[proofplan] We construct a non-measurable set using an equivalence relation on $[0,1]$ that identifies points differing by a rational. The Axiom of Choice selects one representative from each equivalence class, forming a set $V$. We then show that $V$ cannot be Lebesgue measurable: if it were, the countable additivity and translation invariance of Lebesgue measure would force $\mathcal{L}^1([0,1])$ to be either $0$ or $\infty$, contradicting $\mathcal{L}^1([0,1]) = 1$. [/proofplan]

[step:Define an equivalence relation on $[0,1]$ by rational translation] Define a relation $\sim$ on $[0, 1]$ by \begin{align*} x \sim y \iff x - y \in \mathbb{Q}. \end{align*} This is an equivalence relation: reflexivity ($x - x = 0 \in \mathbb{Q}$), symmetry ($x - y \in \mathbb{Q} \implies y - x = -(x-y) \in \mathbb{Q}$), and transitivity ($x - y \in \mathbb{Q}$ and $y - z \in \mathbb{Q}$ imply $x - z = (x-y) + (y-z) \in \mathbb{Q}$) all hold. The equivalence classes are the cosets $[x] = (x + \mathbb{Q}) \cap [0,1]$, and they partition $[0,1]$.[/step]

[guided]The equivalence classes are the sets of the form $\{y \in [0,1] : y - x \in \mathbb{Q}\}$. Each class is countable (since $\mathbb{Q}$ is countable and each class is a subset of $x + \mathbb{Q}$), but the number of distinct classes is uncountable (since $[0,1]$ is uncountable and each class is countable, there must be uncountably many classes to cover $[0,1]$).[/guided]

[step:Apply the Axiom of Choice to select one representative per class] By the Axiom of Choice, there exists a set $V \subset [0, 1]$ containing exactly one element from each equivalence class of $\sim$. That is, for every $x \in [0,1]$, the intersection $V \cap [x]$ is a singleton. [/step]

[step:Construct a countable family of translates of $V$ that covers $[0,1]$] Let $(q_k)_{k \in \mathbb{N}}$ be an enumeration of $\mathbb{Q} \cap [-1, 1]$. For each $k$, define \begin{align*} V_k := \{v + q_k : v \in V\} \cap [0, 1] = (V + q_k) \cap [0, 1]. \end{align*} We will instead work with translates modulo $1$ to avoid boundary effects. Define the "wrap-around" translate \begin{align*} W_k := \{v + q_k \mod 1 : v \in V\}, \end{align*} where $t \mod 1$ denotes the fractional part of $t$ (the unique element of $[0,1)$ with $t - (t \mod 1) \in \mathbb{Z}$). Equivalently, we work on the circle $\mathbb{R}/\mathbb{Z}$ with addition modulo $1$. Alternatively, we use the direct approach without modular arithmetic. Enumerate $\mathbb{Q} \cap (-1, 1)$ as $(q_k)_{k \in \mathbb{N}}$ and define $V_k := V + q_k = \{v + q_k : v \in V\}$. We establish two properties: **The $V_k$ are pairwise disjoint.** Suppose $v + q_j = w + q_k$ for $v, w \in V$ and $j \neq k$. Then $v - w = q_k - q_j \in \mathbb{Q}$, so $v \sim w$. Since $V$ contains exactly one representative per class, $v = w$, hence $q_j = q_k$, contradicting $j \neq k$ (assuming the enumeration is injective). Therefore $V_j \cap V_k = \varnothing$ for $j \neq k$. **The union covers $[0,1]$.** Let $x \in [0,1]$. Since $V$ contains a representative of $[x]$, there exists $v \in V$ with $x - v \in \mathbb{Q}$. Since $x, v \in [0,1]$, we have $x - v \in [-1, 1] \cap \mathbb{Q}$, so $x - v = q_k$ for some $k$. Then $x = v + q_k \in V_k$. Therefore $[0,1] \subset \bigcup_{k=1}^\infty V_k$. **The union is contained in $[-1, 2]$.** Since $v \in [0,1]$ and $q_k \in (-1,1)$, we have $v + q_k \in (-1, 2)$, so $\bigcup_{k=1}^\infty V_k \subset (-1, 2)$. [/step]

[step:Derive a contradiction from Lebesgue measurability of $V$]Suppose for contradiction that $V$ is Lebesgue measurable. Each $V_k = V + q_k$ is a translate of $V$, hence also Lebesgue measurable with $\mathcal{L}^1(V_k) = \mathcal{L}^1(V)$ (by translation invariance of Lebesgue measure). Since the $V_k$ are pairwise disjoint and $[0,1] \subset \bigcup_{k=1}^\infty V_k \subset (-1, 2)$, countable additivity gives \begin{align*} 1 = \mathcal{L}^1([0,1]) \le \mathcal{L}^1\!\left(\bigcup_{k=1}^\infty V_k\right) = \sum_{k=1}^\infty \mathcal{L}^1(V_k) = \sum_{k=1}^\infty \mathcal{L}^1(V) \le \mathcal{L}^1((-1, 2)) = 3. \end{align*} The middle sum $\sum_{k=1}^\infty \mathcal{L}^1(V)$ is a series with all terms equal to $\mathcal{L}^1(V) \ge 0$: - If $\mathcal{L}^1(V) = 0$, the sum equals $0$, contradicting $\sum_{k=1}^\infty \mathcal{L}^1(V) \ge 1$. - If $\mathcal{L}^1(V) > 0$, the sum equals $+\infty$, contradicting $\sum_{k=1}^\infty \mathcal{L}^1(V) \le 3$. Both cases yield a contradiction. Therefore $V$ is not Lebesgue measurable.[/step]

[guided]The argument rests on three properties of Lebesgue measure: 1. **Translation invariance:** $\mathcal{L}^1(A + q) = \mathcal{L}^1(A)$ for every measurable $A$ and every $q \in \mathbb{R}$. 2. **Countable additivity:** $\mathcal{L}^1(\bigsqcup_{k=1}^\infty A_k) = \sum_{k=1}^\infty \mathcal{L}^1(A_k)$ for pairwise disjoint measurable sets. 3. **Normalisation:** $\mathcal{L}^1([0,1]) = 1$. If $V$ were measurable, all three would apply simultaneously to the family $(V_k)_{k \in \mathbb{N}}$. The pairwise disjointness (verified above) and the inclusion $[0,1] \subset \bigcup V_k \subset (-1,2)$ yield the chain of inequalities \begin{align*} 1 \le \sum_{k=1}^\infty \mathcal{L}^1(V) \le 3. \end{align*} But the left side is a constant series. The only constant $c \ge 0$ for which $\sum_{k=1}^\infty c$ lies in $[1,3]$ would require $c > 0$ (to get at least $1$) and $c = 0$ (to avoid $+\infty$). No such $c$ exists. This contradiction shows that $V$ cannot be Lebesgue measurable. The role of the Axiom of Choice is precisely in the selection of $V$. Without it, we cannot form a set containing exactly one element from each equivalence class. In ZF alone (without choice), it is consistent that every subset of $\mathbb{R}$ is Lebesgue measurable (Solovay's model, assuming the consistency of an inaccessible cardinal).[/guided]