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Identify the vertices of the Boolean hypercube $Q_n$ with subsets of $[n]$, adjacent when their symmetric difference has size one. Write $N(\mathcal{A})$ for the closed graph neighbourhood, including $\mathcal{A}$ itself. In simplicial order on the discrete cube, smaller sets come first, and equal-sized sets are ordered by lexicographic order: the least element of the symmetric difference belongs to the earlier member. Let $I_m$ denote the first $m$ vertices. The vertex-isoperimetric inequality in the discrete cube statesSubtracting $|\mathcal{A}|$ gives the equivalent assertion for the external vertex boundary.

We first check that $N(I_m)$ is also an initial segment in simplicial order on the discrete cube. Apart from the empty and full cases, $I_m$ consists of complete lower layers and a lexicographic initial segment $\mathcal{F}$ in its top rank $r$. Its closed graph neighbourhood contains all ranks at most $r$ and the upper shadow $\nabla \mathcal{F}$ in rank $r+1$. This upper shadow is lexicographic initial: an $(r+1)$-subset $C$ belongs to it precisely when its first $r$-subset in lexicographic order, obtained by deleting its largest element, belongs to $\mathcal{F}$. Taking the first $r$ elements preserves weak lexicographic order, proving the assertion. This also handles $r=0$, where $N(\{\varnothing \})$ consists of ranks zero and one.

We prove the vertex-isoperimetric inequality in the discrete cube by mathematical induction on $n$, with $n=0,1$ immediate. Fix a coordinate $i$ and write ${\mathcal{A}}_0,{\mathcal{A}}_1\subseteq Q_{n-1}$ for the sections without and with $i$. The two sections of its closed graph neighbourhood areA simplicial section compression replaces ${\mathcal{A}}_j$ by the initial segment $I_{a_j}$ of size $a_j=|{\mathcal{A}}_j|$ in the other coordinates. By the preceding observation, both $N(I_{a_j})$ and $I_{a_{1-j}}$ are initial segments, so their union has size $\max \{|N(I_{a_j})|,a_{1-j}\}$. The mathematical induction hypothesis givesAdding the two section bounds proves that simplicial section compression cannot increase the closed graph neighbourhood.

Apply changing simplicial section compressions until none remains. The sum of the global simplicial order on the discrete cube positions strictly decreases at each step: within a section the order is exactly the restriction of the global order. Thus the procedure terminates at a set family $\mathcal{B}$ every coordinate section of which is initial. We now classify the possible terminal families for simplicial section compression, rather than assuming this terminal set family must be globally initial.

If $X<Y$, $X\notin \mathcal{B}$ and $Y\in \mathcal{B}$, then $X,Y$ cannot agree in any coordinate, since that coordinate section would have an omitted earlier member and an included later one. Therefore $Y=X^c$. Let $X$ be the first omitted vertex and $Y$ the last included vertex. If $\mathcal{B}$ is noninitial, $X<Y$ and $Y=X^c$. Any vertex $Z$ strictly between them would have to be included, since an omitted $Z$ paired with $Y$ would have to equal $Y^c=X$. But it would also have to be omitted, since an included $Z$ paired with $X$ would have to equal $X^c=Y$. Thus there is no such $Z$: $X,Y$ are consecutive complementary vertices. Everything before $X$ is included and everything after $Y$ omitted. Hence $\mathcal{B}$ is obtained from the initial segment of its size by replacing $X$ with its immediate successor $X^c$.

There are only two forms of this exception. If $n=2r+1$, consecutive complementary vertices must straddle ranks $r,r+1$; otherwise an intermediate rank or a further vertex in one of their ranks separates them. They are consequentlywith $X$ the last $r$-subset and $Y$ the first $(r+1)$-subset. The corresponding initial segment is all ranks at most $r$. For $r\ge 1$, the omitted $X$ remains in $N(\mathcal{B})$ through a lower neighbour, and every $(r+1)$-subset has at least two $r$-subsets, only one of which was omitted. Thus $N(\mathcal{B})$ contains every rank at most $r+1$, namely $N(I_{|\mathcal{B}|})$. For $r=0$ both singleton closed graph neighbourhoods are the whole $Q_1$.

If $n=2r$, the two complementary vertices must both have rank $r$. The complement map reverses the lexicographic order in that rank, so a consecutive complementary pair must be its central pair. Exactly half the $r$-subsets contain $1$, and all of those precede those missing $1$. Therefore the pair isHere $I_{|\mathcal{B}|}$ comprises all ranks below $r$ and all $r$-subsets containing $1$. For $r\ge 2$, $N(\mathcal{B})$ contains every rank at most $r$, because all lower ranks were retained. Every $(r+1)$-subset containing $1$ has $r\ge 2$ different $r$-subsets containing $1$, so deletion of the one member $X$ removes no vertex of the upper shadow. Consequently $N(\mathcal{B})\supseteq N(I_{|\mathcal{B}|})$. For $r=1$, the two closed graph neighbourhoods both have size three in $Q_2$. These cases exhaust the exceptions. Since the successive simplicial section compressions never increased the closed graph neighbourhood, this proves the vertex-isoperimetric inequality in the discrete cube.

The same argument gives all integer radii. Write $N^t$ for $t$ repetitions of the closed graph neighbourhood. Its application to an initial segment is again initial; the sizes of these initial segments are monotone in the original size. Inducting on $t$ therefore gives

For the concentration question, use uniform probability measure on a finite connected graph and normalize its graph distance by its graph diameter. A sequence is a Lévy family of graphs if, for every fixed $\epsilon >0$ and $\delta >0$, the radius-$\epsilon$ neighbourhood of every set of measure at least $\delta$ has measure tending to one, uniformly over those sets. One may equivalently require this only for sets of measure at least one half. To see the reverse implication, first the radius-$\epsilon /2$ neighbourhood of every positive-density set must eventually have measure greater than one half. Otherwise its complement has measure at least one half, while the radius-$\epsilon /4$ neighbourhood of that complement misses the original positive-density set, contradicting the half-measure condition. Apply the half-measure condition once more to the radius-$\epsilon /2$ neighbourhood, enlarging it by another $\epsilon /2$, to obtain the full claim. The metric normalization is essential here.

For $Q_n$, the graph diameter is $n$. A uniformly chosen vertex has size $Z_n$ with binomial distribution $\mathrm{Bin}(n,1/2)$, expectation $n/2$ and variance $n/4$. Fix $\delta >0$ and choose $C>0$ so that $1/(4C^2)<\delta$. By Chebyshev's inequality,For sufficiently large $n$, put $k_n=\lfloor n/2-C\sqrt{n}\rfloor \ge 0$. Every initial segment of measure at least $\delta$ contains the complete ranks through $k_n$. Its radius-$t$ graph neighbourhood therefore contains every rank at most $k_n+t$. Taking $t=\lfloor \epsilon n\rfloor$, the all-radius vertex-isoperimetric inequality in the discrete cube gives, uniformly for $|\mathcal{A}|\ge \delta 2^n$,The last Chebyshev's inequality is used only when its denominator has positive square root. These are precisely the radius-$\epsilon$ neighbourhoods in the normalized graph distance. The discrete cubes form a Lévy family. In fact, taking $t=\lfloor n^{3/4}\rfloor$ in the same bound makes the omitted measure $O(n^{-1/2})$ for each fixed $\delta$: a radius that is $o(n)$ already suffices.

For fixed $d\ge 1$, the nearest-neighbour grid $[n{]}^d$ has graph distance ${\sum }_i|x_i-y_i|$ and graph diameter $d(n-1)$. Take the slab ${\mathcal{A}}_n=\{x:x_1\le \lceil n/2\rceil \}$, whose uniform measure is at least one half. Its integer radius-$t$ graph neighbourhood is exactly $\{x:x_1\le \min (n,\lceil n/2\rceil +t)\}$: changing the first coordinate alone attains the minimum distance to the slab. For $0<\epsilon <1/(2d)$ and $t=\lfloor \epsilon d(n-1)\rfloor$, the omitted proportion tends toThus there is failure of concentration in fixed-dimensional grids, and the fixed-dimensional grids do not form a Lévy family with normalized graph distance.