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By A. Jaffe (Chief Editor)

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1. Let us remark that for any k ≥ 1 we have µ(τU = k) = µ(U ∩ {τU > k − 1}). (3) Since {τU > k} = T −1 (U c ∩ {τU > k − 1}) by the invariance of µ we get that µ(τU > k) = µ(τU > k − 1) − µ(U ∩ {τU > k − 1}), whence the result. Next, for all k > 0 we have µ(τU > k) = µ(τU > k − 1) − µ(U )µU (τU > k − 1) = µ(τU > k − 1) − µ(U )[µ(τU > k − 1) + c(k, U )] = µ(τU > k − 1)[1 − µ(U )] − µ(U )c(k, U ). Then it follows by an immediate induction that k k c(j, U )(1 − µ(U ))k−j . µ(τU > k) = (1 − µ(U )) − µ(U ) j =1 Hence for all t ≥ 0, putting kt = [t/µ(U )], we have kt µ(τU > kt ) − (1 − µ(U ))kt ≤ µ(U ) |c(j, U )| ≤ tc(U ).

Let U ⊂ X a measurable set. The following estimate holds: c(U ) ≤ inf { aN (U ) + bN (U ) + N µ(U )| N ∈ N}, where the quantities are defined by N T −j U ) = µU (τU ≤ N ), aN (U ) = µU ( j =1 bN (U ) = sup |µU (T −N V ) − µ(V )| V ∈U∞ with U = {U, U c }, Un = n−1 −k U k=0 T and U∞ = ∪n σ (Un ). Statistics of Return Times 39 Proof. Let N ∈ N. If k < N, we just bound c(k, U ) by |µU (τU > k) − µ(τU > k)| = |µU (τU ≤ k) − µ(τU ≤ k)| ≤ |µU (τU ≤ k)| + |µ(τU ≤ k)| ≤ aN (U ) + kµ(U ) ≤ aN (U ) + N µ(U ).

Since τU takes only integer values, the distribution for t = kµ(U ) and t = (k + 1/2)µ(U ) is the same, then εk,U ≥ |exp(−kµ(U )) − exp(−(k + 1/2)µ(U ))| ≥ exp(−kµ(U ))(1 − e−µ(U )/2 ) ≥ e−kµ(U ) µ(U ). 1. Let us remark that for any k ≥ 1 we have µ(τU = k) = µ(U ∩ {τU > k − 1}). (3) Since {τU > k} = T −1 (U c ∩ {τU > k − 1}) by the invariance of µ we get that µ(τU > k) = µ(τU > k − 1) − µ(U ∩ {τU > k − 1}), whence the result. Next, for all k > 0 we have µ(τU > k) = µ(τU > k − 1) − µ(U )µU (τU > k − 1) = µ(τU > k − 1) − µ(U )[µ(τU > k − 1) + c(k, U )] = µ(τU > k − 1)[1 − µ(U )] − µ(U )c(k, U ).

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