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  "body": "\n## Counterexample\n\nOn $[0, 1]$, let $x \\sim y$ mean that $x - y$ is a rational number.\nThis is an [[equivalence relation|0G8VEZ3WTK]].\nBy the [[axiom of choice|0HTGB13FE2]], there is a set $V \\subseteq [0, 1]$ that contains exactly one element of each equivalence class.\nSuch a set $V$ is called a Vitali set.\nNo Vitali set is measurable for [[Lebesgue measure|2Z2QVWZ262]] $\\lambda$.\n\n## Remarks\n\nLet $q_1, q_2, \\ldots$ list the [[countable set|01TQB88H4C]] $\\Q \\cap [-1, 1]$, and put $V_k = V + q_k = \\set{v + q_k : v \\in V}$.\nThe sets $V_k$ are pairwise disjoint: if $v + q_k = w + q_j$ with $v, w \\in V$, then $v \\sim w$, so $v = w$ and $q_k = q_j$.\nEvery $x \\in [0, 1]$ is equivalent to some $v \\in V$, and then $x - v \\in \\Q \\cap [-1, 1]$; so $[0, 1] \\subseteq \\bigcup_k V_k \\subseteq [-1, 2]$.\nIf $V$ were measurable, each $V_k$ would be measurable with $\\lambda(V_k) = \\lambda(V)$, by translation invariance.\nCountable additivity would give $1 \\le \\sum_{k} \\lambda(V) \\le 3$.\nThis is impossible: the sum is $0$ if $\\lambda(V) = 0$ and $\\infty$ otherwise.\nThe same argument shows that there is no translation-invariant, countably additive measure on all subsets of $\\R$ that gives $[0, 1]$ a finite positive value.\nSome form of choice is needed.\nSolovay showed that, if an inaccessible cardinal is consistent with ZFC, then so is ZF with dependent choice and all sets of reals Lebesgue measurable.\n\n## Explanations\n\n### L1 human\n\nGroup the numbers of $[0, 1]$ into classes, putting two numbers in the same class when they differ by a fraction.\nPick one number from each class to form the set $V$.\nCountably many shifted copies of $V$ fill $[0, 1]$ without overlap and stay inside $[-1, 2]$.\nIf $V$ had a length, these equal lengths would have to add up to a number between $1$ and $3$.\nBut countably many equal numbers add up to $0$ or to $\\infty$.\nSo $V$ cannot be given a length at all.\n",
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