No bigeodesics in planar first-passage percolation

Proof. Every edge weight is strictly positive almost surely. By dominated convergence, choose c>0 such that a=4\EE e^{-ct_e}<1. For a fixed vertex x and a fixed K<\infty, the probability that some simple path of j edges from x has cost at most K is at most e^{cK}a^j. The sum over j is finite. Consequently only finitely many simple paths from x have cost at most K, almost surely. Take the countable intersection over vertices and positive integer K. Every finite walk can be shortened to a simple path by erasing loops. Comparing with any fixed path between two vertices now shows that the infimum is attained among a finite collection of paths.

Two different simple paths with the same endpoints have an edge in their symmetric difference. Conditional on all other weights, equality of their costs prescribes one value of the independent nonatomic weight on that edge and thus has probability zero. There are only countably many pairs of finite paths.□

Proof. The sequence

is decreasing, by the triangle inequality and calibration of the segments of p, and is bounded below by -T(p_0,x). Write a_\infty(x) for its limit. Then B_p(x,y)=a_\infty(x)-a_\infty(y). Additivity, antisymmetry, and the metric bound follow immediately. For j\ge i and k\ge j, the geodesic property gives T(p_i,p_k)-T(p_j,p_k)=T(p_i,p_j), proving calibration. Tail invariance and the coalescence assertion follow from the defining limit. Finally,

so the left side of (3) is a_j(x)-a_\infty(x).□

Proof. Suppose a bigeodesic has a forward end of label \rho. The forward ray starting at any of its backward vertices has that same label by tail invariance. Fixed-label uniqueness identifies all of these rays with g_\rho(x). The middle vertex would belong to this family from infinitely many starting vertices, contrary to backward finiteness. The final assertion is a countable intersection.□

Proof. The extended shape theorem [AH19, Proposition 3.4] states that for each \eps>0, almost surely there is an N<\infty such that

Given x,y, take z to be the endpoint with larger norm and v the displacement to the other endpoint. Then |z|\ge (|x|+|y|)/2 and \max\{|z|,|v|\}\le |x|+|y|. Symmetry of T and m proves the claim.□

Proof. Loop erasure allows the passage-time infimum to be taken over simple paths. Each such path uses every changed edge at most once, so its cost changes by at most K. Taking infima gives the distance bound. For a retained ray p, passing to its defining Busemann limits gives

Linear asymptotics along x=nv, for two independent lattice vectors v, imply equality of the linear parts. Tail invariance handles the stated extension. Under an increase, every competitor is at least as expensive as before, whereas each segment of the retained path has unchanged cost.□

Proof. The spaces \Omega, (\Z^2)^{\mathbb N}, and (\Z^2)^{\Z} are standard Borel spaces. Nearest-neighbor adjacency and simplicity are countable coordinate tests. Each finite segment minimizes exactly when its cost is no greater than that of every finite competitor with the same endpoints; there are countably many competitors. Equivalently, the passage distance is a countable infimum of Borel finite-path costs. Limits defining Busemann values are countable tests, as are asymptotic linearity with rational tolerances and integer radii. Coalescence is expressed by

The standard analytic projection and Jankov–von Neumann selection theorems give the existence and selection assertions; see [Kec95, Theorems 29.7 and 29.9]. To test that the label count is at least j, project the Borel condition for j path witnesses with pairwise distinct labels.□

Proof. For vertices p_i^+ and p_j^- on opposite arms, optimality and the triangle inequality give

Subtract the right-hand terms and let i,j\to\infty. This proves (6). Additivity and asymptotic linearity imply that its left-hand side is

Putting w=z+ne_k and w=z-ne_k, dividing by n, and letting n\to\infty, for k=1,2, proves that the sum of the functionals vanishes. The assertion about moving the base point follows from tail invariance in Lemma 2.2.

The exclusion of a deterministic countable set is Corollary 2.4.□

Proof. Let K=\{x:m(x)\leq1\}, and let K^*=\{\rho:\rho(x)\leq m(x)\text{ for all }x\} be the dual unit ball. For \rho\in\partial K^* the exposed face

is a point or a line segment. Its radial image is A(\rho). If this image contains two directions, it is a nondegenerate closed circular arc. The relative interiors of these arcs are disjoint for distinct \rho: a direction in such an interior corresponds to the relative interior of a nondegenerate face of K, whose supporting line, and hence its normalized functional, is unique. Each nondegenerate arc contains a rational angle. Thus only countably many functionals have more than one contact direction.

There are also only countably many members of C that are isolated on at least one circular side. Indeed, they are endpoints of components of \partial K^*\setminus C, with the finite-set cases included. The open components are disjoint and each contains a point of a fixed countable dense subset of the circle.

Consider a remaining \rho\in C, so that A(\rho)=\{d\} and C accumulates at \rho from both circular sides. A lattice rotation makes d_2>0. The contact correspondence is upper semicontinuous: if \rho_n\to\rho and d_n\in A(\rho_n) with d_n\to d', then m(d')=\rho(d'). Compactness of the unit circle therefore supplies an open boundary neighborhood of \rho on which every contact has vertical coordinate at least some c>0.

Choose once and for all a countable dense subset D_0 of C. In the neighborhood just obtained, choose one point of D_0 on either side of \rho. The compact boundary arc joining them through \rho can be taken inside that neighborhood. There are only countably many pairs from D_0 and finitely many choices of lattice coordinates, so all such arcs form a countable family. For D take the union of D_0, the two countable exceptional sets described above, and their negatives.

For completeness, suppose two chart points \rho,\lambda had the same horizontal coordinate and \lambda(e_2)>\rho(e_2). Choosing any d\in A(\rho), for which d_2>0, would give \lambda(d)>\rho(d)=m(d), contrary to \lambda\in K^*. The same argument shows that each chart point is the top endpoint of the vertical section of K^* at its horizontal coordinate. The continuous horizontal coordinate is thus injective on the arc and strictly monotone along it. Orient J so that it increases. On the opposite lower arc, the horizontal coordinate again increases from left to right; negation reverses the order of the corresponding labels.

Finally, Lemma 3.1 excludes D simultaneously and identifies the opposite end. The contact restriction implies that rays of the chosen upper label tend to height +\infty, and those of the negative label tend to height -\infty. This gives the claimed orientation.□

Proof. If two rays meet infinitely often, choose a shared vertex beyond their finite initial portions. The geodesic segments from that vertex to every later common vertex agree by finite uniqueness. The later common vertices escape to infinity, so the two tails agree. Tail invariance of the Busemann function then gives equality of their labels. Thus distinct labels have only finitely many common vertices.

Suppose the tails of p and q are disjoint. Choose an integer height h above both starting sites, above all intersections, and above finite prefixes preceding the disjoint tails. After their respective last visits to row h, both rays stay strictly above that row. For sufficiently large N, their first visits to row N occur after those last visits, since the prefixes ending at the last visits are finite. The intervening arcs are disjoint strip crossings. Their order on row N therefore agrees with the fixed order of the last visits to row h. This proves stabilization.

Assume this stable order is p to the left of q. For a positive integer n, set u=(-n,0) and v=(n,0). Let \gamma_j be the concatenation of the finite geodesic from v to p_j and the tail of p after p_j. This concatenation is used as a walk; it need not be a ray or be simple. Define

Since B_p(v,p_j)=B_p(v,p_0)+T(p_0,p_j), the definition of B_p implies \delta_j\to0 for this fixed v. Every vertex w of \gamma_j satisfies

On the initial geodesic this follows by decomposing its cost and the Busemann increment at w, with the remaining deficit nonnegative. On the tail it follows from the triangle inequality and exact calibration of p[p_j,w]. Define \xi_j in the same way by joining u to q_j and then following q, and write its error as \delta'_j\to0.

We next control all row-zero visits of these walks, including visits very far from the origin. The horizontal directions are outside both contact sets, so

Write a_p(x)=B_p(0,x)-\rho_p(x), and define a_q similarly. Given \eps>0, the uniform metric estimate and these two rooted asymptotics give a finite K_\eps such that, for all lattice sites x,y,

For w=(t,0) with d=|w-v|\geq n/2, we have |v|+|w|\leq2n+d\leq5d. Additivity therefore gives the explicit bound

The identical estimate holds with u,q in place of v,p. Choose \eps<\gamma/40 first, and then n so large that 3K_\eps\leq\gamma n/8. The right-hand side of (10) is at least \gamma d/2\geq\gamma n/4. For this fixed n, choose j so large that both approximate-calibration errors are smaller than \gamma n/4. It follows that every row-zero visit of \gamma_j has horizontal coordinate in (n/2,3n/2), whereas every such visit of \xi_j lies in (-3n/2,-n/2).

Both walks eventually follow their exact upward tails. Their last row-zero visits thus exist and have the order \xi_j,\gamma_j. Choose a row N above the finite joining paths, the discarded original prefixes p[p_0,p_j] and q[q_0,q_j], and the prefixes ending at these last row-zero visits, and high enough for the stable order of p,q. The first visits of the two walks to this row occur on their exact tails and have the opposite order \gamma_j,\xi_j. Their portions from the last row-zero visits to these first row-N visits are strip crossings. The planar fact gives a common vertex w_j.

At that vertex, domination by the metric and (9) give

Let j\to\infty with n fixed. We obtain B_p(u,v)\leq B_q(u,v). Finally, the rooted asymptotics at u and v give

Division by 2n proves (8). If the labels are distinct, injectivity of r makes the order strict. Reflection in the horizontal axis proves the lower-ray version; this reflection preserves e_1, so it also preserves the horizontal parameter used in that version.□

Proof. Let \Lambda=\{\rho_\ell,\rho_r,-\rho_\ell,-\rho_r\}. For each of these four deterministic functionals, Theorem 2.3(iii) provides a unique ray g^\lambda_z from every vertex z and a common Busemann function B_\lambda. Let \sigma_\lambda=1 for the upper endpoints and \sigma_\lambda=-1 for the lower endpoints. Define

Each supremum is finite. Indeed, the limit directions of this endpoint ray lie strictly inside its cone of slope 2s, and translating the origin to its fixed starting point does not change its limit directions. Outside a finite prefix the displayed positive part is therefore zero.

We prove (12) first for these endpoint rays. Put a_\lambda(x)=B_\lambda(0,x)-\lambda(x)=o(|x|). If w\in g^\lambda_z, calibration gives

Fix L and \eta>0. Whenever |z|\leq Ln and d=w-z satisfies |d|\geq\eta n, we have

The uniform metric approximation, the four rooted asymptotics, and (14) thus imply

To make the uniformity explicit, bound each rooted error by \eps|x|+K_\eps and the metric error by \eps(|z|+|w|)+K_\eps. The quotient in (15) is then at most 2\eps(1+2L/\eta)+3K_\eps/(\eta n); take n\to\infty and then \eps\downarrow0.

For each \lambda\in\Lambda, the continuous function m(d)-\lambda(d) has a strictly positive minimum on

because all its zeros lie strictly inside the cone of slope 2s. Taking the minimum over the four functionals preserves positivity. It follows from (15) that, for all sufficiently large n, every endpoint-ray displacement with |d|\geq\eta n lies in the unshifted cone of slope s. For the remaining displacements,

Consequently the maximum in (12) is at most 1+(s+1)\eta n eventually. Letting \eta\downarrow0 proves that equation for the H(z) in (13).

It remains to trap every intermediate ray with exactly this offset. Fix z and an upper ray p with r(\rho_\ell)<r(\rho_p)<r(\rho_r). Write a=g^{\rho_\ell}_z, b=g^{\rho_r}_z, and Q=Q^+(z,H(z)). The bounding rays lie in Q. Finite uniqueness implies that they have a common finite prefix ending at a vertex v, and then disjoint tails. Indeed, a reunion after their first split would give two finite geodesics from z to the reunion vertex. Their common prefix is finite because their labels differ.

Suppose, for a contradiction, that p visits a vertex w\notin Q. Choose an integer N above the heights of the entire common prefix of a,b, above the finite prefix of p ending at w, and large enough for the three pairwise orders in Lemma 3.3 to have stabilized. Let a_N,b_N,p_N denote their first arrivals on row N. They are strictly ordered as

Each first-arrival path lies strictly below that row until its endpoint. The two arcs a[v,a_N] and b[v,b_N], together with the horizontal segment [a_N,b_N], form a simple closed polygonal curve \mathcal J. Let \mathcal D be its bounded Jordan domain. The curve lies in Q, including its top segment by convexity. Since a bounded Jordan domain is contained in the convex hull of its boundary, \mathcal D\subset Q.

The common prefix a[z,v]=b[z,v] need not lie in \mathcal D; viewed from the Jordan curve it may form a slit on either side. The following first-arrival argument treats that possibility explicitly; see Figure 1. The last edge into p_N is vertical from below. Since p_N is in the relative interior of the top segment, the part of that edge sufficiently close to p_N lies inside \mathcal D. Starting at w\notin Q, the ray therefore passes from outside \mathcal D to its inside before its first row-N arrival. It cannot enter through the top segment, since that would already be a visit to row N. Hence, after w, it meets one of the two bounding arcs at a lattice vertex u, possibly u=v. Both its prefix p[z,u] and the corresponding bounding prefix are finite geodesics. Uniqueness makes them identical. The bounding prefix is contained in Q, whereas p[z,u] contains w\notin Q, a contradiction. This argument also rules out departures made before the bounding rays split.

Thus every intermediate upper ray is in Q. Rays with an endpoint label are the unique endpoint rays. Reflection proves the same statement for lower rays. The reasoning was deterministic on the single event already specified and applied to an arbitrary ray, so it establishes the simultaneous all-ray assertion.□

Proof. For upper rays put h=\max(x_2,y_2). Both rays have a last visit to row h, because they tend upwards. Call the two last vertices \widehat x,\widehat y. The cone bounds give

Since (h-x_2)+(h-y_2)=|x_2-y_2|, (16) implies \widehat x_1<\widehat y_1. If the rays did not intersect, their portions from these last visits to their first visits to a sufficiently high row would be disjoint strip crossings. Their order there would remain p,q, contradicting Lemma 3.3 and the strict reverse order of their labels. For lower rays use h=\min(x_2,y_2) and the last visits before departure towards height -\infty. The same estimates and the lower form of Lemma 3.3 apply.□

Proof. Take the finite geodesics from x to p_k. Since p is simple, |p_k|\to\infty, and their lengths tend to infinity. Local finiteness and a diagonal subsequence give a local limit q from x. Each finite prefix of q is eventually a prefix of these finite geodesics, so q is a simple infinite geodesic.

By Lemma 2.2,

If w lies on the geodesic from x to p_k, then

because T(w,p_k)\ge B(w,p_k). For each fixed vertex of the local limit, let k tend to infinity along the chosen subsequence. This proves (17).

For any w\in\Z^2, the same metric bound and (17) give

Consequently B_q(x,w)\le B(x,w). Write \sigma=\rho_q and \rho=\rho_p. Asymptotic linearity, with the fixed basepoint x, and the choices w=ke_i and w=-ke_i, i=1,2, imply respectively \sigma(e_i)\le\rho(e_i) and \sigma(e_i)\ge\rho(e_i). Thus \sigma=\rho.□

Proof. We prove the two assertions in the definition separately.

Branching. Fix a vertex x and two different first edges e,f incident to x. There are at most two labels in I that are realized both by a ray from x beginning with e and by one beginning with f. Suppose, to the contrary, that there are three such labels \rho_1,\rho_2,\rho_3, with r(\rho_1)<r(\rho_2)<r(\rho_3). Choose the six corresponding rays. The geodesics from x form a tree: once two of them diverge, they cannot meet again, by uniqueness of finite geodesics. In particular the two first-edge branches are disjoint away from x.

Choose a horizontal row sufficiently high that its six first-arrival vertices are distinct and have the label order supplied by Lemma 3.3. Such a row exists: rays of different labels have disjoint tails, and rays in the two first-edge branches never meet away from x. The endpoints occur as three consecutive groups, one for each label, and each group contains one endpoint of each branch. Color the endpoints according to the first edge e or f. The resulting six-letter word is a concatenation of three two-letter words, each containing both colors. It has at least three changes of color, and hence contains an alternating four-letter subsequence.

Connect the two e-colored endpoints of this subsequence by the unique arc in the e branch, and connect the two f-colored endpoints by the corresponding arc in the f branch. Both arcs avoid x, since the two root paths defining each arc share their first edge. Their interiors are strictly below the chosen row, since all six root paths were stopped on their first visit to that row. The arcs are disjoint, but their endpoints alternate along the row. This is impossible: the first arc together with the row segment between its ends is a Jordan curve, and the second arc must cross it away from the row. This proves the claim for x,e,f.

Two distinct rays from a common vertex have a first vertex of divergence. Applying the preceding claim there, and taking the countable union over vertices and first-edge pairs, proves countability for condition (i).

Busemann nonuniqueness. First consider two vertices x,y, with y sufficiently far to the right of x that the crossing conclusion of Lemma 3.5 applies to every pair of upper rays from these vertices with reversed strict label order. If B,B' are Busemann functions of upper rays with labels \rho,\sigma satisfying r(\rho)<r(\sigma), then

Indeed, use Lemma 4.1 to take a B-calibrated ray from y and a B'-calibrated ray from x. Their labels are \rho and \sigma, respectively, so they intersect at a vertex w. Calibration at w and the metric bounds on Busemann increments give

For such a pair x,y, let

These sets are bounded by T(x,y) in absolute value. If r(\rho)<r(\sigma) and both sets are nonempty, (19) states that

Thus the nonempty open intervals (\inf S_\rho(x,y),\sup S_\rho(x,y)) are pairwise disjoint. Each contains a rational number, so only countably many labels can have more than one possible value at this pair.

It remains to ensure that these comparisons detect every possible disagreement of Busemann functions. Fix arbitrary vertices u,v. Proposition 3.4 supplies finite offsets controlling all upper rays from u and from v. For y_N=(N,0) it also supplies offsets H_N=o(N) controlling all upper rays from y_N. The height differences between y_N and u,v are fixed. For all sufficiently large integers N, the horizontal separations therefore dominate the offsets and height differences as required by Lemma 3.5. Choose any such N. Both pairs (u,y_N) and (v,y_N) then satisfy (20) for all labels and all their Busemann functions.

If two upper-ray Busemann functions B,B' of the same label disagree at (u,v), additivity in

forces a disagreement at one of those two comparison pairs. The labels producing a disagreement at (u,v) are consequently contained in a union of two countable sets. Taking a countable union over u,v\in\Z^2 proves countability for condition (ii). The choice of N may depend on the environment and on u,v; it uses the all-ray cone bounds, and no common rate of Busemann convergence over labels. Reflection of the vertical coordinate proves the lower-sector assertion.□

Proof. Use the standard Borel configuration space \Omega before completing its probability measure, and the path space \mathcal P=(\Z^2)^{\mathbb N_0}. Let \mathsf R(\omega,p,\rho) be the assertion that p is a simple nearest-neighbor ray in \omega, that all its Busemann limits exist, and that its Busemann function is asymptotically linear to \rho. For a linear functional \rho, this is a Borel relation by Lemma 2.7. Define Busemann evaluations to be zero where their limits do not exist; the relation \mathsf R explicitly requires existence.

For paths p_1,p_2 satisfying \mathsf R(\omega,p_i,\rho), the exceptionality witness is the Borel disjunction

The common-tail test for paths q_1,q_2 is Borel as well:

Its negation tests noncoalescence.

Now define the bad set J\subset\Omega\times\Omega by the existence of one label \rho\in I and four paths p_1,p_2,q_1,q_2 such that

1. (a)

   \mathsf R(v,p_i,\rho) holds for i=1,2, and (21) holds in v;

2. (b)

   \mathsf R(w,q_i,-\rho) holds for i=1,2, and (22) fails.

This is directly the projection of a Borel relation on \Omega^2\times I\times\mathcal P^4. In particular, J is analytic and is measurable in the completion of \PP\otimes\PP.

For every v in the full-measure countability event of Proposition 4.2, its section is

For this fixed v, each label in the countable union is an ordinary fixed parameter. Symmetry gives -\rho\in C, and Theorem 2.3 makes the corresponding analytic event in w null. Hence \PP(J_v)=0 for almost every v. Completed-product Fubini gives (\PP\otimes\PP)(J)=0. Equivalently, replace J by a Borel set equal to it outside a Borel product-null set and apply ordinary Fubini; the sections of that null set are null for almost every v. This argument needs no measurable enumeration of \mathcal E_I(v).

No universal all-ray condition is inserted into the four-path relation. Such regularity conditions are imposed outside the projection, as specified after Lemma 2.7.□

Proof. First take a common event on which all-ray asymptotics, finite geodesic existence and uniqueness, and the uniform metric estimate hold in U and in all V_n,W_n. This is a countable intersection of events of probability one.

A single subsequence for the moving cuts. For any metric T put

For a fixed typical metric, E_n(T)\to0. Indeed, for any \eps>0, Lemma 2.5 bounds the numerator by \eps(|x|+|y|) outside a finite set of pairs. The maximum error on the remaining finite set, divided by n, tends to zero. Since each V_n has the iid law, E_n(T_{V_n})\to0 in probability. Choose an increasing deterministic sequence n_k such that

After a further probability-one restriction, Borel–Cantelli gives

This is the only subsequence used in the proof.

Fix now an arbitrary upper U-ray p, with label \rho\in I. Write

All contacts of \rho are strictly upper, so compactness gives

Let K be finite and enlarge it to contain p_0. Set R_K=\max_{x\in K}|x|. For n\ge2R_K+1, x\in K, and z\in H_n, the vector z-x points into the lower half-plane and

For n=n_k, (25) therefore yields

The metric error is o(|z-x|) uniformly over these x,z, by (26) and (29). The same is true of |a(z)|: |z|\ge n\to\infty uniformly on the cut, and |z|\le2|z-x|. Finally, the finitely many |a(x)| are bounded. Thus, for all sufficiently large k, uniformly for x\in K and z\in H_{n_k},

The same estimate holds with V_{n_k} replaced by U, because E_n(T_U)\to0.

The last-cut comparison. The heights of p_j tend to +\infty, so the entire ray lies strictly above H_n for all sufficiently large n. If a finite walk from x to p_j, with both endpoints above the cut, visits H_n, let z be its last cut vertex. Its suffix after z lies strictly above the cut and uses only edges in E_n^+. Its cost in V_n is consequently at least

The suffix bound uses only B_p^U(z,p_j)\le T_U(z,p_j) and its unchanged edge costs. Repeated visits to the cut and repeated edges in the walk cause no problem: all occur either in the prefix or in a suffix whose cost still bounds the corresponding distance. The same comparison holds in U.

For x=p_0, the original segment of p has cost B_p^U(p_0,p_j) and is unchanged. By (31) every cut-touching competitor is more expensive; a competitor avoiding the cut has its original U cost. Hence every prefix of p minimizes in V_{n_k} for all sufficiently large k, and p is a ray there.

For any fixed x\in K, (18), applied in U, says that

Fix a sufficiently large k. Eventually this defect is smaller than \gamma n_k/4. By the U version of (32), a U-geodesic to p_j must then avoid the cut. It provides an unchanged competitor in V_{n_k}. The V_{n_k} version excludes every cut-touching walk that could improve that competitor, and every walk avoiding the cut has its U cost. It follows that

Because K is finite, one target-index threshold works for all its vertices. Subtracting two such distance equalities and taking the limit preserves every B_p(x,y) with x,y\in K.

Exact identification of the label at a fixed cut. Preservation of a path alone does not identify both coordinates of its Busemann asymptote. Fix one n=n_k for which p was preserved, and henceforth regard U,V_n, and the cut as fixed. Fix 0<c<1, take x in the sector x_2\ge c|x|, and let z\in H_n. With d=z-x we have

Consequently

This bound holds uniformly along the entire infinite cut, including cut vertices with arbitrarily large horizontal coordinate. Since d_2<0, (28) gives

For every \eps>0 there is C_\eps<\infty such that |a(w)|\le\eps|w|+C_\eps for every w. The fixed-metric uniform estimate, separately in U and V_n, and (27) now give, for |x| sufficiently large in the sector and every z\in H_n,

Choose \eps<\gamma\kappa/8 and then increase the lower bound on |x|. In both metrics we obtain the uniform barrier

In particular, the error at the moving basepoint x, as well as that at z, has been included explicitly.

For each such x, its calibration defect tends to zero as the target runs out along p. The last-cut argument just used proves T_{V_n}(x,p_j)=T_U(x,p_j) for all sufficiently large j. Meanwhile the preserved root p_0 satisfies T_{V_n}(p_0,p_j)=T_U(p_0,p_j) for every j, whether or not p_0 is in the strict sector. Thus

Take c=1/2, and use first x=ke_2 and then x=k(e_1+e_2). Dividing (36) by k and using asymptotic linearity in the two fixed environments shows that their labels agree on e_2 and e_1+e_2. These vectors are independent, so the labels are exactly equal.

All steps so far are pathwise on the common event already chosen. For a finite collection of rays, apply them to each ray and take the largest of the finitely many thresholds. This proves part (i), with the stated quantifiers.

Lower tails for a fixed W_n. Fix n and an arbitrary lower U-ray q of label \lambda\in-I. Use

Choose c\in(0,1/2] small enough that eventually q_{j,2}\le-c|q_j|. This is possible because the limit directions of q lie in the strict lower sector of the chart. If x_2\le-c|x|, |x|\ge2n/c, and z\in H_n, then

The displacement z-x points upward. Applying the fixed-metric uniform estimate to U,W_n and bounding both errors a_-(x),a_-(z) exactly as above gives, for all sufficiently large |x| in this lower sector, uniformly for z\in H_n,

Only the usual uniform estimate in this particular fixed W_n has been used.

Choose a vertex a=q_L sufficiently far along q that (37) applies to a and the whole remaining tail lies strictly below H_n. For a walk from a to a later q_j that touches the cut, take its last cut vertex z. After this last visit its first edge goes downward and every later vertex is strictly below the cut. Every edge of this suffix is outside E_n^+ and hence is shared by U and W_n. Horizontal edges on the cut, which are not shared, are part of the prefix. The analogue of (32), now using (37), shows that this walk is more expensive than the unchanged segment of q. Walks that avoid the cut have their U costs. Thus the tail from a is a W_n-ray, and its distances from a to all later q_j are unchanged.

Finally take any sufficiently far x in the strict lower sector. Its U calibration defect relative to q_j tends to zero, so (37) and the last-cut argument give eventual equality of distances from x in the two environments. Using the preserved anchor a yields equality of the two tail Busemann functions at (a,x). The sequences x=-ke_2 and x=-k(e_1+e_2) belong to this sector for our choice c\le1/2. Asymptotic linearity along them identifies the label of the W_n-tail as exactly \lambda.

Since all W_n were included in the initial countable probability-one event, the argument applies to every fixed n and every lower ray, with its own tail index. There is no estimate on the rate of convergence of the sequence of metrics W_n, and no uniformity of thresholds over labels is required.□

Proof. Use the coupling (24). For every n, Lemma 4.3 applies to (V_n,W_n). Take their countable intersection, together with the probability-one event in Lemma 4.4.

On this event, fix any \rho\in\mathcal E_I(U) and two upper rays witnessing its exceptionality. If they witness branching, their distinct paths and common starting vertex are preserved by Lemma 4.4(i) for every sufficiently deep subsequence cut. If instead they witness different Busemann functions, choose one pair x,y at which those functions differ and include x,y in the finite witness set of that lemma. At every sufficiently deep subsequence cut the two paths retain both their exact label \rho and those unequal evaluations. In either case choose one such cut n; then \rho\in\mathcal E_I(V_n).

Now let q,q' be any two lower U-rays of label -\rho. Lemma 4.4(ii), for this fixed n, supplies tails of both which are W_n-rays of the same exact label -\rho. They coalesce by the property of (V_n,W_n). Equality of their tails is equality of their vertex sequences and thus also means that the original U-rays coalesce.

The argument made arbitrary pathwise choices of \rho and its witnesses after the common event was fixed. It therefore proves the simultaneous assertion in the coupling, and hence for its U marginal. Its failure is itself the analytic four-witness event with both environments set equal to U, so it is measurable in the completed marginal space. Reflection gives the reversed statement; a countable intersection gives both statements in all charts.□

Proof. Put d=y_1-k_1. If a vertex w belongs to the ball of radius R about k, then

Membership in either comparator cone would therefore require R\ge s(d-R)-H_c. Thus

makes both base comparator rays avoid the entire ball. Their edge weights are unchanged in T^k. Since every other edge weight only increases, each comparator remains a geodesic ray in T^k. Lemma 2.6 preserves its label.

Increase the separation further, if necessary, to apply Lemma 3.5 with offsets H_p,H_c. The upper marked ray is to the left at its starting site but has label strictly east of \alpha. The same ordering holds for the lower rays, since

The upper comparator therefore meets p^+, and the lower comparator meets p^-. Call intersection vertices w^+ and w^-, respectively. By (41), each intersection is outside the ball. Consequently it occurs after the whole prescribed initial gain portion of the corresponding marked arm. This remains true if an arm leaves and later reenters the ball, since every vertex of its prescribed initial portion lies in the ball.

Let g^+ and g^- be the increases on those portions, so g^++g^-\ge b. Writing \operatorname{cost}_T for the cost of a specified finite path in the base weights, domination by passage time and then by a Busemann increment gives

The marked segment from w^- through k to w^+ is minimizing in T^k. A competing finite walk follows the lower comparator backwards from w^- to y, then follows the upper comparator to w^+. The cost of this walk in T^k equals its base cost, namely

No disjointness between the two comparator portions is required: a finite walk is an admissible competitor and its cost counts every traversal. Comparison with the marked segment yields

Additivity cancels the increments ending at the intersection vertices and gives exactly (39). Both the avoidance bound and the separation in Lemma 3.5 have the asserted dependence on the offsets and R.□

Proof. Set w=r(\beta)-r(\alpha)>0 and, for j\ge0, write

The sign in this formula uses \alpha(e_1)+(-\beta)(e_1)=r(\alpha)-r(\beta)=-w. The two fixed-label asymptotics give e_j=o(jl) as j\to\infty. Define

Then E_N/(Nl)\to0 almost surely. To see the needed uniformity, for each \delta>0 choose J so that |e_j|\le\delta jl when j\ge J. The earlier errors have a finite maximum C_J, and E_N\le\max\{C_J,\delta Nl\}. Let N\to\infty and then \delta\downarrow0. This argument controls random first and last marks without replacing them by deterministic endpoints.

Let m_N be the number of marks in \{k_0,\ldots,k_N\}. If m_N\ge2, list their indices as i_1<\cdots<i_{m_N}. Summing the comparison over consecutive marks and using additivity gives

If m_N=0 or 1, the same upper bound applies to b(m_N-1)_+. Therefore, in all cases,

The expectation step uses only the deterministic bounds

Fatou’s lemma applied to the nonnegative variables 2/l-m_N/(Nl) gives the bounded reverse Fatou inequality

Equal marginal probabilities imply \EE m_N=(N+1)q_{\alpha,\beta} by linearity of expectation. Hence q_{\alpha,\beta}/l\le w/b, as asserted. In particular, this proof takes no expectation of either a Busemann error or a Busemann increment.□

Proof. Let a be the bottom of the support of G. It is finite, and G(\{a\})=0. As b\downarrow a with b>a, the probability G([0,b)) tends to zero and is positive. Choose b_0>a so that this probability is less than 1/64, and choose a_0\in(a,b_0). The definition of a gives p_a>0, proving (44).

The edges of a simple path are distinct. For a specified path of length j, independence and a union bound over subsets of its edges bound the probability that at least j/2 of them are unusable by

This also bounds the probability of fewer than j/2 usable edges. There are at most 4^j nearest-neighbor paths of length j from a given vertex, so the union bound over simple paths is at most \theta^j. Summing over j\ge n_0 proves (45); the further union bound over the (6n+1)^2 starting vertices proves (46). Finally, the events in (45) decrease as n_0\to\infty, and their probabilities tend to zero. Countability gives the last assertion simultaneously at all vertices. These estimates use no moment of an individual edge weight.□

Proof. Fix a unit vector v and let e_k=\{k,k+v\} for sites k on a horizontal grid whose spacing will be chosen below. Use one iid base environment t and independent samples \widehat t_{e_k} with law G. The individual environment t^k equals t off e_k and equals \widehat t_{e_k} on e_k. Each individual environment has precisely the iid law. We take the common full-measure event for the base and all these countably many individual environments, including the fixed-label exclusions at every endpoint in D_I and -D_I.

For a bin (\alpha,\beta)_I, let A_e^{\alpha,\beta}(H) denote the event in an iid environment that e is usable, \mathcal R_e(H) holds, and a bigeodesic whose upper label is in this bin uses e. Mark k in this bin when

The original variable t_{e_k} is independent of the entire individual environment t^k: that environment consists of the independent sample \widehat t_{e_k} and the other base coordinates. Consequently the common marginal marking probability is exactly

where e is any fixed translate of \{0,v\}. The regularity test is measurable in the individual environment, so it does not alter this independence calculation. Dependence between marks at different sites is immaterial.

On a marking event, the one changed edge is increased by at least b_0-a_0. Every witnessing bigeodesic passes through k, and the edge e_k is the first edge of one of its arms based there. Use that edge as the gain portion of that arm and use the empty portion of the other arm. Thus the modification and both gain portions fit in a ball of radius one, and the total gain is at least

The regularity event in t^k confines both marked arms with offset H.

We must also establish confinement of the base comparator rays at every marked site; base regularity was not assumed in (49). Let q be any base ray from k with label in I or -I. A simple path from k can traverse e_k only as its first edge. Indeed, a later traversal in either direction would revisit k. If q avoids e_k, all its weights are unchanged and it remains a ray in t^k by monotonicity. Its label is unchanged by Lemma 2.6, so \mathcal R_{e_k}(H) in t^k puts it in the appropriate cone at k.

If q uses e_k first, its tail from k+v avoids e_k and remains a ray in t^k, with the same label. That tail lies in the corresponding cone at k+v with offset H. For either choice of sign, the inequality

implies

Including the initial vertex k, and then the initial edge by convexity, confines the whole base ray with offset at most H+1+s. This proves the needed base comparator control at both fixed labels \alpha and -\beta, uniformly over all bins. It explains why regularity was required at both endpoints of e_k.

Choose an integer grid spacing l=l(H) large enough for Lemma 5.1 with R=1, H_p=H, and H_c=H+1+s. It may be chosen so that

for a chart-dependent constant C_0, uniformly in the unit vector v. The lower bound makes the edges e_k at distinct grid sites distinct. Any two different marked sites are now sufficiently separated. Applying Proposition 5.2 and (50) to each bin yields

For the finite partition \Pi_j, let K_j(e,H) count its occupied bins among the labels counted by N_e(H). Linearity of expectation and (38) give

On the event excluding D_I as bigeodesic labels, refinement never loses a counted label, so K_j(e,H) is nondecreasing. Every finite collection of distinct counted labels is eventually in distinct bins. Thus

also when the latter is infinite: apply the separation assertion to arbitrarily large finite collections. Monotone convergence in (52), followed by (51), proves (48). Translation invariance and the finite number of edge directions give one constant for all e.□

Proof. The counting argument groups forks by label and extracts disjoint forward branches within each group. Positive fork density would force a cubic expected number of usable edge–label pairs in large boxes; uniform confinement and Proposition 5.4 give an arbitrarily small cubic upper bound.

It suffices first to exclude upper forks in one chart. Work on the simultaneous event of Proposition 4.5. Every upper fork label is exceptional by the first test in Proposition 4.2, since two different rays of that label start at its split vertex. Consequently all lower rays with the opposite label coalesce, simultaneously for every label arising in this way. In particular, a finite collection of bigeodesics with one such upper label has a common backwards tail. A vertex sufficiently far along that common tail is a common past root for the entire finite collection.

Let F_n be the number of vertices in [-n,n]^2\cap\Z^2 that are upper forks for this chart, and let

The chart and the fork definition are translation invariant, so

Fix \eps>0 and put H_n=\lceil\eps n\rceil. Let G_n be the intersection of the event U_n in Lemma 5.3 and the event that \mathcal R_e(H_n) holds for every edge with both endpoints in [-2n,2n]^2. Uniform all-ray confinement in Proposition 3.4 gives the latter event eventually almost surely: all its endpoint sites lie in a box of radius a fixed multiple of n, and their required offsets are o(n). Therefore, for this fixed \eps,

No rate uniform in \eps is asserted or needed.

We claim the following deterministic counting inequality on G_n, for n\ge2:

Choose a witnessing pair at each of the finitely many fork sites counted by F_n, and group the choices according to their exact upper label. These finite choices are only used to prove a pathwise inequality; no measurable enumeration of labels is required.

Consider a group containing q\ge1 distinct fork sites. There are at most 2q witnessing bigeodesics in this group. Their lower rays coalesce by Proposition 4.5. Pairwise coalescence of a finite collection gives a common tail by going beyond all their finitely many merging locations. Choose a vertex o on this tail, earlier than every selected split, and orient each witness forwards from o toward its upper end.

The union of these forward paths is a rooted tree. To justify this directly, the initial path from o to any vertex z on a witness is a finite geodesic. If two witnesses visit z, finite uniqueness makes their entire paths from o to z equal. They therefore cannot diverge and later meet, and common edges cannot carry conflicting forward orientations. Each visited vertex other than o has one parent, and the resulting graph has no cycle.

Let \mathcal T be the finite subtree obtained by taking the union of the root paths to the two selected children immediately after every selected split. Each of the q split vertices has at least two children in \mathcal T. For any finite rooted tree with L leaves,

where d^+_{\mathcal T}(u) is its number of children. Indeed, the sum of the outdegrees is the number of edges, one less than the number of vertices; subtracting the number of nonleaves gives (56). Thus L\ge q+1.

Every leaf of \mathcal T is an endpoint of one of the root-to-child paths defining \mathcal T. Otherwise a defining path containing that leaf would continue to another vertex of \mathcal T, contradicting its being a leaf. In particular, every leaf is a selected child of a fork site, and hence lies in [-n-1,n+1]^2. The root paths are allowed to leave and reenter both boxes; only their terminal children determine this location of the leaves. Selected split vertices may also be ancestors of one another, but each is still a distinct vertex with at least two children and contributes at least one to (56).

For each leaf choose a witnessing bigeodesic containing its defining root-to-child path, and continue forwards along that witness. These continuations are pairwise vertex disjoint. In fact, if two met at w, uniqueness of the finite geodesic from o to w would put both leaves on a single root path. They would be comparable in the ancestor order, contrary to being distinct leaves of \mathcal T. The same reasoning rules out an intersection between one continuation and the root path to another leaf: its own leaf would then be an ancestor of the other one. An intersection with an earlier point of its own root path contradicts simplicity. These conclusions hold throughout the plane, including parts of \mathcal T outside either box.

Each leaf has infinity-norm distance at least n-1 from the boundary of [-2n,2n]^2. Its continuation eventually reaches this boundary, since an upper ray in the chart has height tending to +\infty. Stop the continuation at its first boundary visit. The resulting path is simple, all its edges have both endpoints in the outer box, and its length is at least n-1\ge n/2. Its starting point is in [-3n,3n]^2, so U_n supplies at least half its edges as usable. It therefore supplies at least n/4 usable edges inside the outer box. Every such edge lies on the witnessing bigeodesic and satisfies \mathcal R_e(H_n) on G_n. It contributes the group’s label to N_e(H_n).

For this group the leaf continuations are disjoint, so no edge is credited twice with that label. Groups with different labels credit different pairs (e,\rho), even if they use the same geometric edge. Thus the count is of distinct edge–label pairs. Summing over the groups credits at least

pairs, proving (55).

We now take expectations, retaining the scale of the bad-event loss. By (53),

The loss after multiplication by n/4 is at most

There are 2(4n)(4n+1)\le40n^2 edges with both endpoints in the outer box. From (55) and Proposition 5.4,

Divide by n^3 and let n\to\infty for this fixed \eps. Equations (54) and (58) yield p_{\mathrm{fork}}\le40C\eps. Sending \eps\downarrow0 gives p_{\mathrm{fork}}=0. Countability of vertices excludes every upper fork in this chart almost surely. Reversing upper and lower gives the same conclusion for lower forks, and the countable union over charts proves the proposition.□

Proof. Take the common full-measure event of Propositions 4.5 and 6.1 and the structural properties. Let the two bigeodesics have upper label \rho. If their upper Busemann functions differ, \rho passes the second exceptional label test of Proposition 4.2. The simultaneous conclusion of Proposition 4.5 makes their lower rays coalesce. Choose a vertex o sufficiently far along their common lower tail. The forward portions from o are upper geodesic rays. They cannot have the same tail, since tail invariance in Lemma 2.2 would make their upper Busemann functions equal. They must therefore have a first split. Finite uniqueness prevents any later meeting, so this split is an upper fork with a common past, contrary to Proposition 6.1.

If the lower Busemann functions differ, the lower label -\rho is exceptional by the same second test. Apply Proposition 4.5 with upper and lower reversed to coalesce the upper rays. Reversing the preceding argument then produces a forbidden lower fork. This proves both equalities in (59).

All these deductions are pathwise on the stated common event. They do not take an uncountable union of fixed-label null events, nor assert coalescence of every one-sided family at random labels. Absolute continuity transfers the event to each individual changed environment; countable intersection gives the final assertion.□

Proof. The law of \widehat X_e, for h(e)=h, has density relative to G

The two indicated sets are disjoint. Direct integration gives \int\lambda_h\,dG=1 and

For any deterministic finite set of replaced edges, independence of the triples makes the full density the product of these one-edge densities; the exterior has its original iid law. This proves absolute continuity for both the individual and union environments.

To bound the product for an individual region, let N_h count its edges of height h. At positive height h, horizontal edges lie in the two rows of heights h,-h, and vertical edges lie in the two strips adjoining those rows toward height zero. Each row or strip contains at most 2(1+(h+H_0)/s)+1 admissible edges. At height zero there is only one horizontal row. Thus, for example,

It follows from (65) that

The series is finite; its terms of leading order are 1/(h\log^2h). It defines the required constant C. Finally, since \EE_\mu L_{k,R}=1, Cauchy–Schwarz gives

All modified sets are finite. The infinite series has been used only to bound these finite products uniformly.□

Proof. For e\in A_k(R), put Y_e=\widehat X_e and h=h(e). As measures in t, the construction gives

For the first identity multiply the low-input probability p_a, the replacement probability \delta_h/p_a, and the conditional law of Z_e. In the second identity the original sample is retained, except for the removed mass on [0,a_0]. Dividing the first density by their sum (64) proves (66).

The pairs (\xi_e,Y_e) are independent across A_k(R). Hence their conditional kernel given the finite output vector is the product of the kernels just computed. The entire collection of exterior outputs (Y^{k,R}_e:e\notin A_k(R))=(X_e:e\notin A_k(R)) is independent of the inside triples. Testing the product kernel against bounded functions of finitely many exterior coordinates therefore preserves the conditional-expectation identity. A monotone-class argument extends it to the sigma-field of all exterior coordinates. This establishes the claimed law given the full environment.

For the last assertion, the path space is standard Borel. A selector measurable in the completion of the output law has a Borel version under that law. Given the full output, a finite prefix of this version is a fixed list of distinct edges. Restricting the product kernel to that list proves the assertion. One can equally partition by the countably many possible finite prefixes to verify the conditional identity. No output at another site is included in this conditioning.□

Proof. Fix an output in \mathcal E_{k,R}, and denote the successive edges of the selected upper arm by e_1,e_2,\ldots. For j\leq R they belong to A_k(R) and have h(e_j)\leq j, as shown in (62). Put

The a_j and the selected edges are fixed under this conditioning, and A_d\geq d/2 for every d\geq n_0. The decreasing sequence f_j=((j+2)\log(j+2))^{-1} satisfies

Indeed each usable edge contributes the probability in (66), at least \eta f_j/(p_b+\delta_0). Summation by parts, followed by the prefix bounds, gives for d\geq n_0

The last expression tends to infinity. Let m_d be \gamma times this expression, a deterministic increasing lower bound for \mu_d.

The selected path is simple, so Lemma 7.2 gives

Choose R_0\geq n_0 with m_{R_0}\geq8. For every R\geq R_0, Chebyshev’s inequality gives

This choice is independent of R and of the requested gain M. For that gain put K_M=\lceil M/g\rceil, and choose R=R(M)\geq R_0 so large that m_R\geq\max\{2K_M,32\}. Again by Chebyshev,

At least K_M successful raises give total increase at least gK_M\geq M on the prefix. The union bound for the two failure events proves the stated probability 3/4.□

Proof. Take R=R(M) from Lemma 7.3. On \mathcal E_{k,R} use the already selected path, and mark when both success requirements of that lemma hold. Together with (69), the lemma gives marking probability at least (p/2)(3/4)=3p/8, proving (73). All the asserted path properties are part of the selection or the marking test. The region A_k(R) lies in a ball about k of radius at most

and the prescribed gain portion lies in the ball of radius R.

Horizontal translations preserve the distribution of the triples and their height indices. Since the selection rule and all tests are translated in the same way, the marks have equal marginals; in fact this construction is stationary under horizontal translations. Marks at different sites may be dependent.□

Proof. For each edge of p_k of height at most R, (62) puts both endpoints within the width bound of A_k(R). Its individual weight is already \widehat X_e, so passage to the union leaves that weight unchanged. For an edge of greater height neither region changes its base weight. This proves the edgewise assertion along all of p_k, including any initial upper-arm excursion below row zero or lower-arm excursion above it. The same argument applies to p_y. All other weights weakly increase in passing from either individual environment to the union. Every segment of its selected bigeodesic therefore still minimizes: its cost is unchanged and no competitor becomes cheaper.

Only finitely many edges differ between an individual environment and the union. The sum of their finite increases bounds the change of any passage distance, by comparison along a simple minimizing path in the smaller metric. The labels of the retained rays are consequently unchanged by Lemma 2.6.

If e belongs to the first R_0 edges of p_y, both its endpoints satisfy |w_1-y_1|\leq R_0 and h(e)\leq R_0. For every such endpoint,

where (75) was used. Thus e\notin A_k(R). If it is raised in T^{y,R}, its increase in the union over T^{k,R} is at least g. Such an edge cannot lie on p_k: every edge of p_k in this height slab already belongs to A_k(R).□

Proof. Work on the common event specified above. Suppose two marked sites k\ne y have \rho_k=\rho_y. By Lemma 7.5 both selected bigeodesics persist with this label in T^*=T^{k,y,R}, where R=R(M). By Corollary 6.2 their upper and lower Busemann functions in this metric agree; denote the common pair by B^+,B^-. For a vertex u on p_y, the bigeodesic inequality of Lemma 3.1, based first at k on p_k and then at u on p_y, gives

Additivity and antisymmetry therefore imply

Take a raised edge e=\{u,v\} among the first R_0 upper-arm edges of p_y, and orient it toward the upper end, from u to v. Write t_e^* for its weight in the union. Calibration along the bigeodesic gives

Here the last equality holds because the one-edge segment of p_y is itself minimizing.

Let x_N^-,x_N^+ tend respectively to the lower and upper ends of p_k, with k between them. Set

This L_N^* is the cost of an admissible finite walk, so L_N^*\geq D_N^*. As p_k is a bigeodesic, D_N^*=T^*(x_N^-,k)+T^*(k,x_N^+). The signs in the Busemann definition then give, separately,

It follows that

by (76) and (77).

Choose minimizing finite approach paths in T^* from x_N^- to u and from v to x_N^+, and join them by the specified traversal u\to v of e. Call this finite walk W_N. The approach paths may intersect each other, the selected bigeodesics, or e itself; its cost is L_N^* with every traversal counted. By Lemma 7.5, the weight of e in T^{k,R} is its base weight X_e. Thus

On returning from T^* to T^{k,R} every traversal weakly decreases in cost, and the named middle traversal decreases by \Delta. Consequently

Any further traversal of the same edge can only increase this saving. On the other hand, the segment of p_k between x_N^- and x_N^+ has cost D_N^* in both metrics, again by Lemma 7.5. For sufficiently large finite N, (78) gives L_N^*-D_N^*<\Delta. Then (79) makes W_N strictly cheaper in T^{k,R} than that segment, a contradiction. This is a pathwise argument for every marked pair on the common event, and hence proves the assertion simultaneously for the countably many M.□

Proof of Theorem 1.1. Suppose bigeodesics exist with positive probability. Each bigeodesic contains a lattice vertex. Countability of the vertices and the charts in Lemma 3.2, together with the exclusion of their deterministic exceptional labels, yields a chart in which a bigeodesic through some fixed vertex has positive probability. Translate that vertex to zero and use the lattice coordinates of the chart. By Proposition 3.4, its arms are contained in Q^\pm(0,H) for some finite H almost surely. Taking the countable union over integer H gives a deterministic H_0\geq1 for which the event \mathcal B(0) in Section 7 has probability p>0. Removing the countably many deterministic partition endpoints does not change this probability, by the fixed-label backward-finiteness assertion in Theorem 2.3.

Apply Proposition 7.4 for each positive integer M, with the fixed spacing l from (75). We next impose two filters on its marks. The first supplies the base comparator rays required for Lemma 5.1; the second supplies their required separation within each bin.

For a horizontal site k, let \mathcal R_K(k) be the event that, in the unchanged base environment, every ray from k with label in I lies in Q^+(k,K) and every ray from k with label in -I lies in Q^-(k,K). By Proposition 3.4,

Choose one deterministic K\geq1, independent of M, such that this failure probability is at most p/8. It is the same at all horizontal sites by translation invariance. Retain a mark when

Unconditional subtraction, without any independence assertion, gives

This filter is imposed after selecting the paths. Its complement has an existential ray witness, so \mathcal R_K(k) is a universally measurable environment event by Lemma 2.7. It remains a separate filter; it is not placed inside an existential path projection. In particular no new alternation of path quantifiers is used to define the selected marks. The translated rule continues to give equal marginals along the grid.

The near radius R_0, spacing l, and comparator offset K are fixed independently of M. For each M we choose R(M) and the comparison separation below, refine the partition, and only then increase M. Fix M for the moment. Its radius R(M) is a fixed finite number. The selected arms have offset H_0, the base comparator rays have offset K, the changed region has radius at most D_{R(M)} from (74), and the prescribed gain portion has radius at most R(M). The comparison in Lemma 5.1 therefore holds at a finite deterministic horizontal separation S(M). The same S(M) works for every bin of every partition \Pi_j: the cone bounds control all the relevant comparator labels in the chart, and the geometric bounds on the changed region and gain portion are independent of the bin. Increase S(M) if necessary to be nonnegative. Its possible growth with M has no effect on the already fixed grid spacing l.

For the finite partition \Pi_j, retain a site with \widetilde J_M(k)=1 only if there is no other site y with \widetilde J_M(y)=1 and

whose selected label lies in the same bin as \rho_k. Here k,y lie on the horizontal grid, so the displayed distance is their horizontal separation. Denote this final indicator by J_{M,j}(k) and its common marginal probability by q_{M,j}. The test concerns at most 2\lfloor S(M)/l\rfloor other sites and is measurable from the previously selected labels and marks. At a fixed j, two surviving marks in the same bin have separation strictly greater than S(M), so Lemma 5.1 applies to consecutive surviving marks of that bin with gain b=M.

For completeness, the disappearance of this last filter is a pointwise finite-set argument. On the common full-measure event of Section 7.3, fix an outcome with \widetilde J_M(k)=1. By Proposition 7.6, every marked site y\ne k, and in particular every one of the finitely many retained neighbors within distance S(M), has \rho_y\ne\rho_k. The nested partitions eventually put each such pair in different bins. The maximum of their finitely many separating indices is finite, so k survives the last filter for every subsequent partition. If \widetilde J_M(k)=0, all of its final indicators are zero. We have therefore proved

where the probability convergence follows by bounded convergence. No rate of separation of random labels is required.

For a bin (\alpha,\beta)_I of \Pi_j, write q_{M,j}^{\alpha,\beta} for the probability that a given site survives and its selected label lies in that bin. Proposition 5.2, with b=M, gives

Every selected label avoids all partition endpoints, so each surviving mark belongs to exactly one bin. Summing over the finitely many bins and using their total width W_I yields

The label-budget bound uses only equal marginal marking probabilities and the pathwise comparison. In particular the dependencies introduced by the shared raises and both filters are permitted.

Now send j to infinity for this fixed M. Equations (80), (81), and (82) imply

All of p,l,W_I were fixed before M was chosen. Taking any integer M>8lW_I/p is a contradiction.

The reduction to a chart involved countably many vertices and charts. Within the construction the excluded deterministic labels, integer radii, sites, and union metrics form countable collections; equality of the Busemann pairs and distinctness of marked labels hold pathwise for their random labels. Thus the contradiction excludes the existence event itself and proves almost-sure absence of all bigeodesics, with no prescribed deterministic direction.□

Proof. Choose a vertex a after the last exit of the target ray from H. The segment from a to each later v_j minimizes for both metrics, so T_H(a,v_j)=T(a,v_j). Each of the sequences

is decreasing by the triangle inequality along this calibrated tail. They are bounded in absolute value by T_H(x,a) and T(x,a), respectively. These bounds are finite because the half-plane lattice is connected and all edge weights are finite. Their limits therefore exist and are finite. Their difference is nonnegative, since T_H\geq T, and defines D_{\mathcal G}(x).

Any two rays of the family have a common tail, so changing the target ray leaves the limit unchanged. A translation by kz preserves H and replaces the target ray by another translated ray of the same family. This proves covariance, hence stationarity. No choice of the auxiliary anchor a enters the definition of D_{\mathcal G}. Measurability follows from the measurable ray choice and the countable limits defining the distances. Subtracting the two distance limits at x and y proves (83) and the existence of B_{\mathcal G}^H.□

Proof. Set X_k=\min\{\Delta(k,k+1),1\}. Then

Thus A_n\to0 in probability and \EE A_n\leq\eps+\PP(A_n>\eps)\to\eps for every \eps>0. Stationarity gives \EE A_n=\EE X_0 for all n, so X_0=0 almost surely. Stationarity, countability and additivity prove the assertion for every pair. Neither integrability of \Delta nor ergodicity is needed.□

Proof. Let the two families be \mathcal G,\mathcal G', with common functional \rho. By [AH19, Proposition 4.3], their contact sector is contained in the interior of one of the eight half-planes H. Both families eventually enter every translate of H.

We first recall the geometric part of [AH19, Lemma 4.6, Section 4.4], which does not use its subsequent expectation assertion. There is positive probability that the ray from the origin stays in H: on an eventual-entrance ray, a vertex minimizing the normal coordinate starts a tail contained in its translated half-plane. Finite uniqueness and coalescence identify this suffix with the selected ray from that vertex. Countability of the vertices and covariance give positive probability at a fixed vertex. Boundary translations of an iid environment are ergodic, also for diagonal boundaries. Hence there are such starting vertices in both directions along the boundary.

Choose two of these roots on opposite sides of the origin. The two contained rays coalesce, and their portions up to their first common vertex form a finite simple connecting arc. If the origin lies on this arc, its selected suffix is already contained in H. Otherwise, choose consecutive boundary contacts of the arc on opposite sides of the origin. The intervening subarc, together with the boundary segment, encloses a bounded region adjacent to the origin. Choose a vertex w on the common tail of the two bounding rays and the target ray, outside this region and beyond the connecting arc. Every restricted minimizer from the origin to a target beyond w must meet the separating subarc. From that meeting point, the bounding-ray suffix to the target lies in H and is a full-plane minimizer. Restricted uniqueness forces this suffix, hence passage through w. The prefix to w is then the fixed restricted minimizer to w. This proves stabilization and coalescence with the full-plane family, including when the bounding rays revisit the boundary. Applying the construction at every boundary root gives \mathcal G_H,\mathcal G'_H; their common Busemann functions on the boundary are exactly the restricted limits in Lemma A.1.

Suppose that the families differ with positive probability. Because each restricted family coalesces, the event that the two families coalesce is invariant under boundary translations. If they coalesce, finite uniqueness makes their rays from any common root identical. Boundary ergodicity therefore makes distinctness an almost-sure event. The same invariance and ergodicity make their asymptotic planar order deterministic, as in [AH19, Section 4.5]. Orient the boundary vector z so that the finite crossing comparison in equation (12) of that source gives

For clarity, the comparison chooses a crossing of \mathcal G_H(kz) with \mathcal G'_H(lz) and common-tail vertices x of the primed rays and y of the unprimed rays, both beyond the crossing. Swapping the two pieces at the crossing gives walks \pi_x from kz to x and \pi_y from lz to y, and

Both summands are nonnegative finite excess costs. This identity is purely geometric and does not use expectations.

Put Z(x)=D_{\mathcal G}(x)-D_{\mathcal G'}(x). Equation (83) gives

The first difference divided by n tends to zero almost surely, by the full-plane linearity and the equality of the two functionals. For each \eps>0, finiteness and stationarity give

Consequently \Delta(0,n)/n\to0 in probability. The process \Delta is additive and stationary under boundary shifts, so Lemma A.2 and (84) imply \Delta(k,l)=0 for every pair, almost surely.

The final uniqueness argument is the finite one in [AH19, Section 4.5]. Distinct same-root restricted rays have a finite common initial portion. Choose a deterministic bound on its length that holds with positive probability; boundary ergodicity gives two such roots farther apart than twice this bound. In the crossing comparison just described, at least one swapped walk then differs from the corresponding minimizing path. Indeed, if both swapped walks were their unique geodesics, their crossing vertex would lie in the common initial portion from each root. If these portions have at most m edges, the lattice distance between the roots would be at most 2m, contrary to their choice. Since both nonnegative excesses sum to zero, a distinct swapped walk is also minimizing, contrary to uniqueness. Thus the restricted families agree. Their coalescence with the full-plane families and finite uniqueness imply \mathcal G=\mathcal G'.□