By Anon.

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A. ω. We then have that h ◦ t12 − 1 h ◦ t1 + ρ −1 h = 0, 1−q where h := pφ and where we have written p = p(0, ω) and q = q(0, ω). If we now define h := h ◦ t1 − ρ −1 h, we conclude that for every x ∈ Z, h ◦ tx − h = 0. s. equal to a constant C. Assume that C = 0. Then h = 0 is equivalent to h(t1 ω) = ρ −1 (ω)h(ω). We claim that the only solution in this case is h = 0. Indeed, using induction on n we have that n−1 ρ −1 (tj ω). s. lim h(tn ω) = 0. n→∞ Now integrating with respect to P, using its stationarity and the fact that h(tn ω), n ∈ N, are uniformly integrable, we conclude that h(ω)P(dω) = lim n→∞ h(tn ω)P(dω) = 0, 34 A.

S. R1B = 1B . s. it is true that 1B (ω0 ) = Eν [1B (ω1 ) | ω0 ] = R1B (ω0 ). Since Pν [A] = ν[B], it is then enough to prove that ν[B] ∈ {0, 1}. s. we have that 1B (ω) = R1B (ω) = ω(0, e)1B (te ω). |e|1 =1 Using ellipticity, this implies that P[B] ∈ {0, 1}, which again by part 1 of this theorem implies (26). Proof of parts 3 and 4 Let g be a bounded measurable function on . Let ν be any invariant probability measure for the transition kernel R that is absolutely continuous with respect to P. s.

Mk But for each n ∈ N, we have that − 1 1 1 log ph(mnk) (0, mny) ≤ − log ph(mnk) (0, z) − log ph(mnk) (z, mny). mnk mnk mnk By uniform ellipticity, the first term in the right-hand side of the above inequality tends to 0 as m → ∞. Therefore, I (x, ω) = − lim m→∞ 1 1 log ph(mnk) (0, mny) ≤ − lim inf log ph(mnk) (z, mny). m→∞ mnk mnk On the other hand, 1 log ph(mnk) (z, mny) mnk 1 1 ≤− log ph((m−1)nk) (z, (m − 1)ny + z) − log ph(nk−1) ((m − 1)ny + z, mny). mnk mnk − Now, since z ∈ U , one can check that ph(nk−1) ((m − 1)ny + z, mny) ≥ κ nk−1 , so that the last term of the above inequality tends to 0 when m → ∞.

### Actes Du Congres International Des Mathematiciens 1970 by Anon.

by Ronald

4.5