No temporal distributional limit theorem for a.e. irrational translation
NNO TEMPORAL DISTRIBUTIONAL LIMIT THEOREMFOR A.E. IRRATIONAL TRANSLATION
DMITRY DOLGOPYAT AND OMRI SARIG
Abstract.
Bromberg and Ulcigrai constructed piecewise smoothfunctions on the torus such that the set of α for which the sum (cid:80) n − k =0 f ( x + kα mod 1) satisfies a temporal distributional limit the-orem along the orbit of a.e. x has Hausdorff dimension one. Weshow that the Lebesgue measure of this set is equal to zero. Introduction and statement of main result
Background.
Suppose T : X → X is a map, f : X → R is afunction, and x ∈ X is a fixed initial condition. We say that the T –ergodic sums S n = f ( x ) + f ( T x ) + · · · + f ( T n − x ) satisfy a temporaldistributional limit theorem (TDLT) on the orbit of x , if there exists anon-constant real valued random variable Y , centering constants A N ∈ R and scaling constants B N → ∞ s.t. S n − A N B N −−−→ N →∞ Y in distribution , (1.1)when n is sampled uniformly from { , . . . , N } and x is fixed. Equiva-lently, for every Borel set E ⊂ R s.t. P ( Y ∈ ∂E ) = 0,1 N Card { ≤ n ≤ N : S n − A N B N ∈ E } −−−→ N →∞ P ( Y ∈ E ) . We allow and expect A N , B N , Y to depend on T, f, x .Such limit theorems have been discovered for several zero entropyuniquely ergodic transformations, including systems where the moretraditional spatial limit theorems, with x is sampled from a measureon X , fail [Bec10, Bec11, ADDS15, DS17, PS, DSa]. Of particularinterest are TDLT for R α : [0 , → [0 , , R α ( x ) = x + α mod 1 , f β ( x ) := 1 [0 ,β ) ( x ) − β, Mathematics Subject Classification. a r X i v : . [ m a t h . D S ] M a r DMITRY DOLGOPYAT AND OMRI SARIG because the R α –ergodic sums of f β along the orbit of x represent thediscrepancy of the sequence x + nα mod 1 with respect to [0 , β ) [Sch78,CK76, Bec10]. Another source of interest is the connection to the“deterministic random walk” [AK82, ADDS15].The validity of the TDLT for R α and f β depends on the diophantineproperties of α and β . Recall that α ∈ (0 ,
1) is badly approximable if for some c > | qα − p | ≥ c/ | q | for all irreducible fractions p/q .Equivalently, the digits in the continued fraction expansion of α arebounded [Khi63]. Say that β ∈ (0 ,
1) is badly approximable with respectto α if for some C > | qα − β − p | > C/ | q | for all p, q ∈ Z , q (cid:54) = 0 . If α is badly approximable then every β ∈ Q ∩ (0 ,
1) is badly approximablewith respect to α . The recent paper [BU] shows: Theorem 1.1 (Bromberg & Ulcigrai) . Suppose α is badly approximableand β is badly approximable with respect to α , e.g. β ∈ Q ∩ (0 , .Then the R α -ergodic sums of f β satisfy a temporal distributional limittheorem with Gaussian limit on the orbit of every initial condition. The set of badly approximable α has Hausdorff dimension one [Jar29],but Lebesgue measure zero [Khi24]. This leads to the following ques-tion: Is there a β s.t. the R α –ergodic sums of f β satisfy a temporaldistributional limit theorem for a.e. α and a.e. initial condition? In this paper we answer this question negatively.1.2.
Main result.
To state our result in its most general form, weneed the following terminology.Let T := R / Z . We say that f : T → R is piecewise smooth if thereexists a finite set S ⊂ T s.t. f is continuously differentiable on T \ S and ∃ ψ : T → R with bounded variation s.t. f (cid:48) = ψ on T \ S . Forexample: f β ( x ) = 1 [0 ,β ) ( x ) − β (take S = { , β } , ψ ≡ Theorem 1.2.
Let f be a piecewise smooth function of zero mean.Then there is a set of full measure E ⊂ T × T s.t. if ( α, x ) ∈ E thenthe R α –ergodic sums of f do not satisfy a TDLT on the orbit of x . The condition (cid:82) T f = 0 is necessary: By Weyl’s equidistribution theo-rem, for every α (cid:54)∈ Q , f Riemann integrable s.t. (cid:82) T f = 1, and x ∈ T , S n /N dist −−−→ N →∞ U[0 ,
1] as n ∼ U(1 , . . . , N ). See § f ( x ) = { x } − . Unlike the proof givenbelow, [DSb] does not identify the set of α where the TDLT fails, but itdoes give more information on the different scaling limits for the distri-butions of S n , n ∼ U(1 , . . . , N k ) along different subsequences N k → ∞ .[DSb] also shows that if we randomize both n and α by sampling ( n, α ) O TEMPORAL DLT FOR A.E. IRRATIONAL TRANSLATION 3 uniformly from { , . . . , N } × T , then (cid:0) S n − N (cid:80) Nk =1 S k (cid:1) / √ ln N con-verges in distribution to the Cauchy distribution.The methods of [DSb] are specific for f ( x ) = { x } − , and we do notknow how to apply them to other functions such as f β ( x ) = 1 [0 ,β ) ( x ) − β .1.3. The structure of the proof.
Suppose f is piecewise smoothand has mean zero.We shall see below that if f is continuous, then for a.e. α , f isan R α -coboundry, therefore S n are bounded, hence (1.1) cannot holdwith B N → ∞ , Y non-constant. We remark that (1.1) does hold with B N ≡ A N = f ( x ), Y = distribution of minus the transfer function,but this is not a TDLT since no actual scaling is involved.The heart of the proof is to show that if f is discontinuous, then fora.e. α , the temporal distributions of the ergodic sums have differentasymptotic scaling behavior on different subsequences. The proof ofthis has three independent parts:(1) A reduction to the case f ( x ) = d (cid:80) m =1 b m h ( x + β m ), h ( x ) := { x } − . (2) A proof that if N ⊂ N has positive lower density, then there exists M ≥ , A ( N , M ) := (cid:26) α ∈ (0 ,
1) : ∃ n k ↑ ∞ , r k ≤ M s.t. for all k : r k q n k ∈ N , a n k +1 / ( a + · · · + a n k ) → ∞ (cid:27) . Here a n and q n are the partial quotients and principal denominatorsof α , see § N = N ( b , . . . , b d ; β , . . . , β d ) ⊆ N with positivedensity, s.t. for every α ∈ A ( N , M ) and a.e. x , one can analyzethe temporal distributions of the Birkhoff sums of d (cid:80) m =1 b m h ( x + β m ).1.4. Notation. n ∼ U(1 , . . . , N ) means that n is a random variabletaking values in { , . . . , N } , each with probability N . U[ a, b ] is theuniform distribution on [ a, b ]. Lebesgue’s measure is denoted by mes. N = { , , , . . . } and N = N ∪ { } . If x ∈ R , then (cid:107) x (cid:107) := dist( x, Z )and { x } is the unique number in [0 ,
1) s.t. x ∈ { x } + Z . Card( · ) is thecardinality. If ε >
0, then a = b ± ε means that | a − b | ≤ ε . DMITRY DOLGOPYAT AND OMRI SARIG Reduction to the case f ( x ) = d (cid:80) m =1 b m h ( x + β m )Let h ( x ) = { x } − , and let G denote the collection of all non-identically zero functions of the form f ( x ) = d (cid:80) m =1 b m h ( x + β m ) , where d ∈ N , b i , β i ∈ R . We explain how to reduce the proof of Theorem 1.2from the case of a general piecewise smooth f ( x ) to the case f ∈ G .The following proposition was proved in [DSb]. Let C ( T ) denote thespace of continuous real-valued functions on T with the sup norm. Proposition 2.1. If f ( t ) is differentiable on T \ { β , . . . , β d } and f (cid:48) extends to a function with bounded variation on T , then there are d ∈ N , b , . . . , b d ∈ R s.t. for a.e. α ∈ T there is ϕ α ∈ C ( T ) s.t. f ( x ) = d (cid:88) i =1 b i h ( x + β i ) + (cid:90) T f ( t ) dt + ϕ α ( x ) − ϕ α ( x + α ) ( x (cid:54) = β , . . . , β d ) . The following proposition was proved in [DS17]. Let (Ω , B , µ ) be aprobability space, and let T : Ω → Ω be a probability preserving map.
Proposition 2.2.
Suppose f = g + ϕ − ϕ ◦ T µ -a.e. with f, g, ϕ : Ω → R measurable. If the ergodic sums of g satisfy a TDLT along the orbit ofa.e. x , then so do the ergodic sums of f . These results show that if Theorem 1.2 holds for every f ∈ G , thenTheorem 1.2 holds for any discontinuous piecewise smooth functionwith zero mean. As for continuous piecewise smooth functions withzero mean, these are R α -cohomologous to g ≡ α becausethe b i in Proposition 2.1 must all vanish. Since the zero function doesnot satisfy the TDLT, continuous piecewise smooth functions do notsatisfy a TDLT. 3. The set A has full measure Statement and plan of proof.
Let α be an irrational number,with continued fraction expansion [ a ; a , a , a , . . . ] := a + 1 a + · · · , a ∈ Z , a i ∈ N ( i ≥ a n the quotients of α . Let p n /q n denotethe principal convergents of α , determined recursively by q n +1 = a n +1 q n + q n − , p n +1 = a n +1 p n + p n − and p = a , q = 1; p = 1 + a a , q = a . We call q n the principaldenominators and a i the partial quotients of α . Sometimes – but notalways! – we will write q k = q k ( α ), p k = p k ( α ), a k = a k ( α ). O TEMPORAL DLT FOR A.E. IRRATIONAL TRANSLATION 5
Given
N ⊂ N and M ≥
1, let A = A ( N , M ) ⊂ (0 ,
1) denote the setof irrational α ∈ (0 ,
1) s.t. for some subsequence n k ↑ ∞ , ∃ r k ≤ M s.t. r k q n k ∈ N , a n k +1 ( a + · · · + a n k ) −−−→ k →∞ ∞ . (3.1)The lower density of N is d ( N ) := lim inf N Card(
N ∩ [1 , N ]). Thepurpose of this section is to prove: Theorem 3.1.
If a set N has positive lower density then there exists M such that A ( N , M ) has full Lebesgue measure in (0 , . The proof consists of the following three lemmas:
Lemma 3.2.
For almost all α there is n = n ( α ) s.t. if k ≥ n and a k +1 > k (ln k )(ln ln k ) , then a k +1 / ( a + · · · + a k ) ≥ ln ln k. Lemma 3.3.
Suppose α ∈ (0 , \ Q and ( p, q ) ∈ N × N satisfy gcd( p, q ) = 1 and | qα − p | ≤ qL where L ≥ . Then there exists k s.t. q = q k ( α ) and a k +1 ( α ) ≥ L . Lemma 3.4.
Suppose ψ : R + → R is a non-decreasing function s.t. (cid:88) n nψ ( n ) = ∞ . (3.2) Suppose
N ⊂ N has positive lower density. For all M sufficiently large,for a.e. α ∈ (0 , there are infinitely many pairs ( m, n ) ∈ N × N s.t. n ∈ N , gcd( m, n ) ≤ M , and | nα − m | ≤ nψ ( n ) . Remark 1.
By the monotonicity of ψ , if e k − < n < e k then ψ (cid:0) e k − (cid:1) ≤ ψ ( n ) ≤ ψ (cid:0) e k (cid:1) . Hence (3.2) holds iff (cid:80) ψ ( e k ) = ∞ . Remark 2. If N = N , then Lemma 3.4 holds with M = 1 by theclassical Khinchine Theorem. We do not know if Lemma 3.4 holdswith M = 1 for any set N with positive lower density. Proof of Theorem given Lemmas – . We apply these lemmaswith ψ ( t ) = c (ln t ) (ln ln t ) (ln ln ln t ) and c > / ln( √ ).Fix M > ∃ Ω ⊂ (0 ,
1) of full measure s.t.for every α ∈ Ω there are infinitely many ( m, n ) ∈ N × N as follows.Let m ∗ := m/ gcd( m, n ) , n ∗ := n/ gcd( m, n ) , p := gcd( m, n ), then(1) pn ∗ ∈ N , p ≤ M , | n ∗ α − m ∗ | = | nα − m | p ≤ n ∗ ψ ( n ∗ ) ( ∵ n ∗ ≤ n );(2) ∃ k s.t. n ∗ = q k ( α ) and a k +1 ( α ) ≥ ψ ( q k ) ( ∵ Lemma 3.3). By itsrecursive definition, q k ≥ k -th Fibonacci number ≥ ( √ ) k . Sofor all k large enough, a k +1 ( α ) ≥ ψ ( q k ) > k (ln k )(ln ln k ); DMITRY DOLGOPYAT AND OMRI SARIG (3) a k +1 / ( a + · · · + a k ) ≥ ln ln k → ∞ ( ∵ Lemma 3.2).So every α ∈ Ω belongs to A = A ( N , M ), and A has full measure. (cid:3) Next we prove Lemmas 3.2–3.4.3.2.
Proof of Lemma 3.2.
By [DV86], for almost every α ( a + · · · + a k +1 ) − max j ≤ k +1 a j k ln k → < . So if k is large enough, and a k +1 > k (ln k )(ln ln k ) thenmax j ≤ k +1 a j = a k +1 , a + · · · + a k k ln k ≤ , and a k +1 a + · · · + a k >
18 ln ln k. (cid:3) Proof of Lemma 3.3.
For every ( p, q ) as in the lemma, | qα − p | < q . A classical result in the theory of continued fractions [Khi63, Thm19] says that in this case ∃ k s.t. q = q k ( α ) , p = p k ( α ).To estimate a k +1 = a k +1 ( α ) we recall the following facts, valid forthe principal denominators of any irrational α ∈ (0 ,
1) [Khi63]:(a) | q k α − p k | > q k + q k +1 ;(b) q k +1 + q k < ( a k +1 + 2) q k , whence by (a) a k +1 > q k | q k α − p k | − | q k α − p k | = | qα − p | ≤ q k L , so a k +1 > L − ≥ L . (cid:3) Preparations for the proof of Lemma 3.4.
Let (Ω , F , P ) bea probability space, and A k ∈ F be measurable events. Given D > A k are D -quasi-independent , if P ( A k ∩ A k ) ≤ D P ( A k ) P ( A k ) for all k (cid:54) = k . (3.3)The following proposition is a slight variation on Sullivan’s Borel–Cantelli Lemma from ([Sul82]): Proposition 3.5.
For every D ≥ there exists a constant δ ( D ) > such that the following holds in any probability space:(a) If A k are D -quasi-independent measurable events s.t. lim k →∞ P ( A k ) =0 but (cid:80) k P ( A k ) = ∞ , then P ( A k occurs infinitely often ) ≥ δ ( D ) .(b) The quasi-independence assumption in (a) can be weakened to theassumption that for some r ∈ N , P ( A k ∩ A k ) ≤ D P ( A k ) P ( A k ) for all | k − k | ≥ r. (c) One can take δ ( D ) = D .Proof. Since P ( A k ) → (cid:80) P ( A k ) = ∞ , there is an increasingsequence N j such that lim j →∞ N j +1 (cid:88) k = N j +1 P ( A k ) = 1 D .
O TEMPORAL DLT FOR A.E. IRRATIONAL TRANSLATION 7
Let B j be the event that at least one of events { A k } N j +1 k = N j +1 occurs.Since B j = (cid:85) N j +1 k = N j +1 (cid:0) A k \ (cid:83) k − j = N j +1 A j (cid:1) , P ( B j ) ≥ N j +1 (cid:88) k = N j +1 P ( A k ) − (cid:88) N j +1 ≤ k
Let E k be measurable sets in a finite measure space.If the multiplicity of { E k } is less than K , then mes (cid:32)(cid:91) k E k (cid:33) ≥ K (cid:88) k mes( E k ) . Proof. (cid:83) E i ≥ K (cid:80) E i almost everywhere. (cid:3) Proposition 3.7.
For every non-empty open interval I ⊂ [0 , , Card { ( m, n ) ∈ { , . . . , N } : mn ∈ I , gcd( m, n ) = 1 } ∼ I ) N /π ,as N → ∞ .Proof. This classical fact due to Dirichlet follows from the inclusion-exclusion principle and the identity ζ (2) = π /
6, see [HW08, Theorem459]. (cid:3)
DMITRY DOLGOPYAT AND OMRI SARIG
Proposition 3.8.
Suppose α = [0; a , a , . . . ] and α = [0; a (cid:96) +1 , a (cid:96) +2 , . . . ] .Then the principal convergents p (cid:96) /q (cid:96) of α and the principal convergents p (cid:96) /q (cid:96) of α are related by (cid:18) p l + l p l + l +1 q l + l q l + l +1 (cid:19) = (cid:18) p l − p l q l − q l (cid:19) (cid:18) p l p l +1 q l q l +1 (cid:19) Proof.
Since a = 0, the recurrence relations for p n /q n imply (cid:18) p n p n +1 q n q n +1 (cid:19) = (cid:18) p n − p n q n − q n (cid:19) (cid:18) a n +1 (cid:19) , (cid:18) p p q q (cid:19) = (cid:18) a (cid:19) . So (cid:18) p n p n +1 q n q n +1 (cid:19) = (cid:18) a (cid:19) · . . . · (cid:18) a n +1 (cid:19) . It follows that (cid:18) p l + l p l + l +1 q l + l q l + l +1 (cid:19) = (cid:18) p l − p l q l − q l (cid:19) (cid:18) p l p l +1 q l q l +1 (cid:19) , where p i /q i are theprincipal convergents of α := [0; a l +1 , a l +2 , . . . ]. (cid:3) Proof of Lemma 3.4.
Without loss of generality, lim t →∞ ψ ( t ) = ∞ ,otherwise replace ψ ( t ) by the bigger monotone function ψ ( t ) + ln t .Fix M >
1, to be determined later. LetΩ k := { ( m, n ) ∈ N : n ∈ N , n ∈ [ e k − , e k ] , < m < n, gcd( m, n ) ≤ M } ,A m,n,k := { α ∈ T : | nα − m | ≤ e k ψ ( e k ) } , A k := (cid:91) ( m,n ) ∈ Ω k A m,n,k , A := { α ∈ T : α belongs to infinitely many A k } . The lemma is equivalent to saying that A has full Lebesgue measurefor a suitable choice of M .We will prove a slightly different statement. Fix ε > I ⊂ [ ε, − ε ], letΩ k ( I ) := { ( m, n ) ∈ Ω k : mn ∈ I }A k ( I ) := (cid:91) ( m,n ) ∈ Ω k ( I ) A m,n,k A ( I ) := { α ∈ T : α belongs to infinitely many A k ( I ) } . We will prove that there exists a positive constant δ = δ ( ε, M ) s.t. forall intervals I ⊂ [ ε, − ε ], mes( A ( I ) ∩ I ) ≥ δ mes( I ). It then follows bya standard density point argument (see below) that A ∩ [ ε, − ε ] hasfull measure. Since ε is arbitrary, the lemma is proved. Claim 1.
There exist K = K ( ε ) s.t. for every k > K , the multiplicityof { A m,n,k } ( m,n ) ∈ Ω k ( I ) is uniformly bounded by M . O TEMPORAL DLT FOR A.E. IRRATIONAL TRANSLATION 9
Proof:
Suppose ( m i , n i ) ∈ Ω k ( I ) and A m ,n ,k ∩ A m ,n ,k (cid:54) = ∅ . Thenthere is α s.t. | n i α − m i | ≤ δ k := e k ψ ( e k ) . Choose K = K ( ε ) so largethat k > K ⇒ δ k < ε e k .If k > K , then α ≥ m i n i − δ k > min I − ε > ε . Let r i := gcd( m i , n i ) and( n ∗ i , m ∗ i ) := r i ( n i , m i ). Then | n ∗ i α − m ∗ i | ≤ δ k and m ∗ i ≤ n ∗ i ≤ n i ≤ e k ,so | n ∗ m ∗ − n ∗ m ∗ | = α | m ∗ ( n ∗ α − m ∗ ) − m ∗ ( n ∗ α − m ∗ ) | ≤ e k δ k ε/ < . So n ∗ m ∗ = n ∗ m ∗ . Since gcd( n ∗ i , m ∗ i ) = 1, ( n ∗ , m ∗ ) = ( n ∗ , m ∗ ). It followsthat ( n , m ) ∈ { ( rn ∗ , rm ∗ ) : r = 1 , . . . , M } . So the multiplicity of { A m,n,k } ( m,n ) ∈ Ω k ( I ) is uniformly bounded by M . Claim 2.
Let d ( N ) := lim inf N Card(
N ∩ [1 , N ]) >
0, then thereexists M = M ( N ) and (cid:101) K = (cid:101) K ( ε, N , | I | ) s.t. for all k > (cid:101) K , d ( N )mes( I )4 M ψ ( e k ) ≤ mes( A k ( I )) ≤ I ) ψ ( e k ) . (3.4)In particular, mes( A k ( I )) −−−→ k →∞ (cid:80) mes( A k ( I )) = ∞ . Proof: mes( A m,n,k ) = mes (cid:0)(cid:2) mn − r m,n , mn + r m,n (cid:3)(cid:1) = 2 r m,n where r m,n = ne k ψ ( e k ) . Since n ∈ [ e k − , e k ],Card(Ω k ( I )) M e k ψ ( e k ) ≤ mes (cid:0) A k ( I ) (cid:1) ≤ e Card(Ω k ( I )) e k ψ ( e k ) , (3.5)where the lower bound uses Claim 1 and Proposition 3.6.Card(Ω k ( I )) satisfies the bounds A − B ≤ Card(Ω k ( I )) ≤ A where A := Card { ( m, n ) : n ∈ N , n ∈ [ e k − , e k ] , mn ∈ I } B := Card { ( m, n ) : n ∈ N , n ∈ [ e k − , e k ] , mn ∈ I, gcd( m, n ) ≥ M } . Choose (cid:101) K = (cid:101) K ( ε, N , | I | ) > K ( ε ) s.t. for all k > (cid:101) K (a) Card { n ∈ N : 0 ≤ n ≤ e k } ≥ √ d ( N )(b) Card { n ∈ [ e k − , e k ] ∩ N : p | n } ≤ e k − e k − ) /p for all p ≥ n > e (cid:101) K − , p ≥ np √ I ) ≤ Card { m ∈ N : mn ∈ I, p | m } ≤ np mes( I ) . If k > (cid:101) K , then d ( N ) e k mes( I ) ≤ A ≤ e k mes( I ) and B ≤ ∞ (cid:88) p = M Card { ( m, n ) : n ∈ [ e k − , e k ] , mn ∈ I, p | m, p | n } ≤ ∞ (cid:88) p = M e k − e k − ) p · e k mes( I ) p < e k mes( I ) ∞ (cid:88) p = M p ≤ d ( N ) e k mes( I ) , provided we choose M s.t. ∞ (cid:88) p = M p − < d ( N ).Together we get d ( N ) e k mes( I ) ≤ Card(Ω k ( I )) ≤ e k mes( I ). Theclaim now follows from (3.5). Claim 3.
There exists D = D ( N , M ), r = r ( M ), and ˆ K = ˆ K ( ε, N , I )s.t. for all k , k > ˆ K s.t. | k − k | > r ( M ),mes( A k ( I ) ∩ A k ( I ) | I ) ≤ D mes( A k ( I ) | I )mes( A k ( I ) | I ) (3.6) Proof:
By Claim 2, if k , k are large enough, thenmes( A k ( I ) | I )mes( A k ( I ) | I ) ≥ (cid:18) d ( N )5 M (cid:19) ψ ( e k ) ψ ( e k ) , (3.7)where we put 5 instead of 4 in the denominator to deal with edge effectsarising from mes( A k ( I ) \ I ) = O (cid:16) e k ψ ( e k ) (cid:17) .To prove the claim, it remains to bound mes (cid:16) A k ( I ) ∩ A k ( I ) (cid:12)(cid:12)(cid:12) I (cid:17) from above by const R R , where R i := ψ ( e k i ).A cylinder is a set of the form[[ a , . . . , a n ]] = { α ∈ (0 , \ Q : a i ( α ) = a i (1 ≤ i ≤ n ) } . Equivalently, α ∈ [[ a , . . . , a n ]] iff α has an infinite continued fractionexpansion of the form α = [0; a , . . . , a n , ∗ , ∗ , . . . ].Our plan is to cover A k i ( I ) by unions of cylinders of total measure O (1 /R i ), and then use the following well-known fact: There is a con-stant G > a , . . . , a n , b , . . . , b m ) ∈ N n + m , G − ≤ mes[[ a , . . . , a n ; b , . . . , b m ]]mes[[ a , . . . , a n ]]mes[[ b , . . . , b m ]] ≤ G. (3.8)This is because the invariant measure dx x of T : (0 , → (0 , T ( x ) = { x } (the Gauss map) is a Gibbs-Markov measure, because ofthe bounded distortion of T , see § A k i ( I ) by cylinders, it is enough to cover A m,n,k i by cylindersfor every ( m, n ) ∈ Ω k i ( I ). Suppose α ∈ A m,n,k i . Then r := gcd( m, n ) ≤ M and ( m ∗ , n ∗ ) := r ( m, n ) satisfiesgcd( m ∗ , n ∗ ) = 1 , n ∗ ∈ (cid:91) | k ∗ i − k i |≤ ln M [ e k ∗ i − , e k ∗ i ] , | n ∗ α − m ∗ | < n ∗ R i . O TEMPORAL DLT FOR A.E. IRRATIONAL TRANSLATION 11
Assume k i is so large that R i = ψ ( e k i ) ≥ . Then Lemma 3.3 gives a l i +1 > R i . Thus A k i ( I ) ⊂ C k i ( I, R i ) where C k ( I, R ) := (cid:91) k ∗ ∈ [ k − ln M,k ] α ∈ (0 , \ Q : ∃ (cid:96) s.t. q (cid:96) ( α ) ∈ [ e k ∗ − , e k ∗ ] ,a (cid:96) +1 ( α ) ≥ R/ p (cid:96) ( α ) /q (cid:96) ( α ) ∈ I . This is a union of cylinders, because q (cid:96) ( α ) , p (cid:96) ( α ) , a (cid:96) +1 ( α ) are constanton cylinders of length (cid:96) + 1.We claim that for some c ∗ ( M ) which only depends on M , for all k i large enough, mes( C k i ( I, R i )) ≤ c ∗ ( M )mes( I ) R i . (3.9)Every rational mn ∈ (0 ,
1) has two finite continued fraction expansions:[0; a , . . . , a (cid:96) ] and [0; a , . . . , a (cid:96) − ,
1] with a (cid:96) >
1. We write (cid:96) = (cid:96) ( mn )and a i = a i ( mn ). With this notation C k i ( I, R i ) = (cid:91) k ∗ i ∈ [ k i − ln M,k i ] n ∈ [ e k ∗ i − ,e k ∗ i ] (cid:91) gcd( m,n )=1 m/n ∈ I (cid:91) b>R i / [[ a ( mn ) , . . . , a (cid:96) ( mn ) ( mn ) , b ]] ∪ [[ a ( mn ) , . . . , a (cid:96) ( mn ) ( mn ) − , , b ]] . We have [[ a , . . . , a (cid:96) ]] = ( p (cid:96) + p (cid:96) − q (cid:96) + q (cid:96) − , p (cid:96) q (cid:96) ) or ( p (cid:96) q (cid:96) , p (cid:96) + p (cid:96) − q (cid:96) + q (cid:96) − ), depending on theparity of (cid:96) [Khi63]. Since | p (cid:96) q (cid:96) − − p (cid:96) − q (cid:96) | = 1 and q (cid:96) +1 = a (cid:96) +1 q (cid:96) + q (cid:96) − ,we have mes([[ a , . . . , a (cid:96) , b ]]) = q (cid:96) +1 ( q (cid:96) +1 + q (cid:96) ) = bq (cid:96) + q (cid:96) − )(( b +1) q (cid:96) + q (cid:96) − ) ≤ b ( b +1) q (cid:96) , leading tomes( C k i ( I, R i )) ≤ (cid:88) k ∗ i ∈ [ k i − ln M,k i ] (cid:88) n ∈ [ e k ∗ i − ,e k ∗ i ] (cid:88) gcd( m,n )=1 m/n ∈ I (cid:88) b>R i / n b ( b + 1) ≤ Me k i − − ln M ) R i e ki M (cid:88) n =1 { m ∈ N : mn ∈ I, gcd( m, n ) = 1 } ≤ c ∗ ( M ) R i mes( I )where c ∗ ( M ) only depends on M . The last step uses Prop. 3.7.Next we cover A k ( I ) ∩ A k ( I ) by cylinders. Suppose without loss ofgenerality that k > k . Arguing as before one sees that if k > k + ln M + 1 , (3.10)then A k ( I ) ∩A k ( I ) can be covered by sets [[ a , . . . , a (cid:96) , b, a , . . . , a l , b ]] asfollows: The convergents p i /q i of (every) α in [[ a , . . . , a (cid:96) , b, a , . . . , a l , b ]],(1 ≤ i ≤ l + l + 2), satisfy(a) q l ∈ [ e k ∗ − , e k ∗ ], k ∗ ∈ [ k − ln M, k ], p l /q l ∈ I , b ≥ R / (b) q l + l +1 ∈ [ e k ∗ − , e k ∗ ], k ∗ ∈ [ k − ln M, k ], p l /q l ∈ I , b ≥ R / k ∗ > k ∗ (this is where (3.10) is used).We claim that [[ a , . . . , a (cid:96) , b ]] ⊂ C k ( I ) , (3.11) b ≤ e k ∗ − k ∗ +1 , (3.12)[[ a , . . . , a (cid:96) , b ]] ⊂ (cid:91) | r |≤ C k − k + r − ln b ([0 , , R ) . (3.13)(3.11) follows from (a). Next, e k ∗ ≥ q l +1 ≥ bq l ≥ be k ∗ − proving (3.12).To prove (3.13), let p i /q i , 1 ≤ i ≤ (cid:96) + 2, be the principal convergentsof (every) α ∈ [[ b, a , . . . , a (cid:96) , b ]]. By Prop. 3.8, q l +1+ l = q l − p l +1 + q l q l +1 ,whence q l q l +1 ≤ q l +1+ l ≤ q l q l +1 . Since q l ∈ [ e k ∗ − , e k ∗ ] and q l + l +1 ∈ [ e k ∗ − , e k ∗ ], e k ∗ − k ∗ − ≤ q l +1+ l q l ≤ q l +1 ≤ q l +1+ l q l ≤ e k ∗ − k ∗ +1 . (3.14)Next, let (cid:101) p i / (cid:101) q i (1 ≤ i ≤ l ) denote the principal convergents of (every) (cid:101) α ∈ [[ a , . . . , a (cid:96) , b ]]. Then p l +1 q l +1 = 1 / ( b + (cid:101) p l (cid:101) q l ), so q l +1 = b (cid:101) q l + (cid:101) p l , whence b (cid:101) q l ≤ q l +1 ≤ ( b + 1) (cid:101) q l . Thus (cid:101) q l ∈ [( b + 1) − q l +1 , b − q l +1 ]. It follows thatthe l -th principal convergent of every (cid:101) α ∈ [[ a , . . . , a (cid:96) , b ]] satisfies (cid:101) q l ∈ [ e k ∗ − k ∗ − − ln b , e k ∗ − k ∗ +1 − ln b ] . (3.15)It is now easy to see (3.13).By (3.13), A k ( I ) ∩A k ( I ) ∩ I ⊂ (cid:91) | r |≤ (cid:93) [ a,b ] ⊂C k ( I ) (cid:93) [ a (cid:48) ,b (cid:48) ] ⊂C k − k r − ln b ([0 , [[ a, b, a (cid:48) , b ]] . Now arguing as in the proof of (3.9) and using (3.8) we obtainmes( A k ( I ) ∩ A k ( I ) ∩ I ) ≤≤ (cid:88) k ∗ i ∈ [ ki − ln M,ki ] i =1 , | r |≤ (cid:88) n ∈ [ e k ∗ − ,e k ∗ ] (cid:88) gcd( m,n )=1 m/n ∈ I [exp( k ∗ − k ∗ +1)] (cid:88) b =[ R /
2] 2 G mes( C k − k r − ln b ([0 , ,R )) n b ( b +1) ≤ const mes( I ) R R . (3.16)(3.16) uses the estimate mes( C k − k + r − ln b ([0 , , R )) = O (1 /R ) whichis also valid when k − k + r − ln b is small, provided we choose M large enough so that the asymptotic in Prop. 3.7 holds for all N > M with I = [0 , A k i ( I ) are D -quasi-independent for sufficiently large D , proving Claim 3. O TEMPORAL DLT FOR A.E. IRRATIONAL TRANSLATION 13
Claims 2 and 3 allow us to apply Sullivan’s Borel–Cantelli Lemma(Prop. 3.5). We obtain δ = δ ( M ) s.t. for every interval I ⊂ [ ε, − ε ],mes( A ∩ I ) ≥ δ mes( I ). This means that [ ε, − ε ] \ A has no Lebesguedensity points, and therefore must have measure zero. So A has fullmeasure in [ ε, − ε ]. Since ε is arbitrary, A has full measure. (cid:3) Proof of Theorem f ( x ) := (cid:80) dm =1 b m h ( x + β m ) (cid:54)≡ h ( x ) = { x } − . Without loss ofgenerality, β i are different and b i (cid:54) = 0. Notice that f ( x ) = − d (cid:88) m =1 b m ∞ (cid:88) j =1 sin(2 πj ( x + β m )) πj . Therefore (cid:107) f (cid:107) = 12 π (cid:88) n n D ( β n, . . . , β d n ), where D : T d → R is D ( γ , . . . , γ d ) := (cid:90) (cid:34) d (cid:88) m =1 b m sin(2 π ( y + γ m )) (cid:35) dy. (4.1)Since f (cid:54)≡ D ( β n, . . . , β d n ) > n . Let T denote the closurein T d of O := { ( β n, . . . , β d n ) mod Z : n ∈ Z } . This is a minimal setfor the translation by ( β , . . . , β d ) on T d , so a standard compactnessargument shows that the set N := { n ∈ N : D ( β n, . . . , β d n ) > ε } (4.2)is syndetic: its gaps are bounded. Thus N has positive lower density.By Theorem 3.1, if M is sufficiently large then the set A := A ( N , M )has full measure in T . Let S n ( α, x ) := n − (cid:88) k =0 f ( x + kα ) . The proof of Theorem 1.2 for f ( x ) above consists of two parts: Theorem 4.1.
Suppose α ∈ A , then for a.e. x ∈ [0 , , there exist A k ( x ) ∈ R and B k ( x ) , N k ( x ) → ∞ such that S n ( α, x ) − A k ( x ) B k ( x ) dist −−−→ k →∞ U[0 , , as n ∼ U(0 , . . . , N k ( x )) . Theorem 4.2.
Suppose α ∈ A , then for a.e. x ∈ [0 , , there are no A N ( x ) ∈ R and B N ( x ) → ∞ such that S n ( α, x ) − A N ( x ) B N ( x ) dist −−−→ N →∞ U[0 , , as n ∼ U (0 , . . . , N ) . Preliminaries.Lemma 4.3. S q ( α, · ) : T → R has dq discontinuities.Proof. The discontinuities of S q are preimages of discontinuities of f by R − kα with k = 0 , , . . . , q − . (cid:3) Lemma 4.4.
Let C := sup | f (cid:48) | ≤ | (cid:80) b m | . If x (cid:48) , x (cid:48)(cid:48) belong to samecontinuity component of R rα then | S r ( α, x (cid:48) ) − S r ( α, x (cid:48)(cid:48) ) | ≤ Cr | x (cid:48) − x (cid:48)(cid:48) | . Proof.
Since | S (cid:48) r | = (cid:12)(cid:12)(cid:80) r − k =0 f (cid:48) ( x + kα ) (cid:12)(cid:12) ≤ Cr, the restriction of S r to oneach continuity component is Lipshitz with Lipshitz constant Cr. (cid:3)
Lemma 4.5.
There are constants C , C such that the following holds.Suppose that q n is a principal denominator of α , and q n +1 > cq n with c > . Let µ n ( x ) := S q n ( α, x ) , then mes (cid:26) x : S (cid:96)q n ( α, x ) = (cid:96)µ n ± C (cid:96) c for (cid:96) = 0 , . . . , k (cid:27) > − C kc . (4.3) Proof. If x and x + (cid:96)q n α belong to the same continuity interval of R q n α for all (cid:96) = 0 , . . . , k then we have by Lemma 4.4 that for (cid:96) ≤ k | S (cid:96)q n ( α, x ) − (cid:96)µ n | ≤ (cid:96) − (cid:88) j =0 | S q n ( α, x + jq n α ) − S q n ( α, x ) | ≤ Cq n (cid:96) − (cid:88) j =0 (cid:107) jq n α (cid:107)≤ Cq n q n +1 (cid:96) − (cid:88) j =0 j ≤ C (cid:96) c , where C := C/ . Therefore if S (cid:96)q n ( α, x ) (cid:54) = (cid:96)µ n ± C (cid:96) c for some (cid:96) = 0 , . . . , k , then theremust exist 0 ≤ (cid:96) ≤ k s.t. x, R (cid:96)q n α ( x ) are separated by a discontinuity of S q n ( α, · ). Since dist( x, R (cid:96)q n α ( x )) ≤ (cid:96)/q n +1 , x must belong to a ball withradius k/q n +1 centered at a discontinuity of S q n ( α, · ). By Lemma 4.3,there are dq n discontinuities, so the measure of such points is less than dq n (cid:16) kq n +1 (cid:17) ≤ dkc . The lemma follows with C := 2 d . (cid:3) Lemma 4.6.
There is a constant C = C ( b , . . . , b d ) s.t. for every α = [0; a , a , . . . ] , max {| S r ( α, x ) | : 0 ≤ r ≤ q n − } ≤ C ( a + · · · + a n − ) . Proof.
Let r = (cid:80) n − j =0 b j q j denote the Ostrowski expansion of r . Recallthat this means that 0 ≤ b j ≤ a j and b j = a j ⇒ b j − = 0. So S r = b n − − (cid:88) k =0 S q n − ◦ R q n − kα + b n − − (cid:88) k =0 S q n − ◦ R q n − kα + · · · + b − (cid:88) k =0 S q ◦ R q kα . O TEMPORAL DLT FOR A.E. IRRATIONAL TRANSLATION 15
By the Denjoy-Koksma inequality | S r | ≤ (cid:80) b j V ( f ) ≤ V ( f ) (cid:80) a j where V ( f ) ≤ (cid:80) b i is the total variation of f on T . (cid:3) Lemma 4.7.
There exist positive constants ε , ε such that for every α irrational, if q n is a principal denominator of α and q n r n ∈ N with r n ≤ M then mes { x : | S q n ( α, x ) | ≥ ε } ≥ ε . Proof.
We follow an argument from [Bec94]. Suppose q n is a principaldenominator of α and q n r n ∈ N for some r n ≤ M . Let N = q n r n . Since f ( x ) = − (cid:80) dm =1 b m (cid:80) ∞ j =1 sin(2 πj ( x + β m )) πj , for each j ∈ N (cid:107) S N ( α, · ) (cid:107) L ≥ π j (cid:90) (cid:32) d (cid:88) m =1 b m N − (cid:88) k =0 sin(2 πj ( x + kα + β m )) (cid:33) dx. Using the identities (cid:80) Nk =1 sin( y + kx ) = cos( y + x/ − cos( y +(2 N +1) x/ x/ andcos A − cos B = 2 sin( A + B ) sin( B − A ) we find that (cid:107) S N ( α, · ) (cid:107) L ≥≥ (cid:18) sin( πN jα ) πj sin( πjα ) (cid:19) (cid:90) (cid:32) d (cid:88) m =1 b m sin (cid:16) π (cid:16) jx + j ( N − α (cid:17) + 2 πjβ m (cid:17)(cid:33) dx = (cid:18) sin( πN jα ) πj sin( πjα ) (cid:19) (cid:90) (cid:32) d (cid:88) m =1 b m sin (2 π ( y + jβ m )) (cid:33) dy = (cid:18) sin( πN jα ) πj sin( πjα ) (cid:19) D ( jβ , . . . , jβ m ) with D as in (4.1) . We now take j = N = r n q n . The first term is bounded below because (cid:107) N α (cid:107) ≤ M (cid:107) q n α (cid:107) ≤ Mq n +1 ≤ Ma n +1 q n ≤ M a n +1 N = o ( N ) , so sin( πN α ) πN sin( πNα ) −−−→ n →∞ π − . The second term is bounded below by ε , because N = q n r n ∈ N .It follows that for all n large enough, (cid:107) S r n q n ( α, · ) (cid:107) > √ ε / π .For any L -function ϕ and any ˆ ε > (cid:107) ϕ (cid:107) L ≤ (cid:107) ϕ (cid:107) L ∞ mes { x : | ϕ ( x ) | ≥ ˆ ε } + ˆ ε . Hence mes { x : | ϕ ( x ) | ≥ ˆ ε } ≥ (cid:107) ϕ (cid:107) L − ˆ ε (cid:107) ϕ (cid:107) L ∞ . We just saw that for all n large enough, (cid:107) S r n q n ( α, · ) (cid:107) > √ ε / π , and by the Denjoy-Koksmainequality (cid:107) S r n q n ( α, · ) (cid:107) L ∞ ≤ M V ( f ). So for some ˆ ε > n large enough, mes { x : | S r n q n ( α, x ) | > ˆ ε } ≥ ˆ ε. Looking at the inequality | S r n q n ( α, x ) | ≤ (cid:80) r n − k =0 | S q n ( α, x + kq n α ) | ,we see that if | S r n q n ( α, x ) | ≥ ˆ ε , then | S q n ( α, x + kq n α ) | ≥ ˆ ε/M forsome 0 ≤ k ≤ M −
1. So for all n large enough, mes { x : | S q n ( α, x ) | > ˆ ε/M } ≥ ˆ ε/M. (cid:3) Proof of Theorem 4.1.
Let Ω ∗ ( α ) be the set of x where theconclusion of Theorem 4.1 holds. Ω ∗ ( α ) is R α -invariant and it is mea-surable by Lemma A.1 in the appendix. Therefore to show that Ω ∗ ( α )has full measure, it suffices to show that it has positive measure.Suppose α ∈ A and let n k ↑ ∞ be a sequence satisfying (3.1) with N given by (4.2). There is no loss of generality in assuming that a n k +1 a + · · · + a n k > k . So q n k +1 > k L k q n k , where L k := a + · · · + a n k . Recall that µ n k ( x ) = S q nk ( α, x ). For all k sufficiently large, there isa set A k of measure at least ε / x ∈ A k , S (cid:96)q nk ( α, x ) = (cid:96) (cid:18) µ n k ( x ) ± C (cid:96)k L k (cid:19) for all (cid:96) = 0 , , . . . , kL k (4.4) | µ n k ( x ) | ≥ ε . (4.5)This is because Lemma 4.5 says that the total measure of x for which(4.4) fails is O (1 /k ) while (4.5) holds on the set of measure ε byLemma 4.7.It follows that mes( (cid:84) n> (cid:83) k>n A k ) ≥ ε /
2. Therefore there exists x which belongs to infinitely many A k . After re-indexing n k , we mayassume that (4.4), (4.5) are satisfied for all k ∈ N . Henceforth, we fixsuch an x and work with this x . Let N k ( x ) := kL k q n k , B k ( x ) := kL k | µ n k ( x ) | , A k ( x ) := 12 (sgn( µ n k ( x )) − B k . Any n ≤ N k can be written uniquely in the form n = l ( n ) q n k + r ( n ) with 0 ≤ l ( n ) ≤ kL k and 0 ≤ r ( n ) < q n k . It is easy to see that l ( n ) kL k dist −−−→ k →∞ U[0 ,
1] as n ∼ U(1 , . . . , N k ) . Writing S n ( α, x ) = S l ( n ) q nk ( α, x ) + S r ( n ) ( α, x + αl ( n ) q n k ) we obtainfrom (4.4) and Lemma 4.6 that S n ( α, x ) = l ( n ) µ n k ( x ) + O ( L k ) . So S n ( x ) B k is asymptotically uniform on [0 ,
1] when µ n k >
0, and [ − , µ n k <
0. So S n ( x ) − A k B k −−−→ k →∞ U[0 , n ∼ U(1 , . . . , N k ( x )). (cid:3) Proof of Theorem 4.2.
Let Ω( α ) denote the set of x ∈ T := R / Z for which there are B N ( x ) → ∞ and A N ( x ) ∈ R s.t. S n ( α, x ) − A N ( x ) B N ( x ) dist −−−→ N →∞ U[0 , , as n ∼ U(1 , . . . , N ) . (4.6) O TEMPORAL DLT FOR A.E. IRRATIONAL TRANSLATION 17
This is a measurable set, see the appendix. Assume by way of contra-diction that mes[Ω( α )] (cid:54) = 0 for some α ∈ A .Ω( α ) is invariant under R α ( x ) = x + α mod 1 on T := R / Z . Since R α is ergodic, and Ω( α ) is measurable, mes[Ω( α )] = 1.Since α ∈ A , there is an increasing sequence n k satisfying (3.1)where N is given by (4.2). We can choose n k so that q n k r n k ∈ N for r n k ≤ M , and a n k +1 > k L k where L k := a + · · · + a n k . In particular, q n k +1 > k L k q n k . Recall that µ n k ( x ) := S q nk ( α, x ). By Lemma 4.7 we can choose x suchthat for infinitely many k , | µ n k ( x ) | ≥ ε . We will suppose that µ n k ( x ) > k ; the case where µ n k ( x ) < k is similar. Claim 1.
It is possible to assume without loss of generality that (cid:107) B q nk (cid:107) ∞ := sup x ∈ Ω( α ) | B q nk ( x ) | ≤ C L k for all k where C is theconstant from Lemma 4.6. Proof.
We claim that for every x with (4.6), B q nk ( x ) ≤ C L k forall k large enough. Otherwise, by Lemma 4.6, there are infinitelymany k s.t. B q nk ( x ) > {| S r ( α, x ) | : r = 0 , . . . , q n k − } , whence | S n ( α, x ) /B q nk | ≤ for all 0 ≤ n ≤ q n k −
1. In such circumstances,(4.6) does not hold (the spread is not big enough).Since B q nk ( x ) ≤ C L k for all k large enough, there is no harm inreplacing B q nk ( x ) in (4.6) by min { B q nk ( x ) , C L k } . Claim 2.
Fix
D > C = | (cid:80) b m | , and let E k denote the set of x ∈ Ω( α )s.t. S r ( α, x ) = S r ( α, R (cid:96)q nk α ( x )) ± D(cid:96)q nk +1 for all 0 ≤ (cid:96) ≤ B q nk ( x ) , ≤ r 1. Then mes( E ck ) ≤ C k − . Proof. If x (cid:54)∈ E k , then there are 0 ≤ (cid:96) ≤ B q nk ( x ) , ≤ r < q n k − | S r ( α, x ) − S r ( α, x + (cid:96)q n k α ) | ≥ D(cid:96)q n k +1 . By Lemma 4.4, { x } , { x + (cid:96)q n k α } are separated by a singularity of S r ( α, · ). So x belongs to a ball of radius 2 (cid:107) (cid:96)q n k α (cid:107) centered at one ofthe dq n k discontinuities of S q nk ( α, · · · ). Thus mes( E ck ) ≤ dq n k · (cid:107) (cid:96)q n k α (cid:107) . Now (cid:107) (cid:96)q n k α (cid:107) ≤ (cid:96) (cid:107) q n k α (cid:107) ≤ (cid:107) B qnk (cid:107) ∞ q nk +1 ≤ C L k q nk +1 ≤ C k q nk by our choice of n k . So mes( E ck ) ≤ C /k with C := 6 dC . Claim 3. Let F k denote the set of x ∈ Ω( α ) s.t. S (cid:96)q nk ( α, x ) = (cid:96) (cid:18) µ n k ( x ) ± C (cid:96)k L k (cid:19) for all 0 ≤ (cid:96) ≤ B q nk ( x ) . Then mes( F ck ) ≤ C k − . Proof. This follows from Lemma 4.5.By Claims 2 and 3, and a Borel–Cantelli argument, for a.e. x thereis k ( x ) s.t. x ∈ E k ∩ F k for all k ≥ k ( x ).Suppose k ≥ k ( x ), and let N k := q n k B q nk ( x ). Every 0 ≤ n ≤ N k − n = (cid:96)q n k + r with 0 ≤ (cid:96) ≤ B q nk ( x ) − ≤ r ≤ q n k − 1. Using the bound (cid:107) B q nk (cid:107) ∞ = O ( L k ), we find: S n ( α, x ) − A q nk ( x ) B q nk ( x ) = S (cid:96)q nk ( α, x ) B q nk ( x ) + S r ( α, R (cid:96)q nk α x ) − A q nk ( x ) B q nk ( x )= S (cid:96)q nk ( α, x ) B q nk ( x ) + S r ( α, x ) − A q nk ( x ) + o (1) B q nk ( x ) , because x ∈ E k = (cid:96) ( µ n k ( x ) + o (1)) B q nk ( x ) + S r ( α, x ) − A q nk ( x ) + o (1) B q nk ( x ) , because x ∈ F k . If n ∼ U(0 , . . . , N k − (cid:96), r are independent random variables, (cid:96) ∼ U(0 , . . . , B q nk ( x ) − 1) and r ∼ U(0 , . . . , q n k − (cid:96) ( µ nk ( x )+ o (1)) B qnk ( x ) is close to U[0 , µ n k ( x )], and the distribution of S r ( α,x ) − A qnk ( x )+ o (1) B qnk ( x ) converges to U[0 , 1] (because x ∈ Ω).Taking a subsequence such that µ n k ( x ) → ε > ε we see that the ran-dom variables S n ( α, x ) − A q nk ( x ) B q nk ( x ) , where n ∼ U(0 , . . . , n k − (cid:3) Appendix A. Measurability concerns Let Ω( α ) denote the set of x ∈ T := R / Z such that for some B N ( x ) → ∞ and A N ( x ) ∈ R , S n ( α,x ) − A N ( x ) B N ( x ) dist −−−→ N →∞ U [0 , , as n ∼ U(1 , . . . , N ) and let Ω ∗ ( α ) denote the set of x ∈ T such that along asubsequence N k ( x ) there exist some B N k ( x ) → ∞ and A N k ( x ) ∈ R , S n ( α,x ) − A Nk ( x ) B Nk ( x ) dist −−−→ k →∞ U[0 , , as n ∼ U(1 , . . . , N k ) . We make no assump-tions on the measurability of A N , B N , N k as functions of x . The purposeof this section is to prove: Lemma A.1. Ω( α ) and Ω ∗ ( α ) are measurable. The crux of the argument is to show that A N ( x ) , B N ( x ) can be re-placed by measurable functions, defined in terms of the percentiles ofthe random quantities S n ( x, α ), n ∼ U(1 , . . . , N ). O TEMPORAL DLT FOR A.E. IRRATIONAL TRANSLATION 19 Recall that given 0 < t < 1, the upper and lower t -percentiles of arandom variable X are defined by χ + ( X, t ) := inf { ξ : Pr( X ≤ ξ ) > t } χ − ( X, t ) := sup { ξ : Pr( X ≤ ξ ) < t } (0 < t < . Notice that Pr( X ≤ χ + ( X, t )) ≥ t , Pr( X < χ − ( X, t )) ≤ t , andPr( χ − ( X, t ) < X < χ + ( X, t )) = 0. In case X is non-atomic (i.e. P ( X = a ) = 0 for all a ), we can say more: Lemma A.2. Suppose X is a non-atomic real valued random variable,fix < t < and let χ ± t := χ ± ( X, t ) , then(a) Pr( X < χ + t ) = t and Pr( X < χ − t ) = t ;(b) ∀ ε > , Pr( χ − t − ε < X < χ − t ) , Pr( χ + t < X < χ + t + ε ) are positive;(c) ∃ t < t s.t. χ − t < χ + t and χ − t , χ + t have the same sign.Proof. Since X is non-atomic, Pr( X < χ + t ) = Pr( X ≤ χ + t ) ≥ t andPr( X < χ − t ) ≤ t . If χ + t = χ − t , part (a) holds. If χ + t > χ − t then for all h > χ − t + h < χ + t − h whence0 ≤ Pr( χ − t < X < χ + t ) = lim h → + Pr( χ − t + h < X < χ + t − h )= lim h → + Pr( X < χ + t − h ) − lim h → + Pr( X ≤ χ − t + h ) ≤ t − t = 0 . Necessarily lim h → + Pr( X < χ + t − h ) = t and lim h → + Pr( X ≤ χ − t + h ) = t ,which gives us Pr( X < χ + t ) = t and Pr( X < χ + t ) = Pr( X ≤ χ + t ) = t .For (b) assume by contradiction that Pr( χ − t − ε < X < χ − t ) = 0,then for all χ − t − ε < ξ < χ − t , Pr( X ≤ ξ ) = Pr( X < χ − t ) = t , whence χ − t ≤ χ − t − ε , a contradiction. Similarly, Pr( χ + t < X < χ + t + ε ) = 0 isimpossible.To prove (c) note that since X is non-atomic, either Pr( X > 0) orPr( X < 0) is positive. Assume w.l.o.g. that Pr( X > (cid:54) = 0. Bynon-atomicity, there are positive a < b s.t. Pr( X ∈ (0 , a )) (cid:54) = 0 andPr( X ∈ ( a, b )) (cid:54) = 0. Take t := Pr( X < a ) and t := Pr( X < b ). (cid:3) From now on fix a non-atomic random variable Y , and choose 0 Let S N be (possibly atomic) random variables s.t. forsome A N ∈ R and B N → ∞ , S N − A N B N dist −−−→ N →∞ Y . Then S N − A ∗ N B ∗ N dist −−−→ N →∞ Y ,where A ∗ N , B ∗ N are the unique solution to (cid:40) A ∗ N + B ∗ N χ − ( Y, t ) = χ − ( S N , t ) A ∗ N + B ∗ N χ + ( Y, t ) = χ + ( S N , t ) . (A.1) Proof. Without loss of generality, χ − ( Y, t ) , χ + ( Y, t ) are both positive.We need the following fact (which is not automatic since S N areallowed to be atomic):lim N →∞ Pr( S N < χ − ( S N , t )) = t for all 0 < t < . (A.2)Indeed, given ε > 0, let ξ N := B N χ − ( Y, t − ε ) + A N , then Pr( S N < ξ N ) = Pr (cid:16) S N − A N B N < χ − ( Y, t − ε ) (cid:17) −−−−→ N →∞ Pr( Y < χ − ( Y, t − ε )) = t − ε, by Lemma A.2(a). So for all N large enough, ξ N ≤ χ − ( S N , t ), whencelim inf Pr( S N < χ − ( S N , t )) ≥ lim Pr( S N < ξ N ) = t − ε . Since ε is arbitrary, lim inf Pr( S N < χ − ( S N , t )) ≥ t . The other inequalitylim sup Pr( S N < χ − ( S N , t )) ≤ t is clear since Pr( S N < χ − ( S N , t )) ≤ t for all N .With (A.2) proved, we proceed to prove that A ∗ N − A N B N −−−→ N →∞ B ∗ N B N −−−→ N →∞ . (A.3)It will then be obvious that S N − A N B N dist −−−→ N →∞ Y implies S N − A ∗ N B ∗ N dist −−−→ N →∞ Y .Define two affine transformations, ϕ N ( t ) = t − A N B N and ϕ ∗ N ( t ) = t − A ∗ N B ∗ N .Notice that ( ϕ ∗ N ) − ( t ) = A ∗ N + B ∗ N t , so ( ϕ ∗ N ) − ( χ − ( Y, t )) = χ − ( S N , t ),by (A.1). Since B ∗ N = χ + ( S N ,t ) − χ − ( S N ,t ) χ + ( Y,t ) − χ − ( Y,t ) > ϕ ∗ N is increasing. By(A.2), Pr( ϕ ∗ N ( S N ) < χ − ( Y, t )) = Pr( S N < χ − ( S N , t )) −−−→ N →∞ t . So t = lim N →∞ Pr( ϕ ∗ N ( S N ) < χ − ( Y, t ))= lim N →∞ Pr (cid:0) ϕ N ( S N ) < ϕ N [( ϕ ∗ N ) − ( χ − ( Y, t ))] (cid:1) = lim N →∞ Pr (cid:18) ϕ N ( S N ) < B ∗ N B N (cid:18) χ − ( Y, t ) + A ∗ N − A N B N (cid:19)(cid:19) . We claim that this implies thatlim inf N →∞ B ∗ N B N (cid:18) χ − ( Y, t ) + A ∗ N − A N B N (cid:19) ≥ χ − ( Y, t ) . (A.4) O TEMPORAL DLT FOR A.E. IRRATIONAL TRANSLATION 21 Otherwise, ∃ ε s.t. lim inf N →∞ B ∗ N B N (cid:16) χ − ( Y, t ) + A ∗ N − A N B N (cid:17) < χ − ( Y, t ) − ε , so t = lim inf N →∞ Pr (cid:18) ϕ N ( S N ) < B ∗ N B N (cid:18) χ − ( Y, t ) + A ∗ N − A N B N (cid:19)(cid:19) ≤ lim inf N →∞ Pr (cid:18) ϕ N ( S N ) < χ − ( Y, t ) − ε (cid:19) = Pr( Y < χ − ( Y, t ) − ε )= Pr( Y < χ − ( Y, t )) − Pr( χ − ( Y, t ) − ε ≤ Y < t ) < t , by Lemma A.2, parts (a),(b) . Similarly, one shows thatlim sup N →∞ B ∗ N B N (cid:18) χ + ( Y, t ) + A ∗ N − A N B N (cid:19) ≤ χ + ( Y, t ) . (A.5)It remains to see that (A.4) and (A.5) imply (A.3). First we divide(A.4) by (A.5) to obtainlim sup N →∞ χ − ( Y, t ) + A ∗ N − A N B N χ + ( Y, t ) + A ∗ N − A N B N ≥ χ − ( Y, t ) χ + ( Y, t ) . Since x (cid:55)→ a + xb + x is strictly decreasing on [0 , ∞ ) when a > b > 0, thisimplies that lim sup N →∞ A ∗ N − A N B N ≤ . (A.6)Looking at (A.4), and recalling that χ − ( Y, t ) > 0, we deduce thatlim inf N →∞ B ∗ N B N ≥ . (A.7)Next we look at the difference of (A.4) and (A.5) and obtainlim sup N →∞ B ∗ N B N (cid:0) χ + ( Y, t ) − χ − ( Y, t ) (cid:1) ≤ χ + ( Y, t ) − χ − ( Y, t ) , whence lim sup( B ∗ N /B N ) ≤ 1. Together with (A.7), this proves that B ∗ N /B N −−−→ N →∞ 1. Substituting this in (A.4), gives lim inf A ∗ N − A N B N ≥ A ∗ N − A N B N −−−→ N →∞ 0. This completesthe proof of (A.3), and with it, the lemma. (cid:3) Proof of Lemma A.1 . We begin with the measurability of Ω( α ).Let S N ( x ) denote the random variable equal to S n ( α, x ) with prob-ability N for each 1 ≤ n ≤ N . We will apply Lemma A.3 with Y := U[0 , S N = S n ( x ) and (say) t := , t := . It says thatΩ( α ) = { x ∈ T : S N ( x ) − A ∗ N ( x ) B ∗ N ( x ) dist −−−→ N →∞ U [0 , } , where A ∗ N ( x ) and B ∗ N ( x ) are the unique solutions to (A.1). Since thepercentiles of S N ( x ) are measurable as functions of x , A ∗ N ( x ) , B ∗ N ( x )are measurable as functions of x .We claim that Ω( α ) = Ω ( α ) ∩ Ω ( α ) whereΩ ( α ) := ∞ (cid:92) (cid:96) =1 ∞ (cid:91) M =1 ∞ (cid:92) N = M +1 (cid:40) x ∈ T : 1 N N (cid:88) n =1 (2 , ∞ ) (cid:16)(cid:12)(cid:12)(cid:12) S n ( α,x ) − A ∗ N ( x ) B ∗ N ( x ) (cid:12)(cid:12)(cid:12)(cid:17) < (cid:96) (cid:41) Ω ( α ) = (cid:92) t ∈ Q \{ } (cid:40) x ∈ T : lim N →∞ N N (cid:88) n =1 e it (cid:18) Sn ( α,x ) − A ∗ NB ∗ N ( x ) (cid:19) = E (cid:0) e itY (cid:1)(cid:41) . This will prove the lemma, since the measurability of A ∗ N ( · ) , B ∗ N ( · ) im-plies the measurability of Ω i ( α ).If x ∈ Ω( α ) then x ∈ Ω ( α ) because P [ | S N ( x ) − A ∗ N B ∗ N | > −−−→ N →∞ 0, and x ∈ Ω ( α ) because E (cid:0) e it (cid:0) SN ( x ) − A ∗ NB ∗ N (cid:1)(cid:1) −−−→ N →∞ E (cid:0) e itY (cid:1) pointwise.Conversely, if x ∈ Ω ( α ) ∩ Ω ( α ) then it is not difficult to see that E (cid:0) e it (cid:0) SN ( x ) − A ∗ NB ∗ N (cid:1)(cid:1) −−−→ N →∞ E (cid:0) e itY (cid:1) for all t ∈ R . So x ∈ Ω( α ) by L´evy’scontinuity theorem. Thus Ω( α ) = Ω ( α ) ∩ Ω ( α ), whence Ω( α ) ismeasurable.The proof that Ω ∗ ( α ) is measurable is similar. Enumerate Q \ { } = { t n : n ∈ N } , then α ∈ Ω ∗ ( α ) iff for every (cid:96) ∈ N there exist M ∈ N s.t.for some N > M (1) N (cid:80) Nn =1 (2 , ∞ ) (cid:16)(cid:12)(cid:12)(cid:12) S n ( α,x ) − A ∗ N ( x ) B ∗ N ( x ) (cid:12)(cid:12)(cid:12)(cid:17) < (cid:96) (2) | E (cid:0) e it n (cid:0) SN ( x ) − A ∗ N ( x ) B ∗ N ( x ) (cid:1)(cid:1) − E (cid:0) e it n Y (cid:1) | < (cid:96) for all n = 1 , . . . , (cid:96) These are measurable conditions, because A ∗ N ( · ) , B ∗ N ( · ) are measurable.So Ω ∗ ( α ) is measurable. (cid:3) References [ADDS15] Artur Avila, Dmitry Dolgopyat, Eduard Duryev, and Omri Sarig. Thevisits to zero of a random walk driven by an irrational rotation. IsraelJ. Math. , 207(2):653–717, 2015.[ADU93] Jon Aaronson, Manfred Denker, and Mariusz Urba´nski. Ergodic theoryfor Markov fibred systems and parabolic rational maps. Trans. Amer.Math. Soc. , 337(2):495–548, 1993. O TEMPORAL DLT FOR A.E. IRRATIONAL TRANSLATION 23 [AK82] Jon Aaronson and Michael Keane. The visits to zero of some determin-istic random walks. Proc. London Math. Soc. (3) , 44(3):535–553, 1982.[Bec94] J´ozsef Beck. Probabilistic Diophantine approximation. I. Kronecker se-quences. Ann. of Math. (2) , 140(1):109–160, 1994.[Bec10] J´ozsef Beck. Randomness of the square root of 2 and the giant leap,Part 1. Period. Math. Hungar. , 60(2):137–242, 2010.[Bec11] J´ozsef Beck. Randomness of the square root of 2 and the giant leap,Part 2. Period. Math. Hungar. , 62(2):127–246, 2011.[BU] Michael Bromberg and Corinna Ulcigrai. A temporal central limit theo-rem for real-valued cocycles over rotations. preprint ArXiv:1705.06484.[CK76] Jean-Pierre Conze and Michael Keane. Ergodicit´e d’un flot cylindrique.page 7, 1976.[DSa] Dmitry Dolgopyat and Omri Sarig. Asymptotic windings of horocycles. to appear in Israel Math. J. [DSb] Dmitry Dolgopyat and Omri Sarig. Quenched and annealed temporallimit theorems for circle rotations. Preprint (2017).[DS17] Dmitry Dolgopyat and Omri Sarig. Temporal distributional limit the-orems for dynamical systems. J. Statistical Physics , 166(3-4):680–713,2017.[DV86] Harold G. Diamond and Jeffrey D. Vaaler. Estimates for partial sumsof continued fraction partial quotients. Pacific J. Math. , 122(1):73–82,1986.[HW08] Godfrey Harold Hardy and Edward Maitland Wright. An introductionto the theory of numbers . Oxford University Press, Oxford, sixth edi-tion, 2008. Revised by D. R. Heath-Brown and J. H. Silverman, With aforeword by Andrew Wiles.[Jar29] Vojtech Jarn´ık. Zur metrischen theorie der diophantischen approxima-tionen. Prace Matematyczno-Fizyczne , 36(1):91–106, 1928-1929.[Khi24] Alexander Ya. Khintchine. Einige S¨atze ¨uber Kettenbr¨uche, mit An-wendungen auf die Theorie der Diophantischen Approximationen. Math.Ann. , 92(1-2):115–125, 1924.[Khi63] Alexander Ya. Khintchine. Continued fractions . Translated by PeterWynn. P. Noordhoff, Ltd., Groningen, 1963.[PS] Elliot Paquette and Younghwan Son. Birkhoff sum fluctuations in sub-stitution dynamical systems. arXiv:1505.01428, preprint (2015) .[Sch78] Klaus Schmidt. A cylinder flow arising from irregularity of distribution. Compositio Math. , 36(3):225–232, 1978.[Sul82] Dennis Sullivan. Disjoint spheres, approximation by imaginary qua-dratic numbers, and the logarithm law for geodesics. Acta Math. , 149(3-4):215–237, 1982. Department of Mathematics, University of Maryland at CollegePark, College Park, MD 20740, USA E-mail address : [email protected] Faculty of Mathematics and Computer Science, The Weizmann In-stitute of Science, POB 26, Rehovot, Israel E-mail address ::