Proofs of remaining lemmas

a common exceptional set of ¯ ν λ,ρ -probability 0 for all r ∈ [ λ, ρ] ∩ R. This proves 52. ƒ Proof of Corollary 3.1. i. By Proposition 3.1 cf. [8, 28], d d v [Gh c v − vh c v] = −h c v weakly with respect to v. Thus, setting u = R λ,ρ , we have [Guv, 1 − vuv, 1] − [Guw, 1 − wuw, 1] = Z w v ux, 1d x 70 for all v, w ∈ R. Let a b in R. Setting ̟ = T € τ [N x ] η N s η, ω, τ [N x ] η N s ξ, ω Š we have β N N t θ ′ [N x ],N s η, ξ, ωa, b] = N −1 X [N a]x≤[N b] η N t ̟, θ [N x ],N s ωx = tN t −1 N a t,bt N t θ ′ [N x ],N s ω ′ Thus, by Proposition 3.2 and 70, lim N →∞ β N N t θ ′ [N x ],N s ω ′ a, b] = t Z b t a t ux, 1d x = Z b a ux, td x 71 ¯ ν λ,ρ ⊗ IP-a.s., where the last equality follows from Proposition 3.1. Now 70 implies that the r.h.s. of 30 is a continuous function of v, w, while the l.h.s. is a uniformly Lipschitz function of v, w, since the number of particles per site is bounded. It follows that one can find a single exceptional set of ¯ ν λ,ρ ⊗IP-probability 0 outside which 71 holds simultaneously for all a, b, which proves the claim. ii. Since η λ,ρ has distribution ¯ ν λ,ρ under IP , the statement follows from i with x = s = 0. ƒ

3.4 Proofs of remaining lemmas

Proof of Lemma 3.2. Let ǫ 0. We consider the probability space Ω ′ × Z + Z equipped with the product measure IP ′ ǫ := ¯ ν λ,ρ ⊗ IP ⊗ P ǫ where P ǫ is the product measure on Z whose marginal at each site is Poisson with mean K1 + ǫ. A generic element of this space is denoted by ω ′ , χ, with ω ′ = η, ξ, ω and χ ∈ Z + Z . We first prove 40. Because of 26 that is, coupled configurations are ordered under ¯ ν λ,ρ , by the attractiveness property 15, we may define γ s ω ′ := η s ξ, ω − η s T η, ξ, ω 72 19 for s ≥ 0. Therefore γ-particles represent the discrepancies between the system starting from ξ and the system starting from T η, ξ. We set pz = pz + p−z||b|| ∞ 73 and fix ¯ v such that ¯ v X z∈Z z pz 74 Let v ¯ v. Because γ x = 0 for all x 0, by the definition of current 32, φ v t ξ, ξ, ω = φ v t ω ′ + X y [v t] γ t y Therefore, to prove 40, we want to obtain lim t→∞ t −1 X y ¯ v t γ t ◦ θ ′ [β t],αt ω ′ y = 0, ¯ ν λ,ρ ⊗ IP-a.s. 75 To this end, we follow the proof of [1, Proposition 5], with minor modifications. We emphasize that even if the latter proof corresponds to α = β = 0, we will see that the arguments extend to α, β 6= 0, 0. Each γ -particle and χ-particle is given an integer label j ≤ 0, and we denote by R j resp. Z j the position of the γ -particle resp. χ-particle with label j. That is, R . and Z . must satisfy X j≤0 δ R j = X x∈Z γ xδ x , X j≤0 δ Z j = X x∈Z χxδ x 76 There is a unique way to choose such R . resp. Z . if we impose moreover that R j ≤ R j+1 resp.Z j ≤ Z j+1 for all j 0, which we will assume. By the definition of χ, the number of χ-particles between −n and 0 will be eventually larger than nK. Let W χ := inf n n ∈ Z + : Z j ≥ −[ j K] for every j ≤ −n o By Poisson large deviation bounds, the random variable W is IP ′ ǫ -a.s. finite with exponentially de- caying distribution. Since γ x ≤ K for every x ∈ Z, we have R j ≤ −[ j K] for all j ≤ 0, hence Z j ≥ R j for every j ≤ −W χ. The dynamics of R . t t≥0 is defined as follows. By 72 and the graph- ical construction, at most one discrepancy can jump at each event time of ω. If a γ-particle jumps at time t from x to y, we set R j t = y and R k t = R k t − for all k 6= j, where j is the largest label l such that R l t − = x. The choice of the largest label is arbitrary and could be replaced by any deterministic or random choice procedure, provided the corresponding source of randomness is independent of ω ′ , χ. Note that we do not in general have R j t ≤ R j+1 t for t 0, but this does not matter for our purpose. The dynamics of Z j . is defined as follows: if R j t − = x and, for some z 0 and u ∈ [0, 1], {t, x, z, u, t, x + z, −z, u} ∩ ω 6= ; 77 then Z j t = Z j t − + z. In other words, χ-particles evolve as mutually independent given their initial positions random walks, that jump from y to y + z at rate pz given by 73 for all y ∈ Z, z ≥ 0. Since a jump for R j . from R j t − = x to R j t = x + z is possible only under 77, R j ≤ Z j ⇒ ∀t 0, R j t ≤ Z j t 78 20 In view of 75, 78, we estimate X x ¯ v t γ t x = X j≤0 1 {R j t ω ′ ¯ v t} = X j≤−W χ 1 {R j t ω ′ ¯ v t} + X −W χ j≤0 1 {R j t ω ′ ¯ v t} ≤ X j≤−W χ 1 {Z j t ω ′ ¯ v t} + W χ ≤ ¯ Z t ω, χ + W χ 79 where ¯ Z t := P i≤0 1 {Z i t ¯ v t} is a Poisson random variable with mean IE ′ ǫ ¯ Z t = K1 + ǫ X j≥¯ v t IP ′ ǫ Y t j 80 for Y t a random walk starting at 0 that jumps from y to y + z with rate pz. Since ¯ v satisfies 74, repeating [1, 43–45] gives that lim t→∞ IE ′ ǫ ¯ Z t t = 0 81 Let δ 0. Since ¯ Z t is a Poisson variable, IP ′ ǫ | ¯ Z t − IE ′ ǫ ¯ Z t | δt ≤ IE ′ ǫ ¯ Z t − IE ′ ǫ ¯ Z t 4 δt 4 ≤ [IE ′ ǫ ¯ Z t − IE ′ ǫ ¯ Z t 2 ] 2 δt 4 ≤ IE ′ ǫ ¯ Z t t 2 t 2 δt 4 82 Therefore, by Borel Cantelli lemma lim t→∞ ¯ Z t − IE ′ ǫ ¯ Z t t = 0, IP ′ ǫ -a.s. 83 Since IP is invariant by θ [β t],αt , ¯ Z t θ [β t],αt ω, χ has the same distribution as ¯ Z t ω, χ under IP ′ ǫ . Thus 80–83 still hold with ¯ Z t θ [β t],αt ω, χ instead of ¯ Z t ω, χ, and lim t→∞ t −1 ¯ Z t θ [β t],αt ω, χ = 0, IP ′ ǫ -a.s. 84 Because the random variable W in 79 does not depend on ω ′ , 84 and 79 imply 75. This concludes the proof of 40. If we now define γ s as γ s ω ′ := η s T η, ξ, ω − η s η, ω and replace φ v t ω ′ by −φ v t ω ′ so that the current, which was rightwards, becomes leftwards, then the proof of 41 can be obtained by repeating the same steps as in the previous argument. ƒ Proof of Lemma 3.3. 21

i. Since