2 Preliminaries

In this section, we collect some preliminary results. In the first two lemmas, $\{\widetilde {X}_{n}\}_{n\geq 1}$ is assumed to be a sequence of independent random variables on a probability space $(\widetilde {\Omega }, \widetilde {\mathcal {F}}, \widetilde {\mathbb {P}})$ with $\mathbb {E}_{\widetilde {\mathbb {P}}}(\widetilde {X}_{n})=0$ . For any $n\geq 1$ , write $\widetilde {\sigma }_n^2=\mathbb {E}_{\widetilde {\mathbb {P}}}(\widetilde {X}_{n}^{2}), \widetilde {V}_{n}^2=\sum _{i=1}^{n}\widetilde {\sigma }_i^2$ and $\widetilde {B}_{n}=\widetilde {V}_{n}^{-3}\sum _{i=1}^{n}\mathbb {E}_{\widetilde {\mathbb {P}}}(|\widetilde {X}_{i}|^{3})$ .

Lemma 2.1 [Reference Petrov19, Lemma 5.1].

Let $\{\widetilde {X}_{n}\}_{n\geq 1}, \widetilde {V}_{n}, \widetilde {B}_n$ be defined as above and satisfying $\mathbb {E}_{\widetilde {\mathbb {P}}}(|\widetilde {X}_{n}|^{3})<\infty $ for each $n\geq 1$ . Write $\widetilde {Y}_n=\widetilde {V}_{n}^{-1}\sum _{i=1}^{n}\widetilde {X}_{i}$ . Then

$$ \begin{align*}|\mathbb{E}_{\widetilde{\mathbb{P}}}(e^{it\widetilde{Y}_n})-e^{-t^{2}/2}|\leq 16 \widetilde{B}_{n}|t|^{3}e^{-t^{2}/3} \quad\mbox{for } |t|\leq\frac{1}{4\widetilde{B}_{n}}. \end{align*} $$

Lemma 2.2 (Bernstein inequality, [Reference Petrov19, Theorem 2.8]).

Let $\{\widetilde {X}_{n}\}_{n\geq 1}, \widetilde {\sigma }_{n}, \widetilde {V}_n$ be defined as above and satisfying $\widetilde {\sigma }_n^2<\infty $ for each $n\geq 1$ . Suppose that there exists a constant $H>0$ such that $|\mathbb {E}_{\widetilde {\mathbb {P}}}(\widetilde {X}_n^m)|\leq \tfrac 12m!\widetilde {\sigma }_n^2H^{m-1}$ for all $n\geq 1$ and $m\geq 2$ . Then

$$ \begin{align*}\widetilde{\mathbb{P}}\bigg(\bigg\{\bigg|\!\sum_{i=1}^n \widetilde{X}_i\bigg|\geq a\bigg\}\bigg)\leq 2\exp\bigg[\frac{-a^2}{2(\widetilde{V}_n^2+H a)}\bigg].\end{align*} $$

In the next two lemmas, $(\Omega , \mathcal {F}, \mathbb {P})$ denotes the underlying probability space, $\lambda $ denotes the Lebesgue measure on the unit circle $\mathbb {T}$ and $I_{n}=(\omega _n,\omega _n+\ell _n)$ is the random interval with left endpoint $\omega _n$ and length $\ell _n$ . For a measurable function f on $\Omega \times \mathbb {T}$ , the next lemma allows us to estimate $\mathbb {P}(\{\mathbb {E}_{\lambda }(f)\geq a\})$ by virtue of $\mathbb {P}\times \lambda (\{f\geq b\})$ , which is easier to treat via Markov’s inequality and Fubini’s theorem.

Lemma 2.3 [Reference Ayyer, Liverani and Stenlund3, Section 5].

Let $M>0$ and let $f=f(\omega , x)$ be a measurable function on $\Omega \times \mathbb {T}$ with $0\leq f\leq M$ . Then, for any $\alpha>0$ ,

$$ \begin{align*} \mathbb{P}(\{\mathbb{E}_{\lambda}(f)\geq 2\alpha\})\leq \alpha^{-1}\cdot M\cdot\mathbb{P}\times\lambda(\{f\geq \alpha\}). \end{align*} $$

Proof. Note that $\mathbb {P}\times \lambda (\{f\geq \alpha \})= \mathbb {E}_{\mathbb {P}\times \lambda }(\mathbf {1}_{\{f\geq \alpha \}}) \geq \mathbb {E}_{\mathbb {P}}(\mathbb {E}_{\lambda }(\mathbf {1}_{\{f\geq \alpha \}})\cdot \mathbf {1}_{\{\mathbb {E}_{\lambda }(f)\geq 2\alpha \}})$ . Since $M\cdot \mathbb {E}_{\lambda }(\mathbf {1}_{\{f\geq \alpha \}})+\alpha \geq \mathbb {E}_{\lambda }(f)$ , the bound $\mathbb {E}_{\lambda }(f)\geq 2\alpha $ implies that $\mathbb {E}_{\lambda }(\mathbf {1}_{\{f\geq \alpha \}})\geq \alpha /M$ . Thus, by Fubini’s theorem and Markov’s inequality,

$$ \begin{align*} \mathbb{P}(\{\mathbb{E}_{\lambda}(f)\geq 2\alpha\})\leq \mathbb{P}(\mathbb{E}_{\lambda}(\mathbf{1}_{\{f\geq\alpha\}})\geq\alpha/M) \leq\alpha^{-1}\cdot M\cdot\mathbb{P}\times\lambda(\{f\geq \alpha\}). \end{align*} $$

Lemma 2.4. For any fixed $x, y\in \mathbb {T}$ and any integers $k, m\geq 1$ ,

(2.1) $$ \begin{align} \mathbb{E}_{\mathbb{P}}((\mathbf{1}_{I_{k}}(x)-\ell_k)^m)=\ell_k(1-\ell_k)^m+ (-1)^m\ell_k^m (1-\ell_k), \end{align} $$

(2.2) $$ \begin{align} \mathbb{E}_{\mathbb{P}}(|\mathbf{1}_{I_{k}}(x)-\mathbf{1}_{I_{k}}(y)|^m)=2 \min\{\ell_k, |x-y|\} \end{align} $$

and

(2.3) $$ \begin{align} \mathbb{E}_{\mathbb{P}}((\mathbf{1}_{I_{k}}(x)-\mathbf{1}_{I_{k}}(y))^m) = \begin{cases} 0 & m \ \text{is odd},\\ 2 \min\{\ell_k, |x-y|\} & m \ \text{is even}. \end{cases} \end{align} $$

Proof. By the binomial formula,

$$ \begin{align*} \mathbb{E}_{\mathbb{P}}((\mathbf{1}_{I_{k}}(x)-\ell_k)^m) & = \mathbb{E}_{\mathbb{P}}\bigg(\sum_{j=0}^{m-1}(-1)^j C_{m}^{j}\mathbf{1}_{I_{k}}(x)\cdot\ell_k^j+(-1)^m\ell_k^m\bigg) \\ & = \ell_k(1-\ell_k)^m+ (-1)^m\ell_k^m (1-\ell_k). \end{align*} $$

Notice that $|\mathbf {1}_{I_{k}}(x)-\mathbf {1}_{I_{k}}(y)|$ takes only 0 and 1 as values. For fixed $x, y\in \mathbb {T}$ , the function $|\mathbf {1}_{I_{k}}(x)-\mathbf {1}_{I_{k}}(y)|$ as a function of $\omega $ is the indicator function of the event that $\omega $ belongs to one of the two intervals $(x-\ell _k, x)$ and $(y-\ell _k, y)$ but not the other, that is, to the symmetric difference of the two intervals. There are two cases: when $|x-y|>\ell _k$ , the two intervals are disjoint and the event has probability $2\ell _k$ ; when $|x-y|\leq \ell _k$ , the symmetric difference is the union of two disjoint intervals of length $|x-y|$ and the event has probability $2|x-y|$ . Then (2.2) follows immediately.

If m is even, as we just proved,

$$ \begin{align*}\mathbb{E}_{\mathbb{P}}((\mathbf{1}_{I_{k}}(x)-\mathbf{1}_{I_{k}}(y))^m)=2 \min\{\ell_k, |x-y|\}.\end{align*} $$

If m is odd, by the fact that $\mathbb {E}_{\mathbb {P}}(\mathbf {1}_{I_{k}}(x)\cdot \mathbf {1}_{I_{k}}(y))=\max \{\ell _k-|x-y|, 0\}$ and the binomial formula,

$$ \begin{align*} \mathbb{E}_{\mathbb{P}}((\mathbf{1}_{I_{k}}(x)-\mathbf{1}_{I_{k}}(y))^m) & = \mathbb{E}_{\mathbb{P}}\bigg(\mathbf{1}_{I_{k}}(x)+\sum_{j=1}^{m-1}(-1)^j C_{m}^{j}\mathbf{1}_{I_{k}}(x)\cdot \mathbf{1}_{I_{k}}(y)- \mathbf{1}_{I_{k}}(y)\bigg) \\ & = \ell_k+ \max\{\ell_k-|x-y|, 0\}\sum_{j=1}^{m-1}(-1)^j C_{m}^{j} -\ell_k=0. \end{align*} $$

For the last equality, we have used $\sum _{j=0}^{m}(-1)^j C_{m}^{j}=(1-1)^m=0$ .

3 Proof of Proposition 1.2

Recall that

$$ \begin{align*} Y_n=\frac{\sum_{k=1}^{n}(\mathbf{1}_{I_{k}}(x)-l_{k})}{V_n}\quad\text{and}\quad Z_n=\frac{\sum_{k=1}^{n}(\mathbf{1}_{I_{k}}(x)-\mathbf{1}_{I_{k}}(y))}{V_n}, \end{align*} $$

where $x, y\in \mathbb {T}$ . By Lévy’s continuity theorem, we need to prove that, $\mathbb {P}$ -almost surely, $\mathbb {E}_{\lambda }(e^{itY_{n}})\rightarrow e^{-t^{2}/2}$ for all $t\in \mathbb {R}$ . For simplicity, let

$$ \begin{align*}\xi_n=\xi_n(t)=\mathbb{E}_{\lambda}(e^{itY_{n}}).\end{align*} $$

Since $\mathbb {R}$ can be covered by countably many compact sets, it suffices to prove that

$$ \begin{align*} \lim_{n\rightarrow\infty}\max_{t\in [-T, T]} |\xi_n(t)-e^{-t^{2}/2}|=0, \quad \mathbb{P}\text{-almost surely}, \end{align*} $$

where $T>0$ is a fixed integer. This is implied by

$$ \begin{align*} \lim_{k\rightarrow\infty}\max_{n\in [2^k, 2^{k+1})}\max_{t\in [-T, T]} |\xi_n(t)-e^{-t^{2}/2}|=0,\quad \mathbb{P}\text{-almost surely}. \end{align*} $$

We start by estimating the second moment $\mathbb {E}_{\mathbb {P}}(|\xi _n^2(t)-e^{-t^{2}/2}|^{2})$ . First, we can write

$$ \begin{align*}|\xi_n-e^{-t^{2}/2}|^{2}=|\xi_n|^{2}- 2e^{-t^{2}/2}{\mathrm{Re}}(\xi_n)+e^{-t^{2}}=(|\xi_n|^{2}-e^{-t^{2}})-2e^{-t^{2}/2}({\mathrm{Re}}(\xi_n)-e^{-t^{2}/2}).\end{align*} $$

Note that $|\xi _n|^2=\mathbb {E}_{\lambda \times \lambda }(e^{itZ_{n}})$ by Fubini’s theorem. Hence, taking expectations with respect to $\mathbb {P}$ leads to

$$ \begin{align*} \mathbb{E}_{\mathbb{P}}(|\xi_n(t)-e^{-t^{2}/2}|^{2})= (\mathbb{E}_{\mathbb{P}\times\lambda\times\lambda}(e^{itZ_{n}})-e^{-t^{2}}) -2e^{-t^{2}/2} (\mathbb{E}_{\mathbb{P}\times\lambda}(e^{itY_{n}})-e^{-t^{2}/2}). \end{align*} $$

We apply hypotheses (1) and (2) of the theorem to compute an $L^2$ estimate of the random variable $\xi _n(t)-e^{-t^{2}/2}$ , which gives

(3.1) $$ \begin{align} \mathbb{E}_{\mathbb{P}}(|\xi_n(t)-e^{-t^{2}/2}|^{2})\ll (1+|t|^3) n^{-\sigma}. \end{align} $$

By Chebyshev’s inequality and (3.1), for any $\varepsilon>0$ ,

(3.2) $$ \begin{align} \mathbb{P}(\{|\xi_n(t)-e^{-t^{2}/2}|\geq\varepsilon\})\ll \varepsilon^{-2}(1+|t|^3) n^{-\sigma}. \end{align} $$

Now we choose b with $\max \{{(\tau +1)}/{2}, 1-{\tau }/{2}\}<b<1$ . We divide the interval of integers $[2^k, 2^{k+1})$ into $2^{(1-b)k}$ intervals of length $2^{bk}$ , denoted by $\{I_{k, p}\}_{p=1}^{2^{(1-b)k}}$ , and we divide the interval of real numbers $[-T, T]$ into $2Tk$ intervals of length $k^{-1}$ , denoted by $\{J_{k, q}\}_{q=1}^{2Tk}$ . In this way,

$$ \begin{align*}\{(n, t)\in\mathbb{N}\times\mathbb{R}: \;2^{k}\leq n< 2^{k+1}, \;t\in [-T, T]\}\subset \bigcup_{p, q} I_{k, p}\times J_{k, q},\end{align*} $$

so that

$$ \begin{align*} \max_{n\in [2^k, 2^{k+1})}\max_{t\in [-T, T]} |\xi_n(t)-e^{-t^{2}/2}|\leq \sup_{p, q}\max_{n\in I_{k, p}}\max_{t\in J_{k, q}} |\xi_n(t)-e^{-t^{2}/2}|. \end{align*} $$

In each ‘rectangle’ $I_{k, p}\times J_{k, q}$ , we choose a point $(n_{k, p}, t_{k, q})$ and we compare the maximum on this ‘rectangle’ with the value at the point $(n_{k, p}, t_{k, q})$ . For $(n, t)\in I_{k, p}\times J_{k, q}$ ,

$$ \begin{align*} |\xi_n(t)-e^{-t^{2}/2}|&\leq |\xi_n(t)-\xi_{n_{k, p}}(t)| + |\xi_{n_{k, p}}(t)-\xi_{n_{k, p}}(t_{k, q})|\\ &\quad+|\xi_{n_{k, p}}(t_{k, q})-e^{-t_{k, q}^{2}/2}| + |e^{-t_{k, q}^{2}/2}-e^{-t^{2}/2}|. \end{align*} $$

The third term on the right-hand side of the above inequality does not depend on t but only on $t_{k, q}$ . By using the inequality $|e^{ia}-e^{ib}|\leq |a-b|$ , the first, second and fourth terms can be estimated by

$$ \begin{align*}|\xi_n(t)-\xi_{n_{k, p}}(t)|=|\mathbb{E}_{\lambda}(e^{itY_{n}}-e^{itY_{n_{k, p}}})|\leq \mathbb{E}_{\lambda}(|e^{itY_{n}}-e^{itY_{n_{k, p}}}|),\end{align*} $$

$$ \begin{align*}|\xi_{n_{k, p}}(t)-\xi_{n_{k, p}}(t_{k, q})|=|\mathbb{E}_{\lambda}(e^{itY_{n_{k, p}}}-e^{it_{k, q}Y_{n_{k, p}}})|\leq\mathbb{E}_{\lambda}(|e^{itY_{n_{k, p}}}-e^{it_{k, q}Y_{n_{k, p}}}|)\end{align*} $$

and

$$ \begin{align*}|e^{-t_{k, q}^{2}/2}-e^{-t^{2}/2}|\leq \frac{T}{k}\quad\text{as}\ k\rightarrow\infty.\end{align*} $$

The fourth term tends to zero. So we only need to prove that, $\mathbb {P}$ -almost surely,

(3.3) $$ \begin{align} \lim_{k\rightarrow\infty}\sup_{p, q}\max_{n\in I_{k, p}}\max_{t\in J_{k, q}} \mathbb{E}_{\lambda}(|e^{itY_{n}}-e^{itY_{n_{k, p}}}|)=0, \end{align} $$

(3.4) $$ \begin{align} \lim_{k\rightarrow\infty}\max_{1\leq p\leq 2^{(1-b)k}}\max_{1\leq q\leq 2Tk}|\xi_{n_{k, p}}(t_{k, q})-e^{-t_{k, q}^{2}/2}|=0 \end{align} $$

and

(3.5) $$ \begin{align} \lim_{k\rightarrow\infty}\sup_{p, q}\max_{t\in J_{k, q}}\mathbb{E}_{\lambda}(|e^{itY_{n_{k, p}}}-e^{it_{k, q}Y_{n_{k, p}}}|)=0. \end{align} $$

To prove (3.3)–(3.5), we bound the tail probability of certain events and then apply the Borel–Cantelli lemma. Let $\epsilon>0$ be arbitrarily small. First,

(3.6) $$ \begin{align} & \mathbb{P}(\sup_{p, q}\max_{n\in I_{k, p}}\max_{t\in J_{k, q}} \mathbb{E}_{\lambda}(|e^{itY_{n}}-e^{itY_{n_{k, p}}}|)\geq 2\epsilon) \notag \\ & \leq \sum_{1\leq p\leq 2^{(1-b)k}}\sum_{1\leq q\leq 2Tk}\mathbb{P}(\max_{n\in I_{k, p}}\max_{t\in J_{k, q}}\mathbb{E}_{\lambda}(|e^{itY_{n}}-e^{itY_{n_{k, p}}}|)\geq 2\epsilon) \notag \\ & \leq \sum_{1\leq p\leq 2^{(1-b)k}}\sum_{1\leq q\leq 2Tk} 2 \epsilon^{-1}\mathbb{P}\times\lambda(\{\max_{n\in I_{k, p}}\max_{t\in J_{k, q}}|e^{itY_{n}}-e^{itY_{n_{k, p}}}|\geq\epsilon\}) \notag \quad (\text{by Lemma } 2.3)\\ & \leq \kern-1pt \sum_{1\leq p\leq 2^{(1-b)k}}\sum_{1\leq q\leq 2Tk} \kern-3pt 2\epsilon^{-1} \mathbb{P}\times\lambda\bigg(\bigg\{\max_{n\in I_{k, p}}\max_{t\in J_{k, q}}|Y_n-Y_{n_{k, p}}|\geq\frac{\epsilon}{T}\bigg\}\bigg) \quad (\text{since} \;|e^{ia}-e^{ib}|\leq |a-b|) \notag\\ & \leq \sum_{1\leq p\leq 2^{(1-b)k}}\sum_{1\leq q\leq 2Tk}\sum_{n\in I_{k, p}} 2\epsilon^{-1} \mathbb{P}\times\lambda\bigg(\bigg\{|Y_n-Y_{n_{k, p}}|\geq\frac{\epsilon}{T}\bigg\}\bigg). \end{align} $$

Note that $Y_n-Y_{n_{k, p}}$ is a weighted sum of $\mathbf {1}_{I_{k}}-\ell _{k}$ and that the probability law of $\mathbf {1}_{I_{k}}-\ell _{k}$ is independent of x, so

(3.7) $$ \begin{align} \mathbb{P}\times\lambda\bigg(\bigg\{|Y_n-Y_{n_{k, p}}|\geq\frac{\epsilon}{T}\bigg\}\bigg)=\mathbb{P}\bigg(\bigg\{|Y_n-Y_{n_{k, p}}|\geq\frac{\epsilon}{T}\bigg\}\bigg). \end{align} $$

Write $S_n=\sum _{k=1}^{n}(\mathbf {1}_{I_{k}}-\ell _{k})$ and assume that $n> n_{k, p}$ (the case $n<n_{k, p}$ is the same). Since $Y_n-Y_{n_{k, p}}=(V_n^{-1}-V_{n_{k, p}}^{-1})S_n-V_{n_{k, p}}^{-1}(S_n-S_{n_{k, p}}) = a_n S_n+b_n(S_n-S_{n_{k, p}})$ , say,

(3.8) $$ \begin{align} \mathbb{P}\bigg(\bigg\{|Y_n-Y_{n_{k, p}}|\geq\frac{\epsilon}{T}\bigg\}\bigg)\leq \mathbb{P}\bigg(\bigg\{|S_n|\geq\frac{\epsilon}{2T|a_n|}\bigg\}\bigg) + \mathbb{P}\bigg(\bigg\{|S_n-S_{n_{k, p}}|\geq\frac{\epsilon}{2T|b_n|}\bigg\}\bigg). \end{align} $$

By (2.1), the hypotheses of Lemma 2.2 are satisfied for the sequence $\{\mathbf {1}_{I_{n}}-\ell _{n}\}_{n\geq 1}$ with the constant $H=1$ . Now apply Lemma 2.2 twice to $S_n$ and $S_n-S_{n_{k, p}}$ , which are both sums of independent random variables under the probability $\mathbb {P}$ . This gives

(3.9) $$ \begin{align} \mathbb{P}\bigg(\bigg\{|S_n|\geq\frac{\epsilon}{2T|a_n|}\bigg\}\bigg)\leq 2\exp\bigg[\frac{-\epsilon^2}{8T^2a_n^2 V_n^2+4T|a_n|\epsilon}\bigg] \end{align} $$

and

(3.10) $$ \begin{align} \mathbb{P}\bigg(\bigg\{|S_n-S_{n_{k, p}}|\geq\frac{\epsilon}{2T|b_n|}\bigg\}\bigg)\leq 2\exp\bigg[\frac{-\epsilon^2}{8T^2 b_n^2 (V_n^2-V_{n_{k, p}}^2)+4T\epsilon |b_n|}\bigg]. \end{align} $$

Since $2^k\leq n < 2^{k+1}$ , $|n-n_{k, p}|\leq 2^{bk}$ and $V_n^2\sim {c}n^{1-\tau }/{(1-\tau )}$ , by the mean value theorem,

(3.11) $$ \begin{align} |V_n^2-V_{n_{k, p}}^2|=\sum_{k=n_1+1}^n ck^{-\tau}\ll |n^{1-\tau}-n_1^{1-\tau}|\ll 2^{(b-\tau)k}. \end{align} $$

Recall that $a_n=V_n^{-1}-V_{n_{k, p}}^{-1}$ , $b_n=V_{n_{k, p}}^{-1}$ and $\max \{{(\tau +1)}/{2}, 1-{\tau }/{2}\}< b <1$ . Combining (3.7)–(3.11),

(3.12) $$ \begin{align} \mathbb{P}\times\lambda\bigg(\bigg\{|Y_n-Y_{n_{k, p}}|\geq\frac{\epsilon}{T}\bigg\}\bigg)\ll \exp\bigg[\frac{-\varepsilon^2 2^{(1-b)) k}}{8T^2}\bigg]. \end{align} $$

Substituting (3.12) into (3.6) yields

$$ \begin{align*} \mathbb{P}(\sup_{p, q}\max_{n\in I_{k, p}}\max_{t\in J_{k, q}} \mathbb{E}_{\lambda}(|e^{itY_{n}}-e^{itY_{n_{k, p}}}|)\geq 2\epsilon)\ll Tk2^{k}\exp\bigg[\frac{-\varepsilon^2 2^{(1-b)) k}}{8T^2}\bigg]. \end{align*} $$

Since the sum over k of the terms on the right-hand side of the last inequality is finite, by the convergence part of the Borel–Cantelli lemma, the events

$$ \begin{align*}\{\sup_{p, q}\max_{n\in I_{k, p}}\max_{t\in J_{k, q}}\mathbb{E}_{\lambda}(|e^{itY_{n}}-e^{itY_{n_{k, p}}}|)\geq 2 \epsilon\}\end{align*} $$

happen only finitely many times with probability one. Since $\epsilon>0$ is arbitrarily small, this shows that (3.3) holds $\mathbb {P}$ -almost surely.

Second, for the proof of (3.4), observe that

(3.13) $$ \begin{align} &\;\mathbb{P}(\max_{1\leq p\leq 2^{(1-b)k}}\max_{1\leq q\leq 2Tk}|\xi_{n_{k, p}}(t_{k, q})-e^{-t_{k, q}^{2}/2}|\geq \epsilon)\notag\\ &\quad \leq \sum_{1\leq p\leq 2^{(1-b)k}}\sum_{1\leq q\leq 2Tk}\mathbb{P}(|\xi_{n_{k, p}}(t_{k, q})-e^{-t_{k, q}^{2}/2}|\geq \epsilon)\notag \\ &\quad \ll \varepsilon^{-2}(1+|T|^3)Tk 2^{(1-b-\sigma)k} \quad (\text{using } (3.2), n_{k, p}<2^{k+1}\;\text{and}\;|t_{k, q}|\leq T). \end{align} $$

By the choice of b, we have $\sigma>1-b$ , so the sum over k of the terms in (3.13) is finite. Applying Borel–Cantelli again gives the $\mathbb {P}$ -almost sure convergence of (3.4).

Third, for the proof of (3.5),

$$ \begin{align*} & \mathbb{P}(\sup_{p, q}\max_{t\in J_{k, q}}\mathbb{E}_{\lambda}(|e^{itY_{n_{k, p}}}-e^{it_{k, q}Y_{n_{k, p}}}|)\geq 2 \epsilon) \notag \\ & \leq \sum_{1\leq p\leq 2^{(1-b)k}}\sum_{1\leq q\leq 2Tk}\mathbb{P}(\max_{t\in J_{k, q}}\mathbb{E}_{\lambda}(|e^{itY_{n_{k, p}}}-e^{it_{k, q}Y_{n_{k, p}}}|)\geq 2\epsilon) \notag \\ & \leq \sum_{1\leq p\leq 2^{(1-b)k}}\sum_{1\leq q\leq 2Tk} 2 \epsilon^{-1}\mathbb{P}\times\lambda(\{\max_{t\in J_{k, q}}\mathbb{E}_{\lambda}(|e^{itY_{n_{k, p}}}-e^{it_{k, q}Y_{n_{k, p}}}|)\geq \epsilon\}) \notag \quad (\text{by Lemma } 2.3)\\ & \leq \sum_{1\leq p\leq 2^{(1-b)k}}\sum_{1\leq q\leq 2Tk} 2\epsilon^{-1} \mathbb{P}\times\lambda(\{|Y_{n_{k, p}}|\geq k \epsilon\}) \quad(\text{since} \;|e^{ia}-e^{ib}|\leq |a-b|) \notag\\ & \leq \sum_{1\leq p\leq 2^{(1-b)k}}\sum_{1\leq q\leq 2Tk} 4\epsilon^{-1} e^{-{k^2\epsilon^2}/{4}}\leq 8\epsilon^{-1} Tk 2^{(1-b)k} e^{-{k^2\epsilon^2}/{4}} \quad (\text{by Hypothesis (3)}). \end{align*} $$

Since the sum over k of the terms in the last inequality is finite, the Borel–Cantelli lemma implies the $\mathbb {P}$ -almost sure convergence of (3.5). This completes the proof.

4 Proof of Theorem 1.1 modulo Proposition 1.2

We only need to verify hypotheses (1)–(3) in Proposition 1.2.

Hypothesis (1). For fixed $x\in \mathbb {T}$ and each $k\geq 1$ , the three moments of $\{\mathbf {1}_{I_{k}}(x)-\ell _{k}\}_{k\geq 1}$ are trivially computed: that is,

$$ \begin{align*}\mathbb{E}_{\mathbb{P}}(\mathbf{1}_{I_{k}}(x)-\ell_k)=0, \quad \mathbb{E}_{\mathbb{P}}(|\mathbf{1}_{I_{k}}(x)-\ell_k|^2)=\ell_k(1-\ell_k)\end{align*} $$

and

$$ \begin{align*}\mathbb{E}_{\mathbb{P}}(|\mathbf{1}_{I_{k}}(x)-\ell_k|^3)=\ell_k(1-\ell_k)(1-2\ell_k+2\ell_k^2)<\infty.\end{align*} $$

As $x(1-x)\leq 1/4$ for $1\leq x\leq 1$ , it follows that

$$ \begin{align*}\frac{1}{2V_n}\leq B_{n}:=V_{n}^{-3}\sum_{k=1}^{n}\mathbb{E}_{\mathbb{P}}(|\mathbf{1}_{I_{k}}(x)-\ell_k|^{3})\leq\frac{1}{V_n}.\end{align*} $$

By Lemma 2.1, for any $|t|\leq V_n/4$ ,

(4.1) $$ \begin{align} |\mathbb{E}_{\mathbb{P}}(e^{it Y_{n}})-e^{-t^{2}/2}|\leq \frac{16|t|^{3}e^{-t^{2}/3}}{V_n}. \end{align} $$

Since (4.1) holds uniformly for any fixed $x\in \mathbb {T}$ , we can integrate both sides with respect to x. Hypothesis (1) then follows by Fubini’s theorem and the fact $V_n\asymp n^{(1-\tau )/2}$ .

Hypothesis (2). We also appeal to Lemma 2.1, but, in this case, for the sequence of random variables $\{\mathbf {1}_{I_{k}}(x)-\mathbf {1}_{I_{k}}(y)\}_{k\geq 1}$ for any fixed $x, y\in \mathbb {T}$ . By Lemma 2.4, for any $k\geq 1$ , we have $\mathbb {E}_{\mathbb {P}}(\mathbf {1}_{I_{k}}(x)-\mathbf {1}_{I_{k}}(y))=0$ and

$$ \begin{align*} \mathbb{E}_{\mathbb{P}}(|\mathbf{1}_{I_{k}}(x)-\mathbf{1}_{I_{k}}(y)|^2)=\mathbb{E}_{\mathbb{P}}(|\mathbf{1}_{I_{k}}(x)-\mathbf{1}_{I_{k}}(y)|^3)=2 \min\{\ell_k, |x-y|\}<\infty. \end{align*} $$

Since $\ell _k=c k^{-\tau }$ is strictly decreasing, there exists $k_0=\lfloor c^{1/\tau }|x-y|^{-1/\tau }\rfloor +1$ such that $\ell _k<|x-y|$ if and only if $k> k_0$ . We denote

$$ \begin{align*}\widetilde{V}_n^2:=\sum_{i=1}^{n}\mathbb{E}_{\mathbb{P}}((\mathbf{1}_{I_{k}}(x)-\mathbf{1}_{I_{k}}(y))^{2})=\sum_{i=1}^{k_0}2|x-y|+\sum_{k=k_0+1}^n 2\ell_k\end{align*} $$

and

$$ \begin{align*}\widetilde{B}_{n}=\widetilde{V}_{n}^{-3}\sum_{k=1}^{n}\mathbb{E}_{\mathbb{P}}(|\mathbf{1}_{I_{k}}(x)-\mathbf{1}_{I_{k}}(y)|^{3})=\widetilde{V}_n^{-1}, \quad \widetilde{Z}_n=\widetilde{V}_n^{-1}\sum_{k=1}^{n}(\mathbf{1}_{I_{k}}(x)-\mathbf{1}_{I_{k}}(y)).\end{align*} $$

By Lemma 2.1, for all $t\in \mathbb {R}$ with $|t|\leq {1}/{4\sqrt {2}\widetilde {B}_{n}}$ ,

(4.2) $$ \begin{align} |\mathbb{E}_{\mathbb{P}}(e^{i\sqrt{2}t\widetilde{Z}_n})-e^{-t^{2}}|\leq 32\sqrt{2} \widetilde{V}_n^{-1}|t|^{3}e^{-2t^{2}/3}. \end{align} $$

In addition, the inequality $|e^{ia}-e^{ib}|\leq |a-b|$ implies that

(4.3) $$ \begin{align} |\mathbb{E}_{\mathbb{P}}(e^{i t Z_n})-\mathbb{E}_{\mathbb{P}}(e^{i\sqrt{2}t\widetilde{Z}_n})| \leq |t|\cdot|1/V_n-\sqrt{2}/\widetilde{V}_n|\cdot\mathbb{E}_{\mathbb{P}}\bigg(\bigg|\!\sum_{i=1}^n\mathbf{1}_{I_{k}}(x)-\mathbf{1}_{I_{k}}(y)\bigg|\bigg). \end{align} $$

Now we estimate the right-hand sides of (4.2) and (4.3). For any sufficiently large n, if $x, y\in \mathbb {T}$ satisfies $|x-y|>n^{-\tau /2}$ , then

(4.4) $$ \begin{align} \widetilde{V}_n^2=\sum_{k=1}^{k_0}2|x-y|+\sum_{k=k_0+1}^n 2\ell_k=2 k_0 |x-y|+\sum_{k=k_0+1}^n 2 ck^{-\tau}\asymp |x-y|^{1-1/\tau}+n^{1-\tau}\asymp n^{1-\tau}. \end{align} $$

Since $V_n^2\sim {c}n^{1-\tau }/{(1-\tau )}$ ,

(4.5) $$ \begin{align} |1/V_n-\sqrt{2}/\widetilde{V}_n| \ll n^{-3(1-\tau)/2} \bigg|\!\sum_{i=1}^{i_0}\ell_k-\sum_{i=1}^n \ell_k^2-\sum_{i=1}^{i_0}|x-y|\bigg|\ll n^{-(1+\tau)/2}+n^{-(1-\tau)}. \end{align} $$

In addition, by (2.3), for any fixed $x, y\in \mathbb {T}$ ,

$$ \begin{align*}|\mathbb{E}_{\mathbb{P}}((\mathbf{1}_{I_{k}}(x)-\mathbf{1}_{I_{k}}(y))^m)|\leq\tfrac{1}{2}m!\widetilde{\sigma}_k^2 H^{m-1}\quad \mbox{for all } k\geq 1, m\geq 2,\end{align*} $$

where $H=1$ and $\widetilde {\sigma }_k^2=\mathbb {E}_{\mathbb {P}}((\mathbf {1}_{I_{k}}(x)-\mathbf {1}_{I_{k}}(y))^2)$ . Write $T_n=\sum _{i=1}^n (\mathbf {1}_{I_{k}}(x)-\mathbf {1}_{I_{k}}(y))$ , where $x, y$ satisfy $|x-y|> n^{-\tau /2}$ . Since $|\mathbf {1}_{I_{k}}(x)-\mathbf {1}_{I_{k}}(y)|\leq 2$ , applying Lemma 2.2 to the sequence $\{\mathbf {1}_{I_{k}}(x)-\mathbf {1}_{I_{k}}(y)\}_{k\geq 1}$ with $a=\widetilde {V}_n^{1+\eta }$ gives

$$ \begin{align*} \mathbb{E}_{\mathbb{P}}\bigg(\bigg|\!\sum_{i=1}^n\mathbf{1}_{I_{k}}(x)-\mathbf{1}_{I_{k}}(y)\bigg|\bigg) & = \mathbb{E}_{\mathbb{P}}(|T_n|\cdot\mathbf{1}_{\{|T_n|\geq a\}})+\mathbb{E}_{\mathbb{P}}(|T_n|\cdot\mathbf{1}_{\{|T_n|< a\}}) \\ & \leq 2n\mathbb{P}(\{|T_n|\geq a\}) + a \leq 4n\exp\bigg(\frac{-\widetilde{V}_n^{2\eta}}{4}\bigg)+\widetilde{V}_n^{1+\eta}, \end{align*} $$

where $\eta>0$ is a constant arbitrarily close to 0. Combining this with (4.4) yields

(4.6) $$ \begin{align} \mathbb{E}_{\mathbb{P}}\bigg(\bigg|\!\sum_{i=1}^n\mathbf{1}_{I_{k}}(x)-\mathbf{1}_{I_{k}}(y)\bigg|\bigg) \ll n e^{-n^{\eta (1-\tau)}/4} + n^{(1+\eta)(1-\tau)/2}. \end{align} $$

Substituting (4.5) and (4.6) into (4.3), we obtain

(4.7) $$ \begin{align} |\mathbb{E}_{\mathbb{P}}(e^{i t Z_n})-\mathbb{E}_{\mathbb{P}}(e^{i\sqrt{2}t\widetilde{Z}_n})|\ll |t|(n^{-\tau+\eta} + n^{-(1-\tau)/2+\eta}). \end{align} $$

Combining (4.2), (4.4) and (4.7), we see that, for any $x, y\in \mathbb {T}$ with $|x-y|> n^{-\tau /2}$ ,

$$ \begin{align*} |\mathbb{E}_{\mathbb{P}}(e^{i t Z_n})-e^{-t^2}| & \leq |\mathbb{E}_{\mathbb{P}}(e^{i t Z_n})-\mathbb{E}_{\mathbb{P}}(e^{i\sqrt{2}t\widetilde{Z}_n})|+ |\mathbb{E}_{\mathbb{P}}(e^{i\sqrt{2}t\widetilde{Z}_n})-e^{-t^2}|\nonumber\\ & \ll |t|(n^{-\tau+\eta} + n^{-(1-\tau)/2+\eta}) + |t|^3 n^{-(1-\tau)/2} \end{align*} $$

holds for all $t\in \mathbb {R}$ with $|t|\leq {\widetilde {V}_{n}}/{4\sqrt {2}}$ . Therefore, by Fubini’s theorem,

$$ \begin{align*} &|\mathbb{E}_{\mathbb{P}\times\lambda\times\lambda}(e^{it Z_{n}})-e^{-t^{2}}| \leq \mathbb{E}_{\lambda\times\lambda}(|\mathbb{E}_{\mathbb{P}}(e^{i t Z_n})-e^{-t^2}|)\nonumber\\ &\quad \leq \mathbb{E}_{\lambda\times\lambda}(|\mathbb{E}_{\mathbb{P}}(e^{i t Z_n})-e^{-t^2}|\cdot\mathbf{1}_{\{|x-y|\leq n^{-\tau/2}\}})+\mathbb{E}_{\lambda\times\lambda}(|\mathbb{E}_{\mathbb{P}}(e^{i t Z_n})-e^{-t^2}|\cdot\mathbf{1}_{\{|x-y|>n^{-\tau/2}\}})\nonumber\\ &\quad \ll n^{-\tau/2} + |t|(n^{-\tau+\eta} + n^{-(1-\tau)/2+\eta}) + |t|^3 n^{-(1-\tau)/2} \ll (1+|t|^3) n^{-\sigma}, \end{align*} $$

where $\sigma $ is a constant satisfying $\sigma>\min \{{(1-\tau )}/{2}, {\tau }/{2}\}$ .

Hypothesis (3). We first appeal to Lemma 2.2 for the sequence of random variables $\{\mathbf {1}_{I_{k}}(x)-\ell _{k}\}_{k\geq 1}$ for any fixed $x\in \mathbb {T}$ . From the discussion in Section 3,

$$ \begin{align*}|\mathbb{E}_{\mathbb{P}}((\mathbf{1}_{I_{k}}(x)-\ell_{k})^m)|\leq\tfrac{1}{2}m!\sigma_k^2H^{m-1}\quad\mbox{for all } k\geq 1, m\geq 2\end{align*} $$

with $H=1$ and $\sigma _k^2=\mathbb {E}_{\mathbb {P}}((\mathbf {1}_{I_{k}}(x)-\ell _{k})^2$ . Moreover, $V_n^2=\sum _{k=1}^n \ell _k(1-\ell _k)$ . Applying Lemma 2.2 to $\{\mathbf {1}_{I_{k}}(x)-\ell _{k}\}_{k\geq 1}$ and $a=\varepsilon V_n$ gives

$$ \begin{align*}\mathbb{P}(|Y_{n}|\geq\varepsilon)=\mathbb{P}\bigg(\bigg|\!\sum_{k=1}^n \mathbf{1}_{I_{k}}(x)-\ell_{k}\bigg|\geq \varepsilon V_n\bigg)\leq 2\exp\bigg[\frac{-\varepsilon^2 V_n^2}{2(V_n^2+\epsilon V_n)}\bigg] \leq 2\exp\bigg[\frac{-\varepsilon^2}{4}\bigg],\end{align*} $$

where the last inequality is due to the fact that $\epsilon V_n<V_n^2$ for n large enough. The inequality in Hypothesis (3) follows from Fubini’s theorem.