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Let $A_1,\dots ,A_m$ be random subsets of a common ground set. If every pairwise union $A_i\cup A_j$ takes values in a fixed family $\mathcal{A}$ of sets of cardinality at most $k$, then Shearer's inequality and the fact that a pair of bits with prescribed OR has at most three possibilities give

For a measure-preserving action of $\mathbb{Z}$ and a finite partition $\xi$,The reverse inequality to the interval formula follows from Shearer's inequality applied to translates of $F$ along long intervals; the uncovered boundary has sublinear size.

Use discrete Shannon entropy with base-two logarithms. The standard facts needed are the chain rule for information entropy, conditioning reduces entropy, nonnegativity of conditional entropy, and the fact that a uniform distribution on $M$ outcomes has Shannon entropy ${\log }_{2}M$.

Order the coordinates by their indices and put $h_j=H(X_j\mid X_1,\dots ,X_{j-1})\ge 0$. Applying the chain rule within a coordinate subset $A$ and then adding the missing earlier conditioning coordinates givesOn summing over the family, coordinate $j$ occurs at least $k$ times. Since each $h_j$ is nonnegative,This proves Shearer's inequality without any independence assumption on the coordinates. For countable alphabets, apply the finite-valued assertion to refining finite coordinate partitions; the standard monotone approximation of Shannon entropy gives the same result with extended values. Differential entropy is not the Shannon entropy intended here.

For the graph application, let its bipartition be $U,V$, with degrees $a$ on $U$ and $b$ on $V$, where $a,b\ge 1$. Choose an independent set uniformly from all $i(G)$ such sets, and let $X$ be its vertex-indicator vector. Then $H(X)={\log }_{2}i(G)$. For $v\in V$ put $Y_v=X_{N(v)}$ and $q_v=\mathrm{Pr}(Y_v=0)$.

Each vertex of $U$ belongs to exactly $a$ of these neighbourhoods. Shearer's inequality therefore gives $H(X_U)\le a^{-1}{\sum }_{v\in V}H(Y_v)$. Conditional on $X_U$, a vertex of $V$ is forced absent if any neighbor was selected, and otherwise is a free fair binary choice. There are no edges within $V$, and the conditional independent-set distribution is uniform on all these free choices. Henceand

There are $2^b$ possible neighborhood patterns. Give the zero pattern weight $w_0=2^a$ and every other pattern weight one. For its distribution $(p_y)$, concavity of the logarithm, or Jensen's inequality, givesThis establishes the local weighted-entropy estimate rather than just counting neighborhood patterns.

Edge counting gives $a|U|=b|V|$; together with $|U|+|V|=n$, it yields $|V|/a=n/(a+b)$. Exponentiating the Shannon entropy estimate proves the independent-set bound for biregular graphs:The complete bipartite graph $K_{b,a}$ has exactly $2^a+2^b-1$ independent sets, since one chooses a subset of one side and subtracts the double-counted empty set. Disjoint unions of such components attain equality. If exactly one degree parameter is zero, edge counting forces the graph to be edgeless and the displayed bound equals $2^n$, so it remains valid. If $a=b=0$, the denominator is undefined; the edgeless graph has exactly $2^n$ independent sets.

For a finite measurable partition $\xi$,Its conditional entropy of finite measurable partitions relative to $\eta$ isThe concavity of $-t\log t$, equivalently conditioning reduces entropy, gives

The atoms of $T^{-1}\xi$ are $T^{-1}A$ with the same measures as the atoms $A$ of $\xi$. Hence $\boxed{H_{\mu }(T^{-1}\xi )=H_{\mu }(\xi )}$.

SetThe chain rule and invariance give $a_{N+M}\le a_N+a_M$, so $(a_N)$ is a subadditive sequence. ThereforeThe Kolmogorov-Sinai entropy is $h_{\mu }(T)={\sup }_{\xi }{h}_{\mu }(T,\xi )$ over finite partitions.

Taking $F=\{0,\dots ,N-1\}$ immediately shows that the infimum over arbitrary finite $F$ is at most $h_{\mu }(T,\xi )$. For the reverse inequality, apply Shearer's inequality to translates of a fixed finite $F$ inside a long interval. Every interior coordinate is covered $|F|$ times, while only $O(\max F)$ boundary coordinates are lost. Subadditivity bounds the boundary contribution; division by the interval length and passage to the limit giveTaking the infimum proves

Choose $A$ uniformly from $\mathcal{A}$, and let $X_i=1_{\{i\in A\}}$. The characteristic vector of a set determines $A$, soFor $F\in \mathcal{F}$, the projected vector $X_F$ determines the intersection $A\cap F$ and therefore takes values in the trace of a set family $T_F(\mathcal{A})$. The maximum-entropy bound on a finite set givesEvery coordinate belongs to at least $t$ members of $\mathcal{F}$, so Shearer's inequality givesExponentiation proves the Shearer trace inequality

Shearer's inequality states that if $X=(X_1,\dots ,X_n)$ is a discrete random vector and $\mathcal{F}$ is a collection of subsets of $[n]$ in which every index occurs at least $t$ times, then

Order the coordinates naturally. For every $F\in \mathcal{F}$, the chain rule for information entropy givesRemoving conditioning variables cannot decrease entropy, so every summand is at leastAfter summing over $F$, each index $i$ contributes at least $t$ times. A final application of the chain rule yieldswhich proves the lemma.

For $S_i=[n]\setminus \{i\}$, the chain rule for information entropy givesEach summand is at least $H(X_j\mid X_1^{j-1})$ because conditioning reduces entropy. Summing over $i$, each $j$ occurs $n-1$ times:This is the required special case of Shearer's inequality.

Let $\mathcal{Q}$ be the set of valid quintuples and choose $(A_1,\dots ,A_5)$ uniformly from $\mathcal{Q}$. ThenApply Shearer's inequality to the ten pairs $\{i,j\}\subseteq [5]$. Every index occurs in four pairs, so

Put $U_{ij}=A_i\cup A_j$. Since $U_{ij}\in \mathcal{A}$, it has cardinality $k$. For each element of $U_{ij}$, its membership bits in $(A_i,A_j)$ are one of $(1,0),(0,1),(1,1)$; outside $U_{ij}$ they are forced to be $(0,0)$. Thus at most $3^k$ ordered pairs have any prescribed union. The chain rule for information entropy, conditional entropy, and the support bound for information entropy givebecause every $U_{ij}$ belongs to $\mathcal{A}$. There are ten pairs, henceExponentiating proveswhich is the five-variable case of the entropy bound for pairwise-union tuples.

If every ground element belongs to at least $t$ members of a family $\mathcal{F}\subseteq \mathcal{P}([n])$, thenApply Shearer's inequality to the characteristic vector of a uniformly random member of $\mathcal{A}$.