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On a one-sided sequence space $E^{\mathbb{N}}$ with product measure, the Bernoulli shift isIt preserves the product measure. Every invariant event lies in the tail sigma-algebra of the coordinate process, so the Kolmogorov zero-one law makes the shift an ergodic transformation.

For a probability law $\nu$ on $E$, take the sequence space $E^{\mathbb{N}}$ with product measure ${\nu }^{\otimes \mathbb{N}}$. The coordinate maps are an independent and identically distributed sequence with common law $\nu$, and they generate the product sigma-algebra.

A finite open coordinate set witnesses an increasing event when prescribing those coordinates open forces the event, regardless of the remaining configuration. The disjoint occurrence of increasing events $A\square B$ means that $A$ and $B$ have disjoint finite witnesses. For finite-coordinate events this is the usual disjoint-occurrence definition. The van den Berg-Kesten inequality bounds its probability by $\mathbb{P}(A)\mathbb{P}(B)$ under an independent Bernoulli distribution product measure.

This lattice condition on configuration masses is the hypothesis of the FKG inequality. A product measure on a finite Boolean lattice satisfies it with equality, since in each coordinate the minimum and maximum retain the original two coordinate values. It implies positive association of random variables for increasing functions.

Under a product measure on a finite Boolean lattice, an increasing event $E$ and a decreasing event $G$ obey the displayed inequality. The complement $G^c$ is an increasing event, so Harris' inequality gives $\mathbb{P}(E\cap G^c)\ge \mathbb{P}(E)\mathbb{P}(G^c)$. Subtract from $\mathbb{P}(E)$ to obtain the result. This implication includes degenerate Bernoulli distribution parameters $0$ and $1$.

Write $a=P(A_1)=P(A_2)$. The complements are decreasing events. The Harris-FKG inequality for a product measure givesThe same correlation inequality for two decreasing events follows from that for their increasing complements, since their covariance is unchanged. ThusBoth $1-a$ and the square root are nonnegative, so rearranging gives the square-root bound for increasing events:No independence between the events themselves is asserted; independence is the coordinate property of the underlying product measure.

Write $D=\mathbb{E}{\int }_0^{\infty }{e}^{-bt/R}{Z}_t^{1-1/R}dt$ and use the product measure $d\mathbb{P}dt$. For $0<R<1$, the conjugate exponents are $p=1/(1-R)$ and $q=1/R$. Factor the utility integrand asThe Holder inequality on this product measure givesDivide by $1-R$ to obtain the Hölder bound for discounted CRRA consumption:For a finite right side this is ordinary Hölder; on the infinite measure space one can first truncate time and then pass by monotone convergence theorem. If $D=\infty$ and $X_0>0$, the bound is valid but uninformative. If $X_0=0$, the budget forces $c=0$ almost everywhere, so the utility integral is zero; handle this separately instead of writing an undefined $0\cdot \infty$.

When $D$ is finite and the bound is attained, equality requires full budget use and $c_t=(X_0/D)e^{-bt/R}{Z}_t^{-1/R}$. This is the equality pattern of Hölder, not an assertion that every market can finance that process.

Order configurations coordinatewise. The FKG lattice condition on the masses of a finite Boolean lattice isThe FKG inequality states that under this condition, increasing real-valued functions $f,g$ satisfy $\mathbb{E}[fg]\ge \mathbb{E}[f]\mathbb{E}[g]$. In particular increasing events satisfy $P(A\cap B)\ge P(A)P(B)$. For a product measure with coordinate weights $w_e(0),w_e(1)$, the pair $(\min ({\omega }_e,{\eta }_e),\max ({\omega }_e,{\eta }_e))$ has the same two entries as $({\omega }_e,{\eta }_e)$. Multiplying over coordinates gives equality in the lattice condition, including degenerate Bernoulli parameters without division by zero.

For independent bond percolation on the nearest-neighbor cubic lattice, write ${\theta }_d(p)={\mathbb{P}}_p(|C(0)|=\infty )$. Define the percolation critical probability and the connective constant bywhere $c_n(d)$ counts rooted $n$-step self-avoiding walks. Translation invariance makes the root irrelevant, and part (i) proves existence of the latter limit.

First work on the square lattice, whose dual is another translated square lattice. Put $q=1-p$ and assume $q\mu (2)<1$. If the open cluster at the origin is finite, its exterior edge boundary contains a closed dual simple circuit surrounding the origin. To see this planar fact, surround its finitely many vertices by their unit square cells and follow the exterior boundary: its crossing primal edges are closed. Resolving repeated boundary vertices into simple circuits leaves a circuit separating the origin from infinity.

Let $a_n$ count such dual circuits of length $n$. Every one crosses the positive horizontal ray at distance at most $n$, because it surrounds the origin and its horizontal span is at most its length. Choosing a ray-crossing edge, an orientation and all but the last edge encodes it by one of at most $Cn$ rooted length-$n-1$ self-avoiding walks. Thus $a_n\le Cn{c}_{n-1}(2)$. Choose $b>\mu (2)$ with $qb<1$. The root-count limit gives $c_{n-1}(2)\le C_b{b}^n$, soThis summable tail alone need not make the probability of every enclosing circuit less than one. Choose a large $R$ and let $A_R$ be the absence of enclosing closed dual circuits of length at least $R$, with $P(A_R)>1/2$. Let $B_R$ require every primal edge inside $[-R,R{]}^2$ to be open. It has positive probability. Both events are increasing. The finite-measure Harris-FKG inequality extends to $A_R$ by decreasing limits over finitely many circuit exclusions, so $P(A_R\cap B_R)\ge P(A_R)P(B_R)>0$.

On $B_R$ the origin is connected to every graph vertex of that box. Any closed dual circuit enclosing the origin must then enclose the whole box, hence have length at least $R$. On $A_R\cap B_R$ no such circuit exists, so the origin cluster is infinite. This proves the connective-constant Peierls bound, $p_c(2)\le 1-1/\mu (2)$. Finally the lattice in dimension $d\ge 2$ contains a coordinate copy of the square lattice, whose edge law is unchanged. Percolation in that subgraph implies percolation in the full graph, and hence

A connection to a box boundary can be witnessed by a finite self-avoiding open path stopped at its first boundary hit. On the event of reaching radius $m+n$, split such a path at its first graph vertex $z\in \partial {\Lambda }_n$. Its prefix witnesses connection from zero to $z$ within ${\Lambda }_n$. Its remaining segment reaches a point $y$ with $\Vert y{\Vert }_{\infty }=m+n$, so $\Vert y-z{\Vert }_{\infty }\ge m$; truncate it at its first hit of $z+\partial {\Lambda }_m$. The two witnessing edge sets are disjoint, giving disjoint occurrence of increasing events.

Use the BK inequality: for increasing finite-coordinate events in a Bernoulli product measure, $P(A\square B)\le P(A)P(B)$, where the square means disjoint open-edge witnesses. For each fixed $z$, the first event has probability at most ${\beta }_n$, and the second has probability exactly ${\beta }_m$ by translation invariance. A union bound over $z$ givesThe two boxes can overlap, so replacing BK by an independence claim would be incorrect. Restricting each event to paths stopped at its local boundary makes all the relevant coordinate sets finite, as required by the stated inequality.

Put $x_n=\log {\beta }_n$ and ${\alpha }_n=\log |\partial {\Lambda }_n|$. Here $|\partial {\Lambda }_n|=(2n+1{)}^d-(2n-1{)}^d$, so ${\alpha }_n=O(\log n)=o(n)$. A fixed straight open path of length $n$ gives $p^n\le {\beta }_n\le 1$. Part (a) therefore gives the almost-subadditive percolation decay rateIn particular the limit is finite for the prescribed $0<p<1$.

Give each nearest-neighbor edge of the cubic lattice an independent Bernoulli distribution state, open with probability $p$. The resulting product measure is denoted ${\mathbb{P}}_p$. In this bond percolation model, $C(0)$ is the percolation cluster of the origin, andThe uniform-label monotone coupling of Bernoulli percolation shows that ${\theta }_d$ is increasing.

Let $c_n(d)$ count $n$-step self-avoiding walks starting at the origin, with $c_0(d)=1$. Splitting a walk after $n$ steps, and discarding the avoidance constraint between the two pieces, gives $c_{n+m}(d)\le c_n(d)c_m(d)$. The Fekete lemma therefore gives the connective constantThe locally finite graph structure means that an infinite percolation cluster at the origin supplies an open self-avoiding walk of every length. Each specified walk has $n$ distinct edges and is open with probability $p^n$. The union bound givesFor $p\mu (d)<1$ the right side tends to zero. Hence the connective-constant lower bound for percolation is $p_c(d)\ge \mu (d{)}^{-1}$.

For the upper bound first work on the square lattice. A finite open percolation cluster has an outer boundary containing a simple closed graph cycle of dual edges, all crossing closed primal edges. Such a dual bond percolation circuit separates that cluster from infinity. Write $N_{\ell }$ for the number of simple dual circuits of length $\ell$ surrounding the origin. Each circuit meets the positive horizontal ray at distance at most $\ell$: its bounding box contains the origin and its diameter is bounded by its length. Choose such an intersection as an anchor and orient the circuit. Removing its last edge leaves a rooted self-avoiding walk of length $\ell -1$ in the translated square lattice. Consequently, for an absolute constant $K$,The exact constant and this possible overcount do not matter. If $(1-p)\mu (2)<1$, the root test givesA summable circuit count alone need not give a total sum below one. To use its tail correctly, let $B_R=[-R,R{]}^2\cap {\mathbb{Z}}^2$ and condition every edge internal to $B_R$ to be open. This finite event has positive probability. A closed dual circuit surrounding all of $B_R$ crosses no internal edge of $B_R$, and so its closed-edge probability remains $(1-p{)}^{\ell }$ under this conditioning. Its length tends to infinity with $R$. Choose $R$ so large that the union bound for all such circuits is below one. With positive conditional probability, none occurs.

On that event, $C(0)$ contains $B_R$ and cannot be finite: a finite cluster containing $B_R$ would have an enclosing closed dual circuit. Thus ${\theta }_2(p)>0$ whenever $p>1-\mu (2{)}^{-1}$. This is the connective-constant Peierls bound, proved by excluding short circuits through the open-box conditioning. An embedded coordinate plane in the cubic lattice has exactly the same bond percolation law as the square lattice, so $p_c(d)\le p_c(2)$. Together,There are $2d$ choices for the first step of a self-avoiding walk and at most $2d-1$ thereafter, because immediate reversal is forbidden. Hence $c_n(d)\le 2d(2d-1{)}^{n-1}$ and $\mu (d)\le 2d-1$. In particular $\mu (2)\le 3$. Substituting with the correct directions of the inequalities gives