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Choose a Uniformly random maximal chain in a Boolean lattice by taking a uniformly random permutation of $X$ and successively adjoining its elements. A fixed $r$-element set occurs on this maximal chain in a Boolean lattice with probability $r!(n-r)!/n!={\left(\genfrac{}{}{0ex}{}{n}{r}\right)}^{-1}$. Because an antichain meets every such chain in a partially ordered set at most once, linearity of expectation gives the LYM inequality:Here $L(\mathcal{A})$ is the Lubell mass.

For equality in the LYM inequality, the integer $|\mathcal{A}\cap \mathcal{C}|\le 1$ must equal one for every maximal chain in a Boolean lattice, since every chain has positive probability. Choose $A\in \mathcal{A}$ of rank $r$. If $0<r<n$, take $x\in A$ and $y\notin A$, and choose a permutation with its first $r$ elements equal to $A$, its $r$th element $x$, and its next element $y$. Interchanging those two adjacent elements changes precisely the rank-$r$ member of the corresponding maximal chain in a Boolean lattice, replacing $A$ by $A^{\prime }=A\setminus \{x\}\cup \{y\}$. All other members are unchanged and none belongs to the antichain. The new chain must still meet $\mathcal{A}$, so $A^{\prime }\in \mathcal{A}$.

Successive exchanges connect all $r$-element subsets: exchange an element outside a desired target for an element missing from it, decreasing the number of differences each time. Thus $\mathcal{A}$ contains the whole rank-$r$ level. Every set of another rank is comparable with some set in that level, so the antichain contains nothing else. If $r=0$ or $r=n$, its sole possible member is already the whole corresponding level. Conversely, every full level meets every maximal chain in a Boolean lattice exactly once. Hence

First we need all equality cases, rather than just the examples from part (c). The threshold estimate in part (b), with $t>0$, shows that any maximum-weight k-Sperner family has Lubell mass exactly $k$.

For each $A\in \mathcal{A}$, let $\ell (A)$ be the largest number of members in a strict chain in a partially ordered set within $\mathcal{A}$ ending at $A$. Its values lie in $\{1,\dots ,k\}$. If $A⊊B$ then $\ell (B)\ge \ell (A)+1$, so each height class ${\mathcal{A}}_j=\{A:\ell (A)=j\}$ is an antichain. Its Lubell mass is at most one by the LYM inequality, andConsequently every height class has Lubell mass one. By equality in the LYM inequality, each is a full rank level. Since the classes are disjoint, the maximizing family is a union of exactly $k$ distinct full levels.

Let $t$ again be the $k$th largest $u(i)$, let $h$ ranks have $u(i)>t$, and let $s$ ranks have $u(i)=t$. Equality in the threshold estimate requires all $h$ higher-weight levels, no lower-weight levels, and any $k-h$ of the $s$ tied levels. Thus the number of maximizing weighted k-Sperner families isConversely, every such $h,s$ is realizable. Assign $u(i)=2$ on $h$ chosen ranks, $u(i)=1$ on $s$ further ranks, and $u(i)=1/2$ on all remaining ranks, then set $w(i)=u(i)/\left(\genfrac{}{}{0ex}{}{n}{i}\right)$. These weights are positive and have exactly the stated tie pattern.

Putting $j=k-h$ gives a concise classification of the possible integers:Repeated values in this set are counted only once. The possible counts need not form an interval: for example, when $n=5,k=2$ they are $1,2,3,4,5,6,10,15$.

For positive weights $w(i)$ depending only on rank, put $u_i=\left(\genfrac{}{}{0ex}{}{n}{i}\right)w(i)$. The largest weight of a k-Sperner family in $\mathcal{P}([n])$ is the sum of the $k$ largest $u_i$. Every maximizing family is a union of $k$ full rank levels. To see the equality assertion, its Lubell mass must be $k$; decomposing it into $k$ antichains by chain height and using equality in the LYM inequality forces each part to be a full level.