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  "body": "\n## Proof\n\nWrite $\\lambda$ for Lebesgue measure on $\\R^n$, so that the volume of $S$ is $\\lambda(S)$. The hypothesis speaks of this volume, so $S$ is [[Lebesgue measurable|2Z2QVWZ262]].\n\n1. Put $T = \\frac{1}{2} S = \\set{\\frac{1}{2} s : s \\in S}$.\n2. The map $h(x) = \\frac{1}{2} x$ is a bijection of $\\R^n$. It sends each box to a box whose volume is $2^{-n}$ times as large, and $h^{-1}$ does the reverse.\n3. So $h$ matches the countable covers of a set $A$ by boxes with those of $h(A)$. It multiplies the total volume by $2^{-n}$. Hence $\\lambda^*(h(A)) = 2^{-n} \\lambda^*(A)$ for every $A \\subseteq \\R^n$, where $\\lambda^*$ is Lebesgue outer measure.\n4. The set $S$ satisfies $\\lambda^*(A) = \\lambda^*(A \\cap S) + \\lambda^*(A \\setminus S)$ for all $A$. Apply this to $h^{-1}(A)$ and use step 3. We get the same condition for $T$, so $T$ is measurable and $\\lambda(T) = 2^{-n} \\lambda(S)$.\n5. By hypothesis $\\lambda(S) \u003e 2^n d(L)$, so $\\lambda(T) \u003e d(L)$. Here $d(L)$ is the covolume of the [[lattice|Z4FSES3CZ0]] $L$.\n6. By [[Blichfeldt's theorem|A3FTHW4NES]], there are distinct points $x, y \\in T$ with $x - y \\in L$.\n7. Write $x = \\frac{1}{2} s$ and $y = \\frac{1}{2} s'$ with $s, s' \\in S$.\n8. Since $S$ is symmetric about the origin, $-s' \\in S$.\n9. Since $S$ is [[convex|0K06X9QXDX]], the midpoint $\\frac{1}{2} s + \\frac{1}{2}(-s')$ of $s$ and $-s'$ lies in $S$.\n10. This midpoint is $x - y$. So $x - y$ is a point of $L$ in $S$, and $x - y \\ne 0$ because $x \\ne y$.\n\n## Idea\n\nShrink $S$ by the factor $2$; its volume becomes larger than $d(L)$.\nBlichfeldt's theorem then gives two points of the shrunken set that differ by a lattice vector.\nSymmetry and convexity show that this difference is the midpoint of two points of $S$, so it lies in $S$.\n",
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