[proofplan] Assume, toward a contradiction, that both $\mathrm{AD}$ and full $\mathrm{AC}$ hold. The determinacy [regularity theorem](/theorems/2750) says that every subset of $\mathbb{R}$ is Lebesgue measurable, while full choice gives a Vitali selector for the [equivalence relation](/page/Equivalence%20Relation) $x \sim y$ iff $x-y \in \mathbb{Q}$. We prove directly that such a Vitali selector cannot be Lebesgue measurable, using countable additivity and translation invariance of one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure). This produces a subset of $\mathbb{R}$ that is both Lebesgue measurable and not Lebesgue measurable, the desired contradiction. [/proofplan] [step:Use determinacy to obtain universal Lebesgue measurability] Assume that $\mathrm{AD}$ and $\mathrm{AC}$ both hold. By the determinacy regularity theorem, every subset of $\mathbb{R}$ is Lebesgue measurable (citing a result not yet in the wiki: AD implies every set of reals is Lebesgue measurable). Thus, under the present assumptions, each set $A \subset \mathbb{R}$ is measurable with respect to one-dimensional Lebesgue measure $\mathcal{L}^1$. [/step] [step:Use choice to select one representative from each rational-translation class] Define a relation $\sim$ on $[0,1] \subset \mathbb{R}$ by declaring, for $x,y \in [0,1]$, \begin{align*} x \sim y \iff x-y \in \mathbb{Q}. \end{align*} This is an equivalence relation. By $\mathrm{AC}$, there exists a set $V \subset [0,1]$ containing exactly one representative from each $\sim$-equivalence class. [guided] We need one concrete consequence of full choice. The relation $\sim$ on $[0,1]$ is defined by rational translation: \begin{align*} x \sim y \iff x-y \in \mathbb{Q}. \end{align*} It is reflexive because $x-x=0 \in \mathbb{Q}$, symmetric because $x-y \in \mathbb{Q}$ implies $y-x=-(x-y) \in \mathbb{Q}$, and transitive because $x-y \in \mathbb{Q}$ and $y-z \in \mathbb{Q}$ imply $x-z=(x-y)+(y-z) \in \mathbb{Q}$. Thus $\sim$ partitions $[0,1]$ into equivalence classes. Full $\mathrm{AC}$ lets us choose one element from each equivalence class in this partition. Let $V \subset [0,1]$ denote the resulting choice set. By construction, for every $x \in [0,1]$ there is exactly one $v \in V$ such that $x-v \in \mathbb{Q}$. [/guided] [/step] [step:Translate the selector by rational numbers and prove disjointness] Let $Q := \mathbb{Q} \cap [-1,1]$. For each $q \in Q$, define the translation map \begin{align*} T_q: \mathbb{R} &\to \mathbb{R} \end{align*} by $T_q(x)=x+q$, and define the translated selector \begin{align*} V_q := T_q(V)=\{v+q : v \in V\}. \end{align*} If $q,r \in Q$ and $q \neq r$, then $V_q \cap V_r = \varnothing$. Indeed, if $z \in V_q \cap V_r$, then there exist $v,w \in V$ such that $z=v+q=w+r$. Hence $v-w=r-q \in \mathbb{Q}$, so $v \sim w$. Since $V$ contains exactly one representative from each equivalence class, $v=w$, and then $q=r$, contradicting $q \neq r$. [/step] [step:Place the rational translates between a unit interval and a finite interval] The family $(V_q)_{q \in Q}$ covers $[0,1]$. Indeed, if $x \in [0,1]$, then by the defining property of $V$ there exists $v \in V$ with $x-v \in \mathbb{Q}$. Since $v,x \in [0,1]$, we have $x-v \in [-1,1]$, so $q:=x-v$ belongs to $Q$, and $x=v+q \in V_q$. Also, since $V \subset [0,1]$ and $Q \subset [-1,1]$, every set $V_q$ is contained in $[-1,2]$. Therefore \begin{align*} [0,1] \subset \bigcup_{q \in Q} V_q \subset [-1,2]. \end{align*} [/step] [step:Show that the selector cannot be Lebesgue measurable] Suppose, for contradiction, that $V$ is $\mathcal{L}^1$-measurable. Since each $V_q$ is a translate of $V$, translation invariance of $\mathcal{L}^1$ gives \begin{align*} \mathcal{L}^1(V_q)=\mathcal{L}^1(V) \end{align*} for every $q \in Q$. Let $m:=\mathcal{L}^1(V) \in [0,\infty]$. Because $V \subset [0,1]$, monotonicity gives $m \leq \mathcal{L}^1([0,1])=1$, so $m<\infty$. The set $Q=\mathbb{Q}\cap[-1,1]$ is countably infinite, and the sets $(V_q)_{q \in Q}$ are pairwise disjoint. By countable additivity, \begin{align*} \mathcal{L}^1\left(\bigcup_{q \in Q} V_q\right)=\sum_{q \in Q}\mathcal{L}^1(V_q)=\sum_{q \in Q}m. \end{align*} If $m=0$, then the union has measure $0$, contradicting $[0,1]\subset \bigcup_{q \in Q}V_q$ and $\mathcal{L}^1([0,1])=1$. If $m>0$, then the sum is infinite, contradicting $\bigcup_{q \in Q}V_q\subset[-1,2]$ and $\mathcal{L}^1([-1,2])=3$. Hence $V$ is not Lebesgue measurable. [guided] Now we prove the contradiction inside the measure calculation. Suppose $V$ were measurable with respect to $\mathcal{L}^1$. For each rational $q \in Q$, the set $V_q$ is obtained from $V$ by the translation map $T_q:\mathbb{R}\to\mathbb{R}$, $T_q(x)=x+q$. [Translation invariance of Lebesgue measure](/theorems/4911) gives \begin{align*} \mathcal{L}^1(V_q)=\mathcal{L}^1(V). \end{align*} Define the common value $m:=\mathcal{L}^1(V)$. Since $V \subset [0,1]$, monotonicity of Lebesgue measure gives \begin{align*} 0 \leq m \leq \mathcal{L}^1([0,1])=1. \end{align*} Thus $m$ is finite. The important point is that the rational translates are pairwise disjoint but still cover $[0,1]$. Since $Q=\mathbb{Q}\cap[-1,1]$ is countable, countable additivity applies to the pairwise disjoint measurable family $(V_q)_{q \in Q}$ and gives \begin{align*} \mathcal{L}^1\left(\bigcup_{q \in Q} V_q\right)=\sum_{q \in Q}\mathcal{L}^1(V_q)=\sum_{q \in Q}m. \end{align*} There are now only two possibilities. If $m=0$, then the right-hand side is $0$, so the union has measure $0$. But the union contains $[0,1]$, whose measure is $1$, contradicting monotonicity. If $m>0$, then the sum over the countably infinite set $Q$ diverges to $+\infty$. Hence the union has infinite measure. But the union is contained in $[-1,2]$, whose Lebesgue measure is $3$, again contradicting monotonicity. Both cases are impossible, so the assumption that $V$ is Lebesgue measurable must be false. [/guided] [/step] [step:Conclude that determinacy and full choice cannot both hold] The set $V \subset \mathbb{R}$ was constructed using full $\mathrm{AC}$, and the previous step proves that $V$ is not Lebesgue measurable. But under $\mathrm{AD}$ every subset of $\mathbb{R}$ is Lebesgue measurable. Thus the same set $V$ is both Lebesgue measurable and not Lebesgue measurable, a contradiction. Therefore $\mathrm{AD}$ and full $\mathrm{AC}$ are inconsistent over $\mathrm{ZF}$, equivalently $\mathrm{ZF}$ proves $\mathrm{AD}\implies\neg\mathrm{AC}$. [/step]