$ belongs to $\mathcal{L}$ and satisfies $m(A)=0$. In particular, every subset of $C$ is Lebesgue measurable and null.

Proof. Let $N\in\mathcal{L}$ have $m(N)=0$, and let $A\subseteq N$. Monotonicity of outer measure gives $0\leq m^*(A)\leq m^*(N)=m(N)=0$, so $m^*(A)=0$. Fix any set $T\subseteq\mathbb{R}$. Since $T\cap A\subseteq A$, we have $m^*(T\cap A)=0$. Also $T\setminus A\subseteq T$, so monotonicity gives $m^*(T\setminus A)\leq m^*(T)$. Countable subadditivity, applied to $T=(T\cap A)\cup(T\setminus A)$, gives $$ m^*(T)\leq m^*(T\cap A)+m^*(T\setminus A)=m^*(T\setminus A). $$ Thus $m^*(T)=m^*(T\cap A)+m^*(T\setminus A)$. This holds for every $T$, so the outer-measure criterion shows $A\in\mathcal{L}$. Its Lebesgue measure is $m(A)=m^*(A)=0$. Taking $N=C$, which is measurable and null, proves the final assertion. $\square$

The argument does not require $A$ to be closed, countable, or described by a formula. It requires only the two relevant facts: $A$ lies inside a measurable null set, and Lebesgue measure satisfies the outer-measure criterion. The proof also handles $A=\varnothing$, $A=N$, and the case $T$ has infinite outer measure; the displayed equality follows from monotonicity and subadditivity in those cases as well.

Worked Subsets of the Cantor Set

Worked Example: A Finite Subset

Consider $F=\{0,\frac13,\frac23,1\}\subseteq\mathbb{R}$. These are endpoints retained in the middle-third construction, so $F\subseteq C$. Since $m(C)=0$, the proposition gives $F\in\mathcal{L}$ and $m(F)=0$. We can also check directly: each singleton is a Lebesgue null set, and finite additivity on the disjoint singletons gives $$ m(F)=m(\{0\})+m(\{\tfrac13\})+m(\{\tfrac23\})+m(\{1\})=0+0+0+0=0. $$ The indicator $\chi_F:\mathbb{R}\to\{0,1\}$ is measurable, and $\int_{\mathbb{R}}\chi_F\,dm=m(F)=0$. Check: the calculation gives four singleton contributions, all zero, so the result is exactly zero.

Worked Example: A Countably Infinite Subset

Let $A=\{0\}\cup\{2/3^n:n\geq1\}$. The point $2/3^n$ has a ternary expansion with $n-1$ zero digits followed by a $2$ and then only zero digits. It is therefore in $C$; also $0\in C$, so $A\subseteq C$. The points $2/3^n$ are distinct because $2/3^n>2/3^{n+1}$ for every $n\geq1$. Thus $A$ is countably infinite. By the proposition, $A\in\mathcal{L}$ and $m(A)=0$, and consequently $\int_{\mathbb{R}}\chi_A\,dm=0$.

As an independent check of the measure, enumerate $A$ as the disjoint union of $\{0\}$ and the singletons $\{2/3^n\}$. Every singleton has measure zero, so countable additivity gives $m(A)=0+\sum_{n=1}^{\infty}0=0$. This agrees with the conclusion from completeness.

Worked Example: An Uncountable Subset

Define $$ D=\left\{\sum_{n=1}^{\infty}\frac{2a_n}{3^n}:\ a_1=0,\ a_n\in\{0,1\}\text{ for }n\geq2\right\}. $$ Every point in $D$ has a ternary expansion using only $0$ and $2$, so $D\subseteq C$. The series converges because $0\leq 2a_n/3^n\leq2/3^n$ and $\sum_{n=1}^{\infty}2/3^n=1$.

To verify that $D$ is uncountable, suppose its allowed digit sequences could be listed as $a^{(1)},a^{(2)},\ldots$. Define a new sequence by $b_1=0$ and $b_{k+1}=1-a_{k+1}^{(k)}$ for every $k\geq1$. This is an allowed sequence, and it differs from the $k$-th listed sequence at position $k+1$. The digit sequences represent distinct real numbers: if two sequences first differ at position $r$, the difference contributed there has size $2/3^r$, while the largest possible opposing tail has size $$ \sum_{n=r+1}^{\infty}\frac{2}{3^n}=\frac{1}{3^r}. $$ The first difference therefore cannot be canceled by the tail. The number represented by $b$ is not in the alleged list, a contradiction. Hence $D$ is uncountable.

Despite being uncountable, $D$ is a subset of the null set $C$. Completeness gives $D\in\mathcal{L}$ and $m(D)=0$, hence $\int_{\mathbb{R}}\chi_D\,dm=0$. This conclusion depends on Lebesgue completeness, not on $D$ being countable.

Functions Supported on the Cantor Set

The same reasoning applies to functions with arbitrary real values on $C$. Here “supported on $C$” means that the function is zero at every point outside $C$; no boundedness or continuity is assumed.

Corollary: If $f:\mathbb{R}\to\mathbb{R}$ is zero on $\mathbb{R}\setminus C$, then $f$ is $\mathcal{L}$-measurable and $\int_{\mathbb{R}}|f|\,dm=0$. In particular, $f$ is Lebesgue integrable and $\int_{\mathbb{R}}f\,dm=0$.

Proof. Let $U\subseteq\mathbb{R}$ be open. If $0\notin U$, then $f^{-1}(U)\subseteq C$, because $f(x)=0$ whenever $x\notin C$. The proposition makes this preimage Lebesgue measurable. If $0\in U$, then $\mathbb{R}\setminus f^{-1}(U)=f^{-1}(U^c)\subseteq C$, so the complement of the preimage is measurable by the proposition, and therefore $f^{-1}(U)$ is measurable. Since preimages of open sets are measurable, $f$ is measurable. The nonnegative measurable function $|f|$ vanishes off $C$, so its integral is zero: it is zero almost everywhere because $m(C)=0$. Thus $\int_{\mathbb{R}}|f|\,dm=0<\infty$, which is the condition for integrability of the real-valued function $f$, and $\int_{\mathbb{R}}f\,dm=0$. $\square$

Worked Example: An Unbounded Function on a Null Set

Define $f:\mathbb{R}\to\mathbb{R}$ by $f(0)=0$, $f(2/3^n)=n$ for $n\geq1$, and $f(x)=0$ at all other points. The listed points are distinct and lie in $C$, so $f$ vanishes outside $C$. It is real-valued at every point, although its values on $A$ are unbounded. The corollary gives measurability and $\int_{\mathbb{R}}|f|\,dm=0$. Directly, each set $\{2/3^n\}$ is null, and $f$ is zero outside their countable union together with $\{0\}$, also a null set. Thus $f=0$ almost everywhere, confirming the integral calculation. This example shows why boundedness is not required when a measurable function is supported on a null set.

A Common Mistake and Its Correction

A tempting but invalid argument is: “If $A\subseteq C$, then $m(A)\leq m(C)=0$, so $m(A)=0$.” The problem is that $m(A)$ is not defined until $A$ has been shown measurable. Monotonicity of $m$ applies to measurable sets; it cannot by itself establish that an arbitrary subset is measurable.

The valid correction uses outer measure first: $m^*(A)\leq m^*(C)=0$. This shows $m^*(A)=0$, but the additional step is to verify that $A$ is Lebesgue measurable. Completeness, or the outer-measure criterion proved above, supplies that step; only then may we write $m(A)=0$.

Completeness is a property of the measure space, not a consequence of a set merely having measure zero. For contrast, on $X=C$ let $\mathcal{M}=\{\varnothing,C\}$ and define $\mu(\varnothing)=\mu(C)=0$. This is a measure space, but it is not complete: $C$ is a measurable null set, while a proper subset of $C$ is not in $\mathcal{M}$. Thus the conclusion for subsets of $C$ follows specifically from completeness of $(\mathbb{R},\mathcal{L},m)$.

Finally, “Lebesgue measurable” should not be replaced by “Borel.” The proposition establishes membership in $\mathcal{L}$; it does not establish that every subset of $C$ belongs to the Borel sigma-algebra $\mathcal{B}_{\mathbb{R}}$. A sigma-algebra is closed under countable unions and complements, but those closure rules do not assert closure under arbitrary subsets of its members. Completeness is the extra property that yields the conclusion here.

Check Your Understanding

For each question, distinguish measurability from the measure calculation.

State the definition of a complete measure space. Which part of the proof that $A\subseteq C$ is measurable uses completeness?
Suppose $N\in\mathcal{L}$ and $m(N)=0$. Explain why $m^*(A)=0$ for every $A\subseteq N$, and then give the outer-measure criterion step that proves $A\in\mathcal{L}$.
Why does the fact that $D$ in the worked example is uncountable not prevent $m(D)$ from being zero?
For a real-valued function $f$ that vanishes outside $C$, explain why $f^{-1}(U)$ is measurable both when $0\in U$ and when $0\notin U$.
What is the logical error in writing $m(A)\leq m(C)$ before establishing that $A$ is Lebesgue measurable?