Lebesgue Measurable Sets and Non-Measurable Sets

Theorem 4.1. Every Borel set is Lebesgue measurable.

Proof. The key input is that Lebesgue outer measure is a metric outer measure: if $d(A, B) > 0$ then $m^*(A \cup B) = m^*(A) + m^*(B)$. Subadditivity gives $(\leq)$. For $(\geq)$: given $\varepsilon > 0$, cover $A \cup B$ by cubes of diameter less than $d(A, B)/2$; replacing each covering cube by finitely many subcubes of small diameter covering it changes the total volume by an arbitrarily small amount; so that no cube meets both $A$ and $B$. If $\mathcal{C}_A$ is the subcollection meeting $A$ and $\mathcal{C}_B$ the one meeting $B$,

and $\varepsilon \to 0$ gives equality.

From here, the standard argument makes every closed set Carathéodory-measurable. Let $A$ be closed and let $E$ be any test set; as noted after Theorem 4.8, subadditivity makes the interesting direction $m^*(E) \geq m^*(E \cap A) + m^*(E \setminus A)$, and it suffices to test $E$ with $m^*(E) < \infty$. For $k \geq 1$ let $E_k = \{x \in E \setminus A : d(x, A) \geq 1/k\}$; then $A$ and $E_k$ are separated by $1/k$, so the metric additivity gives

One shows $m^*(E_k) \uparrow m^*(E \setminus A)$: the sets increase to $E \setminus A$, and the tail is controlled by splitting the annuli $R_j = \{x : 1/(j+1) \leq d(x, A) < 1/j\}$ into even- and odd-indexed subfamilies; each subfamily is pairwise separated, so metric additivity makes the sum of its annulus measures at most $m^*(E) < \infty$, forcing both sub-series (hence their sum) to have vanishing tails. Letting $k \to \infty$: $m^*(E) \geq m^*(E \cap A) + m^*(E \setminus A)$. So $A$ is measurable.

Thus all closed sets are measurable; by the σ-algebra property of the measurable sets (Theorem 3.4 of the previous page) their complements; the open sets; are too, and the σ-algebra generated by the open sets; the Borel σ-algebra; is contained in the Lebesgue measurable sets. $\blacksquare$

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).

Proof. First suppose $m^*(A) < \infty$. By definition of outer measure there is an open cover $U \supseteq A$ by cubes with $\sum |C_j| \leq m^*(A) + \varepsilon/2$, so $m^*(U) \leq m^*(A) + \varepsilon/2$. Since $A$ is measurable, Carathéodory’s criterion applied to the test set $U$ splits it:

Subtracting (both sides finite): $m^*(U \setminus A) < \varepsilon$.

For general $A$, write $\mathbb{R}^n = \bigcup_k Q_k$ (cubes of side 1), apply the finite case to $A \cap Q_k$ to get $U_k \supseteq A \cap Q_k$ with $m^*(U_k \setminus (A \cap Q_k)) < \varepsilon/2^k$, and set $U = \bigcup_k U_k$: open, contains $A$, and

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).

Proof. Apply Theorem 4.2 to the measurable complement $A^c$: there is open $U \supseteq A^c$ with $m^*(U \setminus A^c) < \varepsilon$. Set $F = U^c$: closed, and $F \subseteq A$ since $U \supseteq A^c$. Moreover

so $m^*(A \setminus F) < \varepsilon$. $\blacksquare$

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 $m(A \triangle U) < \varepsilon$.

Proof. By Theorem 4.2 take open $V \supseteq A$ with $m(V \setminus A) < \varepsilon/2$. Every open subset of $\mathbb{R}$ is a countable union of disjoint open intervals, $V = \bigsqcup_j I_j$ with $m(V) = \sum_j |I_j|$. Choose $K$ with $\sum_{j > K} |I_j| < \varepsilon/2$ and set $U = \bigcup_{j \leq K} I_j$. Then

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 $m(tA) = t^n m(A)$.

Proof. Outer measure. Translating a cover of $A$ gives a cover of $A + x$ with identical cube volumes, and vice versa: $m^*(A + x) = m^*(A)$. A scaled cover of $A$ covers $tA$ with volumes multiplied by $t^n$: $m^*(tA) = t^n m^*(A)$.

Measurability. Let $\tau(z) = z + x$. For any test set $E$, Carathéodory’s criterion for $A + x$ at $E$ is the criterion for $A$ at $\tau^{-1}(E)$: intersecting with $A + x$ corresponds to translating the intersection with $A$, and outer measure is unchanged by translation. So $A + x$ satisfies the criterion whenever $A$ does; the scaling map is identical with volumes rescaled. $\blacksquare$

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

Proof. Let $N$ be null ($m^*(N) = 0$) and $B \subseteq N$. Then $m^*(B) \leq m^*(N) = 0$ by monotonicity. To see $B$ satisfies Carathéodory’s criterion, let $E$ be any test set. Monotonicity gives $m^*(E \cap B) \leq m^*(B) = 0$, and monotonicity again gives $m^*(E \setminus B) \leq m^*(E)$ (as $E \setminus B \subseteq E$). So

and the reverse inequality is subadditivity. $\blacksquare$

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 $v \in V \subseteq [0, 1]$ and $q \in [-1, 1]$ then $v + q \in [-1, 2]$. Conversely, the union covers $[0, 1]$: given $x \in [0, 1]$, let $v$ be the chosen representative of the class of $x$; then $x - v \in \mathbb{Q}$ and $|x - v| \leq 1$ (both points lie in $[0, 1]$), so $x \in V + q$ for $q = x - v \in \mathbb{Q} \cap [-1, 1]$.

If $V$ were measurable, each $V + q$ would be measurable with $m(V + q) = m(V)$ by translation invariance, and countable additivity would give

The sum is $0$ if $m(V) = 0$, and $\infty$ if $m(V) > 0$. But the union satisfies $[0, 1] \subseteq \bigcup_q (V + q) \subseteq [-1, 2]$, so its measure is squeezed between $1$ and $3$: neither $0$ nor $\infty$. Contradiction. $\blacksquare$

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

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. Let $E = \bigcup_{q \in \mathbb{Q} \cap [-1, 1]} (V + q)$, which satisfies $[0,1] \subseteq E \subseteq [-1,2]$, so $1 \leq m^*(E) \leq 3$. If $V$ satisfied the criterion, then every translate $V + q$ would satisfy it (translation preserves the criterion, Proposition 4.5), the translates would be pairwise disjoint measurable sets, and $m^*$ restricted to measurable sets is countably additive over them; giving

contradicting $1 \leq m^*(E) \leq 3$. The inequality in Carathéodory’s criterion therefore fails strictly for at least one test set. $\blacksquare$

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

Proof. ($\Leftarrow$) Borel sets are measurable (Theorem 4.1), subsets of null sets are measurable (Proposition 4.6), and the measurable sets form a σ-algebra; so $B \cup N$ is measurable.

($\Rightarrow$) Suppose $A$ is measurable, so $A^c$ is too. By Theorem 4.2, for each $k$ choose open $U_k \supseteq A^c$ with $m(U_k \setminus A^c) < 1/k$, and set $G = \bigcap_k U_k$; a $G_\delta$ set, hence Borel, with $A^c \subseteq G$ and, by subadditivity,

so $m(G \setminus A^c) = 0$. Now $F = G^c$ is an $F_\sigma$ set with $F \subseteq A$, and

has measure zero. Therefore $A = F \cup (A \setminus F)$ expresses $A$ as an $F_\sigma$ set - a fortiori a Borel set; unioned with a null set. (Applying the same construction to $A$ itself gives the dual form $A = B \setminus N$ with $B$ a $G_\delta$ set and $N$ null.) $\blacksquare$

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

This is the translation half of Proposition 4.5, restated for emphasis: it is the axiom that makes Lebesgue measure the notion of length, area, and volume; and it is exactly what forces the Vitali set to exist, since invariance plus countable additivity cannot both hold on all of $\mathcal{P}(\mathbb{R})$.

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 $m(A) = \lim_{k\to\infty} m(A_k)$.

Proof. From below. The difference sets $D_k = A_{k+1} \setminus A_k$ are disjoint and

so countable additivity gives

where the telescoping uses $m(A_{K+1}) = m(A_1) + \sum_{k=1}^{K} m(A_{k+1} \setminus A_k)$.

From above. Apply the increasing result to $B_k = A_1 \setminus A_k$, which increases to $B = A_1 \setminus A$:

Subtracting $m(A_1)$ from both sides is legitimate because $m(A_1) < \infty$; this is precisely where the finiteness hypothesis is used. $\blacksquare$

The finiteness hypothesis cannot be dropped: $A_k = (k, \infty)$ decreases to $\bigcap_k A_k = \emptyset$, yet $m(A_k) = \infty$ for every $k$, so $\lim_k m(A_k) = \infty \neq 0 = m(\emptyset)$.

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.

• Lebesgue Outer Measure and Caratheodory Extension: Provides the construction of Lebesgue measure via outer measures and the Caratheodory criterion used throughout this file.

• Measurable Functions: Defines functions that are compatible with the measurable set structure, forming the basis for Lebesgue integration.

• Radon-Nikodym Derivative and Lebesgue Decomposition: Uses the decomposition of measures into absolutely continuous and singular parts, concepts rooted in the measurability theory here.

• Quantum Mechanics

• Graph Theory

• Classical Mechanics

• Electromagnetism

• Statistical Learning

• Statistical Mechanics

Lebesgue measurability determines which sets can be assigned a consistent “size.” The outer measure covers any set with intervals from above, but only measurable sets satisfy the property that their size equals the sizes of their pieces added together. The Vitali construction shows that not all sets are measurable: using the axiom of Choice, you can build a set that is so irregular that no consistent measure can be assigned. The Cantor set shows the opposite extreme, an uncountable set with measure zero, demonstrating that measure and cardinality are unrelated. Lebesgue measurability is the sweet spot: large enough to include all Borel sets and null sets, but small enough to avoid pathological constructions.

Mistake 1: Assuming all subsets of $\mathbb{R}$ are Lebesgue measurable The existence of non-measurable sets (like the Vitali set) depends on the Axiom of Choice. In ZF without Choice, it is consistent that all subsets of $\mathbb{R}$ are measurable. Never assume a set is measurable without verification, especially when constructing sets using Choice-based arguments.

Mistake 2: Confusing Lebesgue measurability with Borel measurability Every Borel set is Lebesgue measurable, but not vice versa. The Cantor set is Borel (closed) and has measure zero, but adding any subset of the Cantor set to a Borel set produces a Lebesgue measurable set that may not be Borel. The Lebesgue $\sigma$-algebra is strictly larger than the Borel $\sigma$-algebra.

Mistake 3: Assuming outer measure is additive Outer measure $m^*$ is subadditive ($m^*(\bigcup A_n) \leq \sum m^*(A_n)$) but not additive. For disjoint non-measurable sets, outer measure can fail to be additive. Additivity holds only for measurable sets. The Vitali construction exploits this failure: the outer measure of the union of translates of $V$ is bounded, but the sum of outer measures is not.

• $L^p$ Spaces
• Fubini and Tonelli Theorems
• Lebesgue Integration
• Lebesgue Outer Measure and Caratheodory Extension