Lebesgue Measurable Sets and Non-Measurable Sets

Theorem 4.1. Every Borel set is Lebesgue measurable.

Theorem 4.2. If $A$ is Lebesgue measurable, then for every $\varepsilon > 0$ there exists an open set $U \supseteq A$ with $m^*(U \setminus A) < \varepsilon$ (outer regularity).

Theorem 4.3. If $A$ is Lebesgue measurable, then for every $\varepsilon > 0$ there exists a closed set $F \subseteq A$ with $m^*(A \setminus F) < \varepsilon$ (inner regularity).

Corollary 4.4 (Approximation by Intervals). If $A$ is Lebesgue measurable, then for every $\varepsilon > 0$ there exists a finite union $U$ of disjoint intervals such that $m(A \triangle U) < \varepsilon$.

Proposition 4.5 (Translation and Scaling). If $A \subseteq \mathbb{R}^n$ is Lebesgue measurable, then for any $x \in \mathbb{R}^n$ and $t > 0$, the translated set $A + x$ and scaled set $tA$ are Lebesgue measurable with $m(A + x) = m(A)$ and $m(tA) = t^n m(A)$.

Proposition 4.6 (Completeness). Every subset of a Lebesgue null set is Lebesgue measurable (and has measure zero). This property makes Lebesgue measure complete.

Theorem 4.7. Assuming the Axiom of Choice, there exists a subset $V \subseteq [0, 1]$ that is not Lebesgue measurable.

Proof sketch. Define an equivalence relation on $[0, 1]$: $x \sim y$ if $x - y \in \mathbb{Q}$. Each equivalence class is $E_x = x + \mathbb{Q}$. By the Axiom of Choice, select one representative from each equivalence class to form a set $V$ (a Vitali set).

Note that for any two distinct rationals $r, s \in [-1, 1]$, the sets $V + r$ and $V + s$ are disjoint (otherwise $(V + r) \cap (V + s) \neq \varnothing$ implies $v_1 + r = v_2 + s$, so $v_1 - v_2 \in \mathbb{Q}$, contradicting distinct representatives).

Now $\bigcup_{q \in \mathbb{Q} \cap [-1, 1]} (V + q) \subseteq [-1, 2]$. If $V$ were measurable, each $V + q$ would be measurable with $m(V + q) = m(V)$ by translation invariance. Then

$m\left(\bigcup_{q}(V+q)\right) = \sum_{q \in \mathbb{Q} \cap [-1,1]} m(V)$

This is $0$ if $m(V) = 0$, or $\infty$ if $m(V) > 0$. But the union is contained in $[-1, 2]$ which has measure $3$. Contradiction. $\blacksquare$

Theorem 4.8 (Carathéodory). A set $A \subseteq \mathbb{R}^n$ is Lebesgue measurable if and only if for every $E \subseteq \mathbb{R}^n$:

$m^*(E) = m^*(E \cap A) + m^*(E \setminus A)$

This criterion provides a definition of measurability that works in any metric space with any outer measure.

Example 4.1. Every interval $I \subseteq \mathbb{R}$ satisfies Carathéodory’s criterion and is therefore measurable. This can be verified by checking the condition for arbitrary $E$.

Example 4.2. The Vitali set $V$ fails Carathéodory’s criterion: there exists a test set $E$ (specifically $E = \bigcup_{q}(V+q) \cap [-1,2]$) such that $m^*(E) < m^*(E \cap V) + m^*(E \setminus V)$.

While the Vitali set is the standard example, other constructions highlight different aspects of non-measurability.

Example 4.3 (Bernstein Set). A Bernstein set $B \subseteq \mathbb{R}$ is a set such that both $B$ and its complement intersect every uncountable closed set. Bernstein sets exist assuming the Axiom of Choice. They are not Lebesgue measurable and, in fact, have inner measure zero and outer measure infinite.

Example 4.4 (Hamel Basis). A Hamel basis of $\mathbb{R}$ over $\mathbb{Q}$ gives another construction. If $H$ is a Hamel basis, then many linear combinations of $H$ yield non-measurable sets. In particular, the set of numbers whose first basis coefficient is positive is not measurable.

Remark. The existence of non-measurable sets is inextricably tied to the Axiom of Choice. Solovay (1970) proved that there exists a model of ZF (without Choice) in which every subset of $\mathbb{R}$ is Lebesgue measurable.

Theorem 4.9 (Decomposition). A set $A \subseteq \mathbb{R}^n$ is Lebesgue measurable if and only if it can be written as $A = B \cup N$ where $B$ is a Borel set and $N$ is a Lebesgue null set.

Equivalently, $A = F \setminus N$ where $F$ is an $F_\sigma$ set (countable union of closed sets) and $N$ is null. This is the Borel approximability property.

Proposition 4.10 (Translation Invariance). The Lebesgue measure is translation-invariant: $m(A + x) = m(A)$ for all measurable $A$ and $x \in \mathbb{R}^n$.

Proposition 4.11 (Continuity from Above/Below). If $\{A_k\}$ is a sequence of measurable sets:

• If $A_k \uparrow A$ (i.e., $A_1 \subseteq A_2 \subseteq \cdots$ and $\bigcup A_k = A$), then $m(A) = \lim_{k\to\infty} m(A_k)$.
• If $A_k \downarrow A$ (i.e., $A_1 \supseteq A_2 \supseteq \cdots$ and $\bigcap A_k = A$) and $m(A_1) < \infty$, then $m(A) = \lim_{k\to\infty} m(A_k)$.

Problem 1. Show that the set $\mathbb{Q} \subseteq \mathbb{R}$ has Lebesgue measure zero.

Solution. $\mathbb{Q}$ is countable: $\mathbb{Q} = \{q_1, q_2, \ldots\}$. For each $\varepsilon > 0$, cover $q_k$ by the interval $(q_k - \varepsilon/2^{k+1}, q_k + \varepsilon/2^{k+1})$. The total length is $\sum_{k=1}^\infty \varepsilon/2^k = \varepsilon$. Hence $m^*(\mathbb{Q}) \leq \varepsilon$ for all $\varepsilon > 0$, so $m^*(\mathbb{Q}) = 0$. $\blacksquare$

Problem 2. Show that the Cantor set $C$ has Lebesgue measure zero.

Solution. The Cantor set $C = \bigcap_{n=0}^\infty C_n$ where $C_0 = [0,1]$ and $C_{n+1}$ is obtained by removing the open middle third of each interval in $C_n$. At stage $n$, $C_n$ consists of $2^n$ intervals each of length $3^{-n}$, so $m(C_n) = (2/3)^n$. Since $C \subseteq C_n$ for all $n$, $m(C) \leq \lim_{n\to\infty} (2/3)^n = 0$. $\blacksquare$

1. Prove that if $A$ and $B$ are measurable then $A \setminus B$ and $A \triangle B$ are measurable.
2. Show that the outer measure of a Vitali set satisfies $m^*(V) > 0$.
3. Prove that every Lebesgue measurable set is the union of an $F_\sigma$ set and a null set.
4. Show that if $m^*(A) = 0$ then $A$ is measurable.
5. Construct a non-measurable set using a Hamel basis approach.