Properties. For any $s\in \mathfrak {F}$ and any $XYZ\in \mathcal {C}_{n}$ , we have

(2.1) $$ \begin{align} s_{X}(YZ)=s_{X}(Y)s_{XY}(Z)\quad\text{and}\quad s_{X_{Y}}(Z)=s_{XY}(Z). \end{align} $$

For any finite measure $\nu \in \mathfrak {F}$ , any cylinder $X\in \mathcal {C}$ , and any integers n and p, we have

(2.2) $$ \begin{align} \langle \nu_{X},\log s_{X} \rangle_{n+p}=\langle \nu_{X},\log s_{X} \rangle_{n}+\sum_{Y\in T^{|X|}\mathcal{C}_{|X|+n}}\nu_{X}(Y)\langle \nu_{XY},\log s_{XY} \rangle_{p}. \end{align} $$

Proof. Since equality (2.2) will be used extensively in what follows, and since it allows one to become familiar with the introduced notation, we provide a proof.

Starting from the definition, we have

$$ \begin{align*} \begin{aligned} \langle \nu_X,\log s_{X} \rangle_{n+p} & = \sum_{|Y|=n+p}\nu_X(Y)\log s_{X}(Y)\\ & = \sum_{|Y|=n,\,|Z|=p}\frac{\nu(XY)}{\nu(X)}\frac{\nu(XYZ)}{\nu(XY)}\log \bigg(\frac{s(XYZ)}{s(XY)}\frac{s(XY)}{s(X)}\bigg)\\ & = \sum_{|Y|=n,\,|Z|=p}\nu_X(Y)\nu_{XY}(Z)\log(s_{XY}(Z)s_X(Y))\\ & = \sum_{|Y|=n,\,|Z|=p}\nu_X(Y)[\nu_{XY}(Z)\log(s_{XY}(Z)) +\nu_{XY}(Z)\log(s_X(Y))]\\ & = \sum_{|Y|=n}\nu_X(Y)\bigg[\!\sum_{|Z|=p}\nu_{XY}(Z)\log(s_{XY}(Z)) +\log(s_X(Y))\sum_{|Z|=p}\nu_{XY}(Z)\bigg]\\ &= \sum_{|Y|=n}\nu_X(Y)[\langle \nu_{XY},\log s_{XY} \rangle_{p} +\log(s_X(Y))\mathfrak{m}_{p}(\nu_{XY})].\\ \end{aligned} \end{align*} $$

Since $\nu $ is a finite measure, we have $\mathfrak {m}_{p}(\nu _{XY})=1$ so that we get the desired equality:

$$ \begin{align*} \begin{aligned} \langle \nu_{X},\log s_{X} \rangle_{n+p}=\langle \nu_{X},\log s_{X} \rangle_{n}+\sum_{Y\in T^{|X|}\mathcal{C}_{|X|+n}}\nu_{X}(Y)\langle \nu_{XY},\log s_{XY} \rangle_{p}. \end{aligned} \end{align*} $$

Proof. We make use of Jensen’s inequality for the function $\log $ , evaluated at $({\nu '(Y)}/{\nu (Y)})_{|Y|=n} $ , and the convex combination defined by the coefficients $({\nu (Y)}/{\mathfrak {m}_{n}(\nu )})_{|Y|=n}$ . We get

$$ \begin{align*} \frac{1}{\mathfrak{m}_{n}(\nu)}\langle \nu,\log \frac{\nu'}{\nu} \rangle_{n}=\sum_{|Y|=n}\frac{\nu(Y)}{\mathfrak{m}_{n}(\nu)}\log\bigg(\frac{\nu'(Y)}{\nu(Y)}\bigg) \le \log\bigg(\!\sum_{|Y|=n}\frac{\nu'(Y)}{\mathfrak{m}_{n}(\nu)}\bigg)=\log\bigg(\frac{\mathfrak{m}_{n}(\nu')}{\mathfrak{m}_{n}(\nu)}\bigg). \end{align*} $$

Multiplying this inequality by $\mathfrak {m}_{n}(\nu )$ , we obtain the result.

Since $\log $ is strictly concave, equality holds if and only if $ {\nu '(Y)}/{\nu (Y)}$ does not depend on Y.

Proof. The constant C will change from line to line, but will stay independent of integers or cylinders, only depending on $\nu $ and s.

For any $\tilde {s}\sim s$ , we have, for any $XY\in \mathcal {C}$ , $-2\log C \le \log \tilde {s}_X(Y)-\log s_X(Y)\le 2\log C$ .

We then deduce, because $\nu $ is a finite measure,

$$ \begin{align*}-2\log C\le\langle \nu_{X},\log \tilde{s}_{X} \rangle_{n}-\langle \nu_{X},\log s_{X} \rangle_{n}\le 2\log C.\end{align*} $$

This tells us that conclusions of the theorem are true for $\tilde {s}$ as soon as they are true for s, and in the following, we focus on s.

Let:

• • for any $X\in \mathcal {C}$ , $g_{n}(X):=\langle \nu _{X},\log s_{X} \rangle _{n},$ with $g_{n}:=g_{n}({\varnothing })$ ;

• • $\alpha _{n}:=\alpha _{n}(\nu ,s)=\sup _k\{\max _{|X|=k}g_{n}(X)-\min _{|X|=k}g_{n}(X)\}$ .

We first note that (2.5.1) and the fact that $\nu $ is a finite measure imply that for any $X\in \mathcal {C}$ , we have

(2.7) $$ \begin{align} |g_{n}(X)|=|\langle \nu_{X},\log s_{X} \rangle_{n}|\le \gamma_{n}, \end{align} $$

an inequality that is also true with (2.5.2) and leads to

(2.8) $$ \begin{align} \alpha_{n}\le 2\gamma_{n}. \end{align} $$

Note also that given any $X\in \mathcal {C}$ , and any integers n and p, we have from (2.2):

(2.9) $$ \begin{align} g_{n+p}(X)=g_n(X)+\sum_{|Y|=n}\nu_{X}(Y)g_{p}(XY). \end{align} $$

From this, we deduce that for any $X ' \in \mathcal {C}_{|X|}$ , we have

(2.10) $$ \begin{align} g_{n+p}(X)-g_{n+p}(X') & = g_{n}(X)-g_{n}(X')+\sum_{|Y|=n}(\nu_{X}(Y)-\nu_{X'}(Y))g_{p}(XY)\nonumber\\ &\quad + \sum_{|Y|=n}\nu_{X'}(Y)(g_{p}(XY)-g_{p}(X'Y)). \end{align} $$

Let $\mathcal {B}^+$ be the subset of $T^{|X|}\mathcal {C}_{|X|+n}$ for which we have $\nu _{X}(Y)-\nu _{X'}(Y)\ge 0$ and let $\mathcal {B}^-$ be its complementary. Then, we have

$$ \begin{align*} \sum_{|Y|=n}(\nu_{X}(Y)-\nu_{X'}(Y))g_{p}(XY) & \le \max_{|Z|=|X|+n}g_{p}(Z)\sum_{Y\in \mathcal{B}^+}(\nu_{X}(Y)-\nu_{X'}(Y))\\ & \quad + \min_{|Z|=|X|+n}g_{p}(Z)\sum_{Y\in \mathcal{B}^-}(\nu_{X}(Y)-\nu_{X'}(Y)). \end{align*} $$

Since $\nu $ is a finite measure, we have $ \sum _{|Y|=n}\nu _{X}(Y)=\sum _{|Y|=n}\nu _{X'}(Y)=1$ .

This implies that $ \sum _{|Y|=n}(\nu _{X}(Y)-\nu _{X'}(Y))=0$ .

Therefore, $ \sum _{Y\in \mathcal {B}^+}(\nu _{X}(Y)-\nu _{X'}(Y))=-\sum _{Y\in \mathcal {B}^-}(\nu _{X}(Y)-\nu _{X'}(Y))$ , which leads to

$$ \begin{align*} \sum_{|Y|=n}(\nu_{X}(Y)-\nu_{X'}(Y))g_{p}(XY) & \le \sum_{Y\in \mathcal{B}^+}(\nu_{X}(Y)-\nu_{X'}(Y)) \bigg(\!\max_{|Z|=|X|+n}g_{p}(Z)-\min_{|Z|=|X|+n}g_{p}(Z)\bigg)\\ & \le \alpha_{p} \sum_{Y\in \mathcal{B}^+}(\nu_{X}(Y)-\nu_{X'}(Y)) \\ & \le \alpha_{p} \sum_{Y\in \mathcal{B}^+}\nu_{X'}(Y)\bigg(\frac{\nu_{X}(Y)}{\nu_{X'}(Y)}-1\bigg) \\ & \le \alpha_{p}\beta_0 \sum_{Y\in \mathcal{B}^+}\nu_{X'}(Y)\quad \text{(because of ASI)} \\ & \le \alpha_{p}\beta_0 \quad\text{(because } \nu \text{ is a finite measure)}. \end{align*} $$

We thus have

(2.11) $$ \begin{align} \sum_{|Y|=n}(\nu_{X}(Y)-\nu_{X'}(Y))g_{p}(XY)\le \unicode{x3bb}\alpha_{p}. \end{align} $$

From ASI, we deduce that

$$ \begin{align*}\nu_{XY}(Z)\le (1+\beta_{|Y|})\nu_{X'Y}(Z),\end{align*} $$

and also

$$ \begin{align*} \log s_{XY}(Z)\le\log s_{X'Y}(Z) +\beta_{|Y|}.\end{align*} $$

This implies the following.

• • If (2.5.1) is fulfilled, then we have $ |\!\log s_{X'Y}(Z)|\le \gamma _{|Z|}$ so that

  $$ \begin{align*} \begin{aligned} \nu_{XY}(Z)\log s_{XY}(Z)&-\nu_{X'Y}(Z)\log s_{X'Y}(Z)\\ &= (\nu_{XY}(Z)-\nu_{X'Y}(Z))\log s_{XY}(Z)+\nu_{X'Y}(Z)\log \frac{s_{XY}(Z)}{s_{X'Y}(Z)}\\ & \le |\nu_{XY}(Z)-\nu_{X'Y}(Z)| \gamma_{|Z|}+\nu_{X'Y}(Z)\bigg|\!\log \frac{s_{XY}(Z)}{s_{X'Y}(Z)}\bigg|\\ \end{aligned}\end{align*} $$
  and since $\nu $ and s fulfill ASI, we get
  $$ \begin{align*} \begin{aligned} & \le \beta_{|Y|}\nu_{X'Y}(Z)\gamma_{|Z|}+\nu_{X'Y}(Z)\beta_{|Y|}\\ & \le \beta_{|Y|}(1+\gamma_{|Z|})\nu_{X'Y}(Z). \end{aligned} \end{align*} $$
  Summing over all cylinders Z of length p and using once again the fact that $\nu $ is a finite measure, we get
  (2.12) $$ \begin{align} g_{p}(XY)-g_{p}(X'Y)\le (1+\gamma_{p})\beta_{|Y|}\le C\gamma_{p}\beta_{|Y|}. \end{align} $$

• • If (2.5.2) is fulfilled, then we have

  $$ \begin{align*} \begin{aligned} & \nu_{XY}(Z)\log s_{XY}(Z)-\nu_{X'Y}(Z)\log s_{X'Y}(Z) \\ & \quad \le \beta_{|Y|}(\nu_{X'Y}(Z)\log s_{X'Y}(Z)+C\nu_{X'Y}(Z)). \end{aligned} \end{align*} $$

  Summing over all cylinders Z of length p and using (2.7), we also get (2.12).

Injecting (2.12) and (2.11) into (2.10) leads to

$$ \begin{align*}g_{n+p}(X)-g_{n+p}(X') \le g_{n}(X)-g_{n}(X')+C\gamma_{p}\beta_{n}+\unicode{x3bb}\alpha_{p},\end{align*} $$

from which we get that

$$ \begin{align*} \alpha_{n+p}\le \alpha_{n}+C\gamma_{p}\beta_{n}+\unicode{x3bb}\alpha_{p},\end{align*} $$

which might be rewritten, for any $p\le n$ , in the following form:

$$ \begin{align*} \alpha_{n}\le \alpha_{p}+C\gamma_{n-p}\beta_{p}+\unicode{x3bb}\alpha_{n-p}\le \alpha_{p}+C\gamma_{n}\beta_{p}+\unicode{x3bb}\alpha_{n-p},\quad(\gamma_{n}) \text{ being increasing.} \end{align*} $$

Using (2.8), we get

$$ \begin{align*} \alpha_{n}\le 2\gamma_{p}+C\gamma_{n}\beta_{p}+\unicode{x3bb} \alpha_{n-p}\le C\max(\gamma_{p},\gamma_{n}\beta_{p})+\unicode{x3bb} \alpha_{n-p}.\end{align*} $$

Let q and $r\le p$ be the quotient and the remainder of the Euclidean division of n by p. By induction, we get

$$ \begin{align*}\begin{aligned} \alpha_{n} & \le C\max(\gamma_{p},\gamma_{n}\beta_{p})\sum_{k=0}^{q-1} \unicode{x3bb}^k+\unicode{x3bb}^q\alpha_{r}\\ & \le C\max(\gamma_{p},\gamma_{n}\beta_{p})+2\unicode{x3bb}^q\gamma_{r}\\ & \le C\max(\gamma_{p},\gamma_{n}\beta_{p})+2\unicode{x3bb}^q\gamma_{p}\\ & \le C\max(\gamma_{p},\gamma_{n}\beta_{p}). \end{aligned} \end{align*} $$

Since $(\gamma _{n})$ and $(\beta _{n})$ are compatible, one may take $p=p_{n}$ to get

$$ \begin{align*} \alpha_{n}\le Cc(\gamma_{n},\beta_{n}),\end{align*} $$

which is the first part of the proposition and tells us that for any X and $X'$ in $\mathcal {C}$ , with $|X|=|X'|$ , we have

$$ \begin{align*} \frac{1}{n}|g_{n(X)}-g_{n(X')}|\le \frac{\alpha_{n}}{n}\le C\frac{c(\gamma_{n},\beta_{n})}{n}\to_{n\to{{{+}}}\infty}0. \end{align*} $$

We show now that we can get rid of the condition $|X|=|X'|$ .

Let $X\in \mathcal {C}$ and let a be a cylinder of length 1. Applying (2.9) with $n=1$ and $p=q$ ,

$$ \begin{align*} g_{q+1}(X)&=g_{1}(X)+\sum_{|b|=1}\nu_{X}(b)g_{q}(Xb)\\ & =g_{q}(Xa)+g_{1}(X)+\sum_{|b|=1}\nu_{X}(b)(g_{q}(Xb)-g_{q}(Xa)). \end{align*} $$

Apply now (2.9) with $n=q$ and $p=1$ :

$$ \begin{align*}g_{q+1}(X)=g_{q}(X)+\sum_{|Y|=q}\nu_{X}(Y)g_{1}(XY).\end{align*} $$

We thus have

$$ \begin{align*}g_{q}(X)-g_{q}(Xa)=g_{1}(X)+\sum_{|b|=1}\nu_{X}(b)(g_{q}(Xb)-g_{q}(Xa))-\sum_{|Y|=q}\nu_{X}(Y)g_{1}(XY).\end{align*} $$

By hypothesis on $\nu $ and s, there exists $C>0$ such that for any cylinder $Z\in \mathcal {C}$ , we have $g_{1}(Z)\le C$ . This leads to

$$ \begin{align*} | g_{q}(X)-g_{q}(Xa)|\le C+\alpha_{q}, \end{align*} $$

and this easily leads, for any Y with $|Y|=p$ , to

(2.13) $$ \begin{align} | g_{q}(X)-g_{q}(XY)|\le (C+\alpha_{q})p\le C\alpha_{q}p. \end{align} $$

Let $X'\in \mathcal {C}_{|X|}$ and $XY\in \mathcal {C}$ . We have

$$ \begin{align*} | g_{q}(X')-g_{q}(XY)|\le | g_{q}(X')-g_{q}(X)|+| g_{q}(X)-g_{q}(XY)|\le \alpha_{q}+Cp\alpha_{q}\le C'p\alpha_{q}\end{align*} $$

and then,

$$ \begin{align*} \frac{1}{q}| g_{q}(X')-g_{q}(XY)|\le C'p\frac{\alpha_{q}}{q}, \end{align*} $$

which finishes the proof.

Proof. By hypothesis, we are in a position to apply Proposition 2.2 with $\nu $ and s, and $\nu _{n}$ and $s_{n}$ , and we use the notation therein.

Let $p\le n$ , with $q\in \mathbb {N}$ such that $qp<n\le (q+1)p$ . Using (2.9), we get

$$ \begin{align*} g_{n}(\varnothing):=g_{n}=g_{pq+r}=g_{pq}+\sum_{|Y|=pq}\nu(Y)g_{r}(Y), \end{align*} $$

from which we get

$$ \begin{align*} g_n-g_{k,n}=g_{pq}-g_{k,pq}+\sum_{|Y|=pq}[\nu(Y)g_{r}(Y)-\nu_k(Y)g_{k,r}(Y)] \end{align*} $$

and also

(2.17) $$ \begin{align} \frac{1}{n}\vert g_n-g_{k,n}\vert\le \frac{1}{n}\vert g_{pq}-g_{k,pq}\vert+\frac{1}{n}\bigg\vert\!\sum_{|Y|=pq}[\nu(Y)g_{r}(Y)-\nu_k(Y)g_{k,r}(Y)]\bigg\vert. \end{align} $$

By hypothesis, we have $|g_r(Y)|\le \gamma _r\le \gamma _p$ and $|g_{k,r}(Y)|\le \gamma _r\le \gamma _p$ , so that

(2.18) $$ \begin{align} \frac{1}{n}\bigg\vert\!\sum_{|Y|=pq}[\nu(Y)g_{r}(Y)-\nu_k(Y)g_{k,r}(Y)]\bigg\vert\le 2\frac{\gamma_p}{n}. \end{align} $$

Using once again (2.9), we get

$$ \begin{align*} g_{pq}=g_{p(q-1)}+\sum_{|Y|=p(q-1)}\nu(Y)g_{p}(Y), \end{align*} $$

so that

$$ \begin{align*} g_{pq}=\sum_{i=0}^{q-1}\sum_{|Y|=ip}\nu(Y)g_{p}(Y) \end{align*} $$

and, thus,

$$ \begin{align*} g_{pq}-g_{k,pq}=\sum_{i=0}^{q-1}\sum_{|Y|=ip}[\nu_k(Y)(g_{p}(Y)-g_{k,p}(Y))+g_{p}(Y)(\nu(Y)-\nu_k(Y))]. \end{align*} $$

Using (2.16), we get

$$ \begin{align*} \vert g_{p}(Y)-g_{k,p}(Y) \vert\le 2\mathrm{sh\,}(p\mathcal{D}(\nu,\nu_k))\gamma_p+p\mathcal{D}(s,s_k), \end{align*} $$

which leads to

(2.19) $$ \begin{align} \bigg\vert\!\sum_{i=0}^{q-1}\sum_{|Y|=ip}\nu_k(Y)(g_{p}(Y)-g_{k,p}(Y))\bigg\vert & \le 2q\mathrm{ sh\,}(p\mathcal{D}(\nu,\nu_k)) \gamma_p+pq\mathcal{D}(s,s_k)\nonumber\\ & \le n\bigg(2\mathrm{sh\,}(p\mathcal{D}(\nu,\nu_k))\frac{\gamma_p}{p}+\mathcal{D}(s,s_k)\bigg). \end{align} $$

However, we have

(2.20) $$ \begin{align} \sum_{i=0}^{q-1}\sum_{|Y|=ip}g_{p}(Y)(\nu(Y)-\nu_k(Y)) & \le \sum_{i=0}^{q-1}\sum_{|Y|=ip}\Big(\nu(Y)\max_{|Z|=ip}g_{p}(Z)-\nu_k(Y)\min_{|Z|=ip}g_{p}(Z)\Big)\nonumber\\ & \le q\Big(\max_{|Z|=ip}g_{p}(Z)-\min_{|Z|=ip}g_{p}(Z)\Big)\nonumber\\ & \le Cqc(\gamma_p,\beta_p)\quad\text{(by Proposition}\ {2.2}, ({2.6})) . \end{align} $$

Inequalities (2.19) and (2.20) lead to

(2.21) $$ \begin{align} \frac{1}{n}\vert g_{pq}-g_{k,pq}\vert \le 2\mathrm{ sh\,}(p\mathcal{D}(\nu,\nu_k))\frac{\gamma_p}{p}+\mathcal{D}(s,s_k)+\frac{\alpha_p(\nu,s)}{p}. \end{align} $$

Using now (2.18), we get

$$ \begin{align*} \frac{1}{n}\vert g_n-g_{k,n}\vert\le 2\mathrm{sh\,}(p\mathcal{D}(\nu,\nu_k)) \frac{\gamma_p}{p}+\mathcal{D}(s,s_k)+C\frac{c(\gamma_p,\beta_p)}{p}+2\frac{\gamma_p}{n}. \end{align*} $$

Let $\varepsilon>0$ .

Fix p big enough to ensure that $ C({c(\gamma _{p},\beta _{p})}/{p})< {\varepsilon }/{3}$ and let $n_{0}\in \mathbb {N}$ such that $ 2({\gamma _{p}}/{n_{0}})<{\varepsilon }/{3}$ .

Finally, let $k_{0}$ such that for any $k\ge k_{0}$ , we have

$$ \begin{align*} 2\mathrm{sh\,}(p\mathcal{D}(\nu,\nu_k))\frac{\gamma_{p}}{p}<\frac{\varepsilon}{3}. \end{align*} $$

Then, for any $k\ge k_{0}$ and any $n\ge n_{0}$ , we have $ | {g_{n}}/{n}-{g_{k,n}}/{n}|<\varepsilon $ , from where the result follows.

Proof. Let us first observe that if the result holds for s, then it also holds for every $\tilde {s} \sim s$ . Indeed,

$$ \begin{align*} \frac{1}{n}\bigg\vert\!\sum_{|Y|=n}\nu(Y)(\log (\tilde{s})-\log (s))\bigg\vert \le \frac{\log(c)}{n}. \end{align*} $$

Given any $n\in \mathbb {N}$ , let $X_n:\mathbb {K}\to \mathcal {C}_n$ , $\mathtt {x}\mapsto X_n(\mathtt {x})$ be the cylinder of size n containing $\mathtt {x}$ . We define $Z_{n}(x):=\log s(X_{n}(x))$ .

If we adopt the notation introduced in Proposition 2.2, with $g_n=\langle \nu ,\log (s) \rangle _{n}$ , the theorem asserts that the sequence $ ({1}/{n})(g_n-Z_n(\cdot ))$ converges to $0 \nu $ -a.s.

For any integer $p>0$ , let $Y_{n,p}:=({1}/{p})(Z_{np}-Z_{(n-1)p})$ . Let also $\mathcal {F}_{n}$ be the $\sigma $ -algebra generated by cylinders in $\mathcal {C}_{n}$ . The idea, once again taken from [Reference Batakis3], is to apply Theorem 2.4 on $(Y_{n,p})_{n}$ , with respect to $(\mathcal {F}_{np})_{n}$ , and to prove that the conditional expectations are asymptotically, on n and p, independent of $\mathtt {x}\in \mathbb {K}$ .

Fix $p>0$ , and $n>p$ and let $n=pq+r$ be the Euclidian division of n by p. We define

$$ \begin{align*} \begin{aligned} {\unicode{x2460}} & = \frac{1}{n}Z_{n}-\frac{pq}{n}\frac{1}{q}\sum_{k=1}^q Y_{k+1,p}\\ {\unicode{x2461}} & = \frac{1}{q}\sum_{k=1}^q[ Y_{k+1,p}-\mathop{\mathbb{E}}\limits_{\nu}[Y_{k+1,p} |\mathcal{F}_{kp}]],\\ {\unicode{x2462}} & = \frac{1}{q}\sum_{k=1}^q \mathop{\mathbb{E}}\limits_{\nu}[Y_{k+1,p} |\mathcal{F}_{kp}]-\frac{1}{qp}g_{qp}(X_{p}),\\ {\unicode{x2463}} & = \frac{1}{n}(g_{qp}(X_{p})-g_{n}(X_{p})),\\ {\unicode{x2464}} & = \frac{1}{n}(g_{n}(X_{p})-g_{n}),\\ \end{aligned} \end{align*} $$

so that

$$ \begin{align*} \frac{1}{n}(Z_n-g_n)={\unicode{x2460}}+\frac{pq}{n}{\unicode{x2461}}+\frac{pq}{n}{\unicode{x2462}}+{\unicode{x2463}}+{\unicode{x2464}}. \end{align*} $$

We treat each of the five terms in the sum separately.

${\text {Term}\ \unicode{x2460}}$ . By definition, we have

$$ \begin{align*} \frac{pq}{n}\frac{1}{q}\sum_{k=1}^q Y_{k+1,p}=\frac{1}{n}Z_{(q+1)p}. \end{align*} $$

Therefore,

$$ \begin{align*} {\unicode{x2460}}=\frac{1}{n}(Z_n-Z_{(q+1)p})=\frac{1}{n}\log\bigg(\frac{s(X_n)}{s(X_{(q+1)p})}\bigg). \end{align*} $$

There exists a cylinder Y, with $\vert Y\vert \le p$ , such that $X_{(q+1)p}=X_nY$ . By hypothesis on s, we thus have

$$ \begin{align*} \bigg\vert\!\log\bigg(\frac{s(X_n)}{s(X_{(q+1)p})}\bigg)\bigg\vert=\vert\!\log(s_{X_n}(Y))\vert \le \gamma_p. \end{align*} $$

This tells us that

(2.22) $$ \begin{align} \vert{\unicode{x2460}}\vert\le \frac{\gamma_p}{n}. \end{align} $$

${\text {Terms}\ \unicode{x2463}\ \text {and}\ \unicode{x2464}}$ . Recall that by (2.9), we have

$$ \begin{align*} g_n(X_p)=g_{pq+r}(X_p)=g_{pq}(X_p)+\sum_{|Y|=pq}\nu_{X_p}(Y)g_{r}(X_pY). \end{align*} $$

Therefore,

(2.23) $$ \begin{align} \vert{\unicode{x2463}}\vert=\frac{1}{n}\bigg\vert\!\sum_{|Y|=pq}\nu_{X_p}(Y)g_{r}(X_pY)\bigg\vert\le \frac{\gamma_r}{n}\le \frac{\gamma_p}{n}. \end{align} $$

However, by (2.13) and Proposition 2.2, we have

$$ \begin{align*} |g_n(X_p)-g_n|\le Cpc(\gamma_n,\beta_n), \end{align*} $$

which allows us to conclude that

(2.24) $$ \begin{align} \vert{\unicode{x2464}}\vert\le Cp\frac{c(\gamma_n,\beta_n)}{n}. \end{align} $$

${\text {Term}\ \unicode{x2462}}$ . By definition, the conditional expectation $\mathbb {E}_{\nu }[Y_{k+1,p} |\mathcal {F}_{kp}]$ is constant on any cylinder of size $kp$ . More precisely, for any $\mathtt {x}\in X_{kp}\in \mathcal {C}_{kp}$ , it is equal to the average of $Y_{k+1,p}$ on $X_{kp}$ , so that we have

$$ \begin{align*} \mathop{\mathbb{E}}\limits_{\nu}\bigg[Y_{k+1,p} |\mathcal{F}_{kp}\bigg](\mathtt{x})=\frac{1}{p}\sum_{|Y|=p}\frac{\nu(X_{kp}Y)}{\nu(X_{kp})}\log \frac{s(X_{kp}Y)}{s(X_{kp})}=\frac{1}{p}g_{p}(X_{kp}). \end{align*} $$

Recall that by (2.9), for any integer k, we have

$$ \begin{align*} g_{kp}(X_p)=g_{(k-1)p}(X_p)+\sum_{|Y|=(k-1)p}\nu_{X_p}(Y)g_{p}(X_pY). \end{align*} $$

Therefore,

$$ \begin{align*} g_{kp}(X_p)-g_{(k-1)p}(X_p)-g_p(X_{kp})=\sum_{|Y|=(k-1)p}\nu_{X_p}(Y)(g_{p}(X_pY)-g_p(X_{kp})). \end{align*} $$

Summing from $1$ to q and dividing by $pq$ , we get

$$ \begin{align*} \frac{1}{pq}g_{qp}(X_{p})-\frac{1}{q}\sum_{k=1}^q\frac{g_{p}(X_{kp})}{p}=\frac{1}{pq}\sum_{k=1}^q\sum_{|Y|=(k-1)p}\nu_{X_p}(Y)(g_{p}(X_pY)-g_p(X_{kp})). \end{align*} $$

Since by Proposition 2.2 for any Y of length $(k-1)p$ , we have $\vert g_p(X_pY)-g_p{X_{kp}}\vert \le Cc(\gamma _p,\beta _p)$ , we get

(2.25) $$ \begin{align} \vert{\unicode{x2462}}\vert=\bigg|\frac{1}{q}\sum_{k=1}^q \mathop{\mathbb{E}}\limits_{\nu}[Y_{k+1,p} |\mathcal{F}_{kp}]-\frac{1}{qp}g_{qp}(X_{p})\bigg|\le C\frac{c(\gamma_p,\beta_p)}{p}. \end{align} $$

${\text {Term}\ \unicode{x2461}}$ . Note that for any n, we have by hypothesis that

$$ \begin{align*} |Y_{k,p}|= \bigg|\!\log\frac{s(X_{kp})}{s(X_{(k-1)p})}\bigg|= |\!\log s_{X_{(k-1)p}}(X_{kp})|\le \gamma_{p}. \end{align*} $$

We may thus apply Theorem 2.4 with $(Y_{n,p})_{n}$ to conclude that p being fixed, $|{\unicode{x2461}}| \nu $ -a.s. converges towards $0$ as $q\to {{{+}}}\infty $ , and we can easily conclude that there exists a set of full measure, denoted by $\mathcal {U}$ , such that it occurs for any p.

We may now conclude the proof.

Let $\mathtt {x}\in \mathcal {U}$ and $\varepsilon>0$ . Choose p big enough to have $ C({c(\gamma _p,\beta _p)}/{p})<{\varepsilon }/{5}$ .

Let $n_{0}\ge p$ be such that for any $n\ge n_{0}$ , we have ${\gamma _p}/{n}<{\varepsilon }/{5}$ .

Since p is fixed, q goes to infinity with n and we may choose $n_0$ big enough to have $\vert {\unicode{x2461}}\vert <{\varepsilon }/{5}$ . Combining this with (2.22), (2.25), (2.23), and (2.24), we deduce that

$$ \begin{align*} \begin{aligned} \frac{1}{n}\vert Z_n(\mathtt{x})-g_n\vert\le \vert{\unicode{x2460}}\vert+\vert{\unicode{x2461}}\vert+\vert{\unicode{x2462}}\vert+\vert{\unicode{x2463}}\vert+\vert{\unicode{x2464}}\vert<\varepsilon. \end{aligned} \end{align*} $$

Proof. We use Proposition 2.1 with $\nu =\mu $ and $\nu '=|\cdot|_{\mathsf {d}}^t$ . We thus have for any n,

$$ \begin{align*} \!\!\!\!\!\!H_{\mu,n}-t\mathcal{X}_{\mu,n}=\frac{1}{n}\bigg\langle \nu,\log \frac{1}{\nu} \bigg\rangle_{n}-\frac{1}{n}\langle \nu,\log \nu' \rangle_{n}=\frac{1}{n}\bigg\langle \nu,\log \frac{\nu'}{\nu} \bigg\rangle_{n}\le \frac{1}{n}\log \mathsf{m}_{n}(\nu'). \end{align*} $$

Proof. The hypothesis allows us to apply Theorem 2.5, with $\nu (X)=\mu (X)$ and $s(X)={1}/{\mu (X)}$ , to conclude that

$$ \begin{align*} \lim_{n\to{{{+}}}\infty} (h_{\mu,n}(\mathtt{a})- H_{\mu,n})=0. \end{align*} $$

Let now $s:X\mapsto {1}/{|{X}|_{\mathsf {d}}}$ .

From the assumption $ \delta ^{\gamma _{|Y|}}\le {|{XY}|_{\mathsf {d}}}/{|{X}|_{\mathsf {d}}}\le 1$ , we deduce that $1\le s_{X}(Y)\le \delta ^{-\gamma _{|Y|}}$ , and we easily conclude that for any $XY\in \mathcal {C}$ , with $|Y|=n$ , we have

$$ \begin{align*} |\!\log s_{X}(Y)|\le -\gamma_n\log d. \end{align*} $$

We may once again apply Theorem 2.5 to conclude.

Proof. From (3.2), we easily get for any $t< t+h$ and any $n\in \mathbb {N}$ ,

$$ \begin{align*}-h\log (1-\delta)+P_{\mu,n}(t+h)\le P_{\mu,n}(t)\le P_{\mu,n}(t+h)-h\log \delta,\end{align*} $$

from which we deduce, with P standing either for or , that we have

$$ \begin{align*}0<-h\log (1-\delta)\le P(t)-P(t+h)\le -h\log \delta,\end{align*} $$

which tells us that $\underline {P_{\mu }}$ and $\overline {P_{\mu }}$ are strictly decreasing and bi-Lipschitz.

Points (1) and (2) are obvious from definitions and hypotheses.

We turn to results about dimensions.

Hypotheses on $\mu $ and $|\cdot|_{\mathsf {d}}$ allow us to apply Theorem 2.5 with, for any $X\in \mathcal {C}$ , $\nu (X)=\mu (X)$ and $s(X)={|{X}|_{\mathsf {d}}^t}/{\mu (X)}$ , as well as Theorem 3.4. In particular, $\mu $ -a.s. we have

(E) $$ \begin{align} \lim_{k\to{{{+}}}\infty} h_{\mu,n_{k}}({\cdot})=h & \Longleftrightarrow \lim_{k\to{{{+}}}\infty} H_{\mu,n_{k}}=h, \end{align} $$

(L) $$ \begin{align} \lim_{k\to{{{+}}}\infty} \chi_{n_{k}}({\cdot})=\chi & \Longleftrightarrow \lim_{k\to{{{+}}}\infty} \mathcal{X}_{\mu,n_{k}}=\chi. \end{align} $$

Since $\underline {P_{\mu }}$ and $\overline {P_{\mu }}$ are strictly decreasing, continuous with positive and negative values, there exists a unique $(\underline {t},\overline {t})\in \mathbb {R}^2$ , with $0\le \underline {t}\le \overline {t}\in \mathbb {R}$ such that $\underline {P_{\mu }}(\underline {t})=0=\overline {P_{\mu }}(\overline {t})$ .

We focus on the result with liminf.

Take a point $\mathtt {a}\in \mathbb {K}$ such that items (E) and (L) are true.

Let $(n_{k})$ be a sequence of integers such that

$$ \begin{align*} \lim_{k\to{{{+}}}\infty} \frac{h_{\mu,n_{k}}(\mathtt{a})}{\chi_{n_{k}}(\mathtt{a})}=\liminf \frac{h_{\mu,n}(\mathtt{a})}{\chi_{n}(\mathtt{a})}=\underline{d}_{\mu}(\mathtt{a}), \end{align*} $$

see (1.2) for the definition of $\underline {d}_{\mu }(\mathtt {a})$ .

Using items (E) and (L), and extracting again if necessary, one may suppose that

(3.3) $$ \begin{align} \lim_{k\to{{{+}}}\infty} H_{\mu,n_k}=\lim_{k\to{{{+}}}\infty} h_{\mu,n_{k}}(\mathtt{a})=h\,\text{ and } \lim_{k\to{{{+}}}\infty} \mathcal{X}_{\mu,n_k}=\lim_{k\to{{{+}}}\infty} \chi_{n_{k}}(\mathtt{a})=\chi. \end{align} $$

By hypothesis on $|\cdot|_{\mathsf {d}}$ , we have $0<\chi <{{{+}}}\infty $ , thus $\underline {d}_{\mu }(\mathtt {a})=\frac {h}{\chi }$ , and we have

$$ \begin{align*} \underline{P_{\mu}}(\underline{d}_{\mu}(\mathtt{a}))\le\lim_{k\to{{{+}}}\infty}( H_{\mu,n_k}-\underline{d}_{\mu}(\mathtt{a})\mathcal{X}_{\mu,n_k})= h-\underline{d}_{\mu}(\mathtt{a})\chi=0, \end{align*} $$

from which we deduce that $\underline {d}_{\mu }(\mathtt {a})\ge \underline {t}$ .

Fix $t>0$ and assume that $n_{k}$ is such that

$$ \begin{align*} \lim_{k\to{{{+}}}\infty} (H_{\mu,n_{k}}-t\mathcal{X}_{\mu,n_{k}})=\underline{P_{\mu}}(t). \end{align*} $$

As before, using items (E) and (L) and extracting a further subsequence if necessary, we may assume that (3.3) is fulfilled. Note that h and $\chi $ denote limits that are different from the previous case. In particular, we can only assert that ${h}/{\chi }\ge \underline {d}_{\mu }(\mathtt {a})$ , since

$$ \begin{align*} \frac{h}{\chi}=\lim_{k\to{{{+}}}\infty}\frac{h_{\mu,n_k}(\mathtt{a})}{\chi_{n_k}(\mathtt{a})}\ge \liminf \frac{h_{\mu,n}(\mathtt{a})}{\chi_{n}(\mathtt{a})}=\underline{d}_{\mu}(\mathtt{a}). \end{align*} $$

We have

$$ \begin{align*} \underline{P_{\mu}}(t)=h-t\chi=\chi\bigg(\frac{h}{\chi}-t\bigg)\ge \chi(\underline{d}_{\mu}(\mathtt{a})-t). \end{align*} $$

We deduce that $\underline {P_{\mu }}(\underline {d}_{\mu }(\mathtt {a}))\ge 0$ so that $\underline {d}_{\mu }(\mathtt {a})\le \underline {t}$ .

This tells us that $\mu $ -a.s. $\underline {d}_{\mu }({\cdot })$ is constant equal to $\underline {t}$ . By classical results on Hausdorff dimension of measures, see [Reference Falconer10, Remark above Proposition 10.5], we can conclude that $\mu $ -a.s.

$$ \begin{align*} HD(\mu)=\underline{d}_{\mu}(\mathtt{a})=\underline{t}. \end{align*} $$

Consider a sequence of integers $(n_k)$ such that

$$ \begin{align*} \lim_{k\to{{{+}}}\infty} ( H_{\mu,n_{k}}-HD(\mu)\mathcal{X}_{\mu,n_{k}})=0. \end{align*} $$

Since $(\mathcal {X}_{\mu ,n_{k}})$ is bounded away from zero, we have

$$ \begin{align*} \liminf \bigg(\frac{H_{\mu,n}}{\mathcal{X}_{\mu,n}}\bigg)\le \lim_{k\to{{{+}}}\infty} \frac{H_{\mu,n_k}}{\mathcal{X}_{\mu,n_k}}=HD(\mu). \end{align*} $$

We now conclude by considering a sequence of integers $(n_k)$ such that

$$ \begin{align*} \liminf \bigg(\frac{H_{\mu,n}}{\mathcal{X}_{\mu,n}}\bigg)= \lim_{k\to{{{+}}}\infty} \frac{H_{\mu,n_k}}{\mathcal{X}_{\mu,n_k}}. \end{align*} $$

Using the fact that $(\mathcal {X}_{\mu ,n_{k}})$ is bounded, we deduce that

$$ \begin{align*} \underline{P_{\mu}}\bigg(\!\liminf \bigg(\frac{H_{\mu,n}}{\mathcal{X}_{\mu,n}}\bigg)\bigg)\le \lim_{k\to{{{+}}}\infty}\bigg( H_{\mu,n_k}-\liminf \bigg(\frac{H_{\mu,n}}{\mathcal{X}_{\mu,n}}\bigg)\mathcal{X}_{\mu,n_k}\bigg)=0, \end{align*} $$

which implies that $ HD(\mu )=\underline {t}\le \liminf ({H_{\mu ,n}}/{\mathcal {X}_{\mu ,n}})$ .

We thus have proved that for any $\mathtt {a}$ in a set of full $\mu $ -measure, we have

$$ \begin{align*} \begin{aligned} \liminf \frac{H_{\mu,n}}{\chi_{\mu,n}}=\liminf \frac{h_{\mu,n}(\mathtt{a})}{\chi_{n}(\mathtt{a})}=HD(\mu). \end{aligned} \end{align*} $$

Proof. For any $X\in \mathcal {C}$ , let

$$ \begin{align*} p(X)=\frac{|{X}|_{\mathsf{d}}^{t}}{\mu(X)}=\frac{s(X)^t}{\mu(X)}\quad\text{and for any } k,\quad p_{k}(X)=\frac{|{X}|_{\mathsf{d}_{\mathsf{k}}}^{t_{k}}}{\mu_{k}(X)}=\frac{s_{k}(X)^{t_{k}}}{\mu_{k}(X)}.\end{align*} $$

Let $C=(1+\sup _{k}\{|t|, |t_{k}|\})$ . For any $XY\in \mathcal {C}$ , we have

$$ \begin{align*} |\!\log p_{X}(Y)|\le|t|\!\log s_{X}(Y)+\log\mu_{X}(Y)\le C\gamma_{|Y|}. \end{align*} $$

In the same way, we check that $|\!\log p_{k_{X}}(Y)|\le C\gamma _{|Y|}.$ Thus, one may take .

We also have for any $XYZ\in \mathcal {C}$ and any $X'\in \mathcal {C}_{|X|}$ ,

$$ \begin{align*} \begin{aligned} \bigg|\frac{p_{XY}(Z)}{p_{X'Y}(Z)}-1| & = \bigg|\frac{\mu_{X'Y}(Z)}{\mu_{XY}(Z)}\bigg(\frac{s_{XY}(Z)}{s_{X'Y}(Z)}\bigg)^t-1\bigg|\\ & = \bigg|\bigg(\bigg(\frac{s_{XY}(Z)}{s_{X'Y}(Z)}\bigg)^t-1\bigg)\frac{\mu_{X'Y}(Z)}{\mu_{XY}(Z)}+\frac{\mu_{X'Y}(Z)}{\mu_{XY}(Z)}-1\bigg|\\ & \le (1+\beta_{0})\bigg|\bigg(\bigg(\frac{s_{XY}(Z)}{s_{X'Y}(Z)}\bigg)^t-1\bigg)\bigg|+\bigg|\frac{\mu_{X'Y}(Z)}{\mu_{XY}(Z)}-1\bigg|\\ & \le 2(1+\beta_{0})^{1+|t|}\beta_{|Y|}\\ & \le C\beta_{|Y|}. \end{aligned} \end{align*} $$

We may thus take . It is easy to prove that and are compatible. We also have that $\mathcal {D}(p,p_{k})\to 0$ , so that we can apply Theorem 2.3. Using notation therein, we have

$$ \begin{align*} \frac{1}{n}g_{n}=H_{\mu,n}-t\chi_{\mu,n}=P_{\mu,n}(t)\quad\text{and}\quad \frac{1}{n}g_{k,n}=H_{\mu_{k},n}-t_{k}\chi_{\mu_{k},n}=P_{\mu_{k},n}(t_{k}), \end{align*} $$

and Theorem 2.3 asserts that

$$ \begin{align*} \lim_{k\to{{{+}}}\infty}\limsup_{n}(P_{\mu,n}(t)-P_{\mu_{k},n}(t_{k}))=0. \end{align*} $$

From this, we conclude that

$$ \begin{align*}\lim_{k\to{{{+}}}\infty} \underline{P}_{\mu_{k}}(t_{k})=\underline{P}_{\mu}(t)\quad\text{and}\quad \lim_{k\to{{{+}}}\infty} \overline{P}_{\mu_{k}}(t_{k})=\overline{P}_{\mu}(t). \end{align*} $$

With $t_{k}={\vphantom {{HD}_{j}}}_{{{{{k}}}}}{HD}_{\, }(\mu _{k})$ and $\delta $ any limit value of $(t_{k})$ , we conclude, using Proposition 3.5, that

$$ \begin{align*}0=\lim_{k\to{{{+}}}\infty} \underline{P}_{\mu_{k}}({\vphantom{{HD}_{j}}}_{{{{{k}}}}}{HD}_{\, }(\mu_{k}))=\underline{P}_{\mu}(\delta), \end{align*} $$

which, by Proposition 3.5, tells us that $\delta =HD(\mu )$ , so that $({\vphantom {{HD}_{j}}}_{ {{{{k}}}}}{HD}_{\, }(\mu _{k}))$ converges towards $HD(\mu )$ .

Proof. First, note that relations (4.1) and (4.2) are relevant only if $\operatorname {\mathrm {diam}} (\mathbb {k})$ is small. Otherwise, it is nothing else than the standard Kœbe distortion theorem. In addition, we assume from now on that $\operatorname {\mathrm {diam}}(\mathbb {k})\le {\delta }/{2}$ .

Let x and y be two points in $\mathbb {k}$ . We apply [Reference Pommerenke17, Corollary 1.5], with $ z=({x-y})/({1-\bar {y}x})$ , we have

$$ \begin{align*} \frac{|f'(x)|}{|f'(y)|}\le \bigg(\frac{1+|z|}{1-|z|}\bigg)^3. \end{align*} $$

Since $\mathbb {k}\subset \{|z|\le 1-\delta \}$ , we have $|z|\le {1}/{\delta }|x-y|\le {1}/{\delta }\operatorname {\mathrm {diam}} \mathbb {k}$ . We also have $|z|<1$ and we can deduce that we have

(4.3) $$ \begin{align} \frac{|f'(x)|}{|f'(y)|}\le 1+C\operatorname{\mathrm{diam}} (\mathbb{k}). \end{align} $$

By exchanging x and y, we also get the opposite inequality.

Let $Z=1-x\bar {y}$ . Then, [Reference Fan11, Theorem 1] asserts that we have

(4.4) $$ \begin{align} \frac{|Z|}{1-|x|^2}\frac{|Z|-|x-y|}{|Z|+|x-y|} \le \frac{|f(x)-f(y)|}{|f'(x)||x-y|}\le \frac{|Z|}{1-|x|^2}\frac{|Z|+|x-y|}{|Z|-|x-y|}. \end{align} $$

Since $Z=1-|x|^2 -x(\overline {y-x})$ , we have

$$ \begin{align*}1-|x|^2-|x-y|\le|Z|\le1-|x|^2+|x-y|,\end{align*} $$

which leads to

$$ \begin{align*} 1-\frac{|x-y|}{1-|x|^2}\le\frac{|Z|}{1-|x|^2}\le 1+\frac{|x-y|}{1-|x|^2} .\end{align*} $$

However, $|x|\le 1-\delta $ , so $\delta \le 1-|x|^2\le 1$ and $|x-y|\le \operatorname {\mathrm {diam}} \mathbb {k}$ , which gives

(4.5) $$ \begin{align} 1-\frac{\operatorname{\mathrm{diam}}\mathbb{k}}{\delta}\le\frac{|Z|}{1-|x|^2}\le 1+\frac{\operatorname{\mathrm{diam}}\mathbb{k}}{\delta}. \end{align} $$

We have assumed that $\operatorname {\mathrm {diam}} \mathbb {k}\le {\delta }/{2}$ ; thus,

$$ \begin{align*}\frac{|Z|+|x-y|}{|Z|-|x-y|}=1+2\frac{|x-y|}{|Z|-|x-y|}\le 1+2\frac{|x-y|}{1-|x|^2-2|x-y|}\le 1+ \frac{2}{\delta(\delta-1)}\operatorname{\mathrm{diam}} \mathbb{k},\end{align*} $$

and we prove in the same way that

$$ \begin{align*}1 -\frac{2}{\delta}\operatorname{\mathrm{diam}} \mathbb{k}\le \frac{|Z|-|x-y|}{|Z|+|x-y|}. \end{align*} $$

Using those two last inequalities and (4.5) in (4.4), we end up with

$$ \begin{align*} \bigg(1-\frac{2}{\delta}\operatorname{\mathrm{diam}} \mathbb{k}\bigg)^2\le \frac{|f(x)-f(y)|}{|f'(x)||x-y|}\le\bigg(1+ \frac{2}{\delta(\delta-1)}\operatorname{\mathrm{diam}} \mathbb{k}\bigg)^2. \end{align*} $$

Using (4.3), with an adjustment of the constant, we may change $|f'(x)|$ with $|f'(z)|$ for any $z\in \mathbb {k}$ , so that for any $x,y$ and z in $\mathbb {k}$ , we have

$$ \begin{align*} 1-C\operatorname{\mathrm{diam}} \mathbb{k}\le \frac{|f(x)-f(y)|}{|f'(z)||x-y|}\le 1+C\operatorname{\mathrm{diam}} \mathbb{k}. \end{align*} $$

We easily conclude that for any $z\in \mathbb {k}$ ,

$$ \begin{align*} (1-C\operatorname{\mathrm{diam}} \mathbb{k})|f'(z)|\le \frac{\operatorname{\mathrm{diam}} f(\mathbb{k})}{\operatorname{\mathrm{diam}} \mathbb{k}}\le (1+C\operatorname{\mathrm{diam}} \mathbb{k})|f'(z)|. \end{align*} $$

Proof. If we apply classical Kœbe distortion estimates with , $\mathbb {k}=\mathbb {X}_{k}$ , and $x=0$ , we get

Thus, s and $\tilde {s}$ are equivalent.

We have

Since and both lie in Y, by (4.1), we have

from where we deduce, using also the fact that by hypothesis, we have $\operatorname {\mathrm {diam}} (Y)\le Cq^{|Y|}$ , that

This can be rewritten as

The same is also true with $X'$ instead of X and this leads to

$$ \begin{align*} \bigg|\frac{\tilde{s}_{XY}(Z)}{\tilde{s}_{X'Y}(Z)}-1\bigg|\le Cq^{|Y|}. \end{align*} $$

One can apply relation (4.2) with , $\mathbb {k}=YZ$ , and also $\mathbb {k}=Y$ and $x\in YZ$ , to get

$$ \begin{align*} \frac{1-C\operatorname{\mathrm{diam}} (YZ)}{1+C\operatorname{\mathrm{diam}} (Y)}\le\frac{\operatorname{\mathrm{diam}} (XYZ)}{\operatorname{\mathrm{diam}} (XY) }\frac{\operatorname{\mathrm{diam}} (Y)}{\operatorname{\mathrm{diam}} (YZ) }\le \frac{1+C\operatorname{\mathrm{diam}} (YZ)}{1-C\operatorname{\mathrm{diam}} (Y)}. \end{align*} $$

We have $\operatorname {\mathrm {diam}} (YZ)<\operatorname {\mathrm {diam}} (Y)$ , and by hypothesis, we know that $\operatorname {\mathrm {diam}} (Y)\le C q^{|Y|}$ . Adjusting the constant C, we easily conclude that we have

$$ \begin{align*}1-C q^{|Y|}\le \frac{\operatorname{\mathrm{diam}} (XYZ)}{\operatorname{\mathrm{diam}} (XY) }\frac{\operatorname{\mathrm{diam}} (Y)}{\operatorname{\mathrm{diam}} (YZ) }\le 1+C q^{|Y|},\end{align*} $$

which is exactly

$$ \begin{align*}1-C q^{|Y|}\le s_{XY}(Z)\frac{\operatorname{\mathrm{diam}} (Y)}{\operatorname{\mathrm{diam}} (YZ) }\le 1+C q^{|Y|}.\end{align*} $$

As a matter of fact, this implies that for any other geometric cylinder $X'\subset \mathbb {X}$ , with $|X'|=|X|$ , we have

$$ \begin{align*} \begin{aligned} \bigg|\frac{s_{XY}(Z)}{s_{X'Y}(Z)}-1\bigg|\le C q^{|Y|}. \end{aligned} \end{align*} $$

Proof. Let . Note that we have .

Let $Xa$ be a cylinder with $|X|=n$ and $|a|=1$ . For any $z\in \mathbb {D}_{\gamma }$ , we have

from where we deduce that

(4.6)

Using Cauchy representation theorem, for any cylinder X with $|X|=k$ , we have

Since, by assumption, , we deduce that

(4.7)

Inequality is true for any cylinder X such that $|X|\le k$ , and stays true if we exchange the role played by and .

Let now $X=YX_{k}$ be any cylinder of length n, with $X_{k}=\varnothing $ if $n\le k$ , and $|X_{k}|=k$ otherwise.

By the chain rule, we have

Therefore, for any cylinder a of length 1, we have

From (4.1) in Proposition 4.1, there exists $C>0$ such that

In addition, from (4.7),

Using the fact that $\text {diam} X_{k}$ and $\text {diam} \tilde {X}_{k}$ are both smaller than $(1-\boldsymbol {\eta })^k$ , and adjusting the constant C, we end up with

(4.8)

Let k be the integer part of , then and , so that we have

Alternatively, we have so that

which leads, with $\alpha ={\log (1-\boldsymbol {\eta })}/{\log (\boldsymbol {\eta }(1-\boldsymbol {\eta }))}$ , to

and gives

This being true for any cylinder X, we conclude that

Proof. The proof relies on standard techniques involving capacity theory and harmonic measure in the context of Cantor-type sets: we hence leave some details to the reader (see [Reference Prats and Tolsa18, §6] and also [Reference Batakis2]). In particular, we use the fact that every point in $\mathbb {X}$ is Wiener regular so that any harmonic functions is uniquely determined by its boundary values.

Let $\Omega =\mathbb {D}_\gamma \setminus \mathbb {X}$ . Recall from (1.1) that for any n, we have $2\le d_n\le D$ .

Let $\mu $ be the uniform probability measure on $\mathbb {X}$ so that for any cylinder $X\in \mathcal {C}_n$ , we have $\mu (X)=1/(d_1\cdot \cdots \cdot d_n)$ .

For each $x\in \Omega $ , let $G^x:\Omega \to \mathbb {R}$ denote the Green function of $\mathbb {D}_{\gamma }$ with pole at x. There exists $C_1$ and $C_2$ , only depending on $\boldsymbol {\eta }$ such that for any $y\in \mathbb {D}$ , we have

$$ \begin{align*} \ln\bigg(\frac{C_1}{|x-y|}\bigg)\le G^x(y)\le \ln\bigg(\frac{C_2}{|x-y|}\bigg). \end{align*} $$

Let $h:\mathbb {D}_{\gamma }\to \mathbb {R}$ , $x\mapsto \int _{\mathbb {X}} G^x(y)\mu (dy)$ .

For $x\in \mathbb {X}$ , if $(X_n)$ , with $|X_n|=n$ , is the sequence of cylinders converging to x, we have

(4.11) $$ \begin{align} h(x)=\sum_{n=1}^{{{{+}}}\infty}\int_{X_n\setminus X_{n+1}}G^x(y)\mu(dy). \end{align} $$

For any n, we have $\mu (X_n\setminus X_{n+1})=({d_{n+1}-1})/({d_1\cdots d_{n+1}})$ , from which we deduce

$$ \begin{align*} \frac{1}{2}D^{-n}\le\mu(X_n\setminus X_{n+1})\le D 2^{-n}. \end{align*} $$

However, for $y\in X_n\setminus X_{n+1}$ , we have $(1-\boldsymbol {\eta })^{-n}\le |x-y|\le \boldsymbol {\eta }^{-n}$ . Consequently, there exist constants $c_1>0$ and $c_2>0$ such that

$$ \begin{align*} c_1n\le \ln \bigg(\frac{C_1}{(1-\boldsymbol{\eta})^n}\bigg)\le G^x(y)\le \ln\bigg(\frac{C_2}{\boldsymbol{\eta}^n}\bigg)\le c_2n. \end{align*} $$

Injecting those inequalities in (4.11) leads to

$$ \begin{align*} 0<c:=\frac{1}{2}c_1\sum_{n=1}^{{{{+}}}\infty}nD^{-n}\le h(x)\le c_2D\sum_{n=1}^{{{{+}}}\infty}n2^{-n}:=C<{{{+}}}\infty. \end{align*} $$

Since the function $x\mapsto \omega (x,\mathbb {X},\mathbb {D}_\gamma \setminus X)$ is constant and equal to $1$ on $\mathbb {X}$ , and since by definition $h(x)=\omega (x,\mathbb {X},\mathbb {D}_\gamma \setminus X)= 0$ for any , we have proved, for any , that we have

$$ \begin{align*} c \omega(x,\mathbb{X},\mathbb{D}_\gamma\setminus \mathbb{X})\le h(x)\le C\omega(x,\mathbb{X},\mathbb{D}_\gamma\setminus \mathbb{X}). \end{align*} $$

By the maximum principle, those inequalities are true for any $x\in \Omega $ .

Clearly, $h(x)\ge c>0$ for and hence the result.

Proof. Let $\Gamma := \{|x|=\gamma \}$ , $u: x\mapsto \omega (x,X,\hat {\mathbb {C}}\setminus \mathbb {X})$ , $\tilde {u}:x\mapsto \omega (x,X,\mathbb {D}_{\gamma X}\setminus X)$ , and $h=u-\tilde {u}$ . The function h is harmonic on $\mathbb {D}_{\gamma X}\setminus X$ , with $h=0$ on X and $h=u$ on so that

Let be such that for any , $u(x)\le u(x_{\gamma })$ , and be such that for any , we have $u(x)\le u(x_{m})$ .

The function $x\mapsto u(x)-u(x_{\gamma })$ is harmonic outside $\mathbb {D}_{X}\cup \mathbb {X}$ , non-positive on , and it is thus negative on $\mathbb {C}\setminus \mathbb {D}_{X}$ and, in particular, we have $u(x_{m})<u(x_{\gamma })$ .

We deduce the following:

By conformal invariance, with $n=|X|$ , we have

with, by definition, . Lemma 4.6 leads to $\tilde {u}(x_{\gamma })\ge u(x_{\boldsymbol {\eta }})c,$ since $\mathbb {X}_{n}$ is the limit Cantor set associated to a NACIFS with the same constant than $\mathbb {X}$ .

We conclude that we have $ 1\le {u(x_{\gamma })}/{\tilde {u}(x_{\gamma })}\le C.$

By Harnack inequality on , we conclude that $ 1\le {u(x)}/{\tilde {u}(x)}\le C$ is true for any , with a different C, still uniquely depending on $\gamma $ .

Proof. Let $u:x\mapsto ({\omega (x,XYZ,\hat {\mathbb {C}}\setminus \mathbb {X})}/{\omega (XYZ)})$ and $v:x\mapsto ({\omega (x,XY,\hat {\mathbb {C}}\setminus \mathbb {X})}/ {\omega (XY)})$ . Note that $u(\infty )=v(\infty )=1$ .

Assume that $Y=y_{1}\cdots y_{n}$ and for any $\ell \in [1,n]$ , let $X_{0}=X$ , $X_{\ell }=Xy_{1}\cdots y_{\ell }$ , $A_{\ell }=\mathbb {D}_{X_{n-\ell }}$ , and $B_{\ell }=\mathbb {D}_{\gamma X_{n-\ell }}$ .

We have $A_{0}\subset B_{0}\subset \cdots \subset A_{n-1}\subset B_{n-1}\subset \hat {\mathbb {C}}\setminus \mathbb {X}$ , and each annulus $B_{i}\setminus A_{i}$ is conformally equal to $\mathbb {D}_{\gamma }\setminus \mathbb {D}$ .

The functions u and v are both harmonic in $\hat {\mathbb {C}}\setminus \mathbb {X}$ , and they vanish on $\mathbb {X}\setminus XY$ .

We are thus in a position to apply Proposition 4.5 to conclude that for any $x\notin \mathbb {X}\cup \mathbb {D}_{\gamma Xy_{1}}$ , and in particular for any ,

(4.13) $$ \begin{align} \bigg|\!\log \bigg(\frac{\omega(x,XYZ,\hat{\mathbb{C}}\setminus\mathbb{X})}{\omega(x,XY,\hat{\mathbb{C}}\setminus\mathbb{X})}\frac{\omega(XY)}{\omega(XYZ)}\bigg)\bigg| =\bigg|\!\log \bigg(\frac{u(x)}{v(x)}\frac{v(\infty)}{u(\infty)}\bigg)\bigg|\le Cq^n. \end{align} $$

Let now $\tilde {u}:x\mapsto ({\omega (x,XYZ,\mathbb {D}_{\gamma X}\setminus X)}/{\omega (XYZ)})$ and $\tilde {v}:x\mapsto ({\omega (x,XY,\mathbb {D}_{\gamma X}\setminus X)}/{\omega (XY)})$ . We have for any $x\in \mathbb {D}_{\gamma X}\setminus X$ ,

and also

We thus have

which may be written as

Since (4.13) is true for any , and by Lemma 4.7, we conclude that there exist $C>0$ and $q<1$ such that for any ,

From this, we conclude that for any , we have

(4.14) $$ \begin{align} \bigg|\!\log\bigg( \frac{\omega(x,XYZ,\mathbb{D}_{\gamma X}\setminus X)}{\omega(x,XY,\mathbb{D}_{\gamma X}\setminus X\bigg)}\frac{\omega(XY)}{\omega(XYZ)}\bigg)\bigg|=\bigg|\!\log \bigg(\frac{\tilde{u}(x)}{\tilde{v}(x)} \bigg)\bigg|\le Cq^n. \end{align} $$

By conformal invariance of harmonic measure, we conclude that for any ,

$$ \begin{align*} \bigg|\!\log \bigg(\frac{\omega(x,YZ,\mathbb{D}_{\gamma}\setminus \mathbb{X}_{k})}{\omega(x,Y,\mathbb{D}_{\gamma}\setminus \mathbb{X}_{k})}\frac{\omega(XY)}{\omega(XYZ)}\bigg)\bigg| \le Cq^n. \end{align*} $$

Applying (4.13) with $\mathbb {X}_{k}$ and $\omega _{k}$ , which is possible since the NACIFS giving birth to $\mathbb {X}_{k}$ is comparable to that giving birth to $\mathbb {X}$ , we get for any ,

(4.15) $$ \begin{align} \bigg|\!\log\bigg(\frac{\omega(x,YZ,\mathbb{D}_{\gamma}\setminus \mathbb{X}_{k})}{\omega(x,Y,\mathbb{D}_{\gamma}\setminus \mathbb{X}_{k})}\frac{\omega_{k}(Y)}{\omega_{k}(YZ)}\bigg)\bigg|\le Cq^n. \end{align} $$

Combining (4.13), (4.14), and (4.15), we end up with

(4.16) $$ \begin{align} \begin{aligned} \bigg|\!\log\bigg( \frac{\omega(XYZ)}{\omega(XY)}\frac{\omega_{k}(Y)}{\omega_{k}(YZ)}\bigg)\bigg|\le Cq^n. \end{aligned} \end{align} $$

Proof. Relation (4.16) asserts that for any $XYZ\in \mathcal {C}$ with $|X|=k$ and $|Y|=n$ ,

$$ \begin{align*} \bigg|\!\log\bigg( \omega_{XY}(Z)\frac{\omega_{k}(Y)}{\omega_{k}(YZ)}\bigg)\bigg|\le Cq^n, \end{align*} $$

from which we get, for any $X'\in \mathcal {C}$ with $|X'|=|X|$ ,

$$ \begin{align*} \begin{aligned} \bigg|\!\log\bigg(\frac{\omega_{XY}(Z)}{\omega_{X'Y}(Z)}\bigg)\bigg| & = \bigg|\!\log\bigg( \omega_{XY}(Z)\frac{\omega_{k}(Y)}{\omega_{k}(YZ)}\bigg)-\log\bigg( \omega_{X'Y}(Z)\frac{\omega_{k}(Y)}{\omega_{k}(YZ)}\bigg)\bigg| \\ & \le \bigg|\!\log\bigg( \omega_{XY}(Z)\frac{\omega_{k}(Y)}{\omega_{k}(YZ)}\bigg)\bigg|+\bigg|\!\log(\omega_{X'Y}(Z)\frac{\omega_{k}(Y)}{\omega_{k}(YZ)}\bigg)\bigg|\\ & \le Cq^n. \end{aligned} \end{align*} $$

From (4.16), we deduce that there exists $C>1$ such that

$$ \begin{align*} \frac{1}{C}\omega_k(Z)\le\frac{\omega(XZ)}{\omega(X)}\le 1. \end{align*} $$

Assume that $Z=z_0\cdots z_{n-1}$ and for any $i\in {[\![0,n-1]\!]}$ , let $Z_i=z_0\cdots z_{i}$ , then

since for any $i\in {[\![1,n-1]\!]}$ , (4.16) tells us that ${1}/{C} \omega _{k+i}(z_{i})\le ({\omega _k(Z_i)}/ {\omega _k(Z_{i-1})})\le 1$ , and we conclude that

From Lemma 4.6, we can deduce that given a NACIFS , there exists a constant $\alpha>0$ , depending only the geometric constants that constrain the geometry of the NACIFS such that for any $\ell \ge 0$ and any $a\in \mathcal {C}_{\ell }$ with $|a|=1$ , we have $\omega _{\ell } (a)\ge \alpha $ , so we conclude that

$$ \begin{align*} \bigg(\frac{\alpha}{C}\bigg)^n\le \frac{\omega(XZ)}{\omega(X)}\le 1, \end{align*} $$

and the result follows.

Proof. Given any $x\in \mathbb {D}\setminus \mathbb {X}$ , let $n(x)=\max \{k\in \mathbb {N}\mid \text { there exists } X\in \mathcal {C}_n,\,x\in \mathbb {D}_X\}$ . Note that $n(x)$ is such that there exists $X\in \mathcal {C}_n$ with $x\in \mathbb {D}_X$ and $x\notin \mathbb {D}_{Xa}$ for any cylinder $Xa \in \mathcal {C}_{n+1}$ .

Using open set and bounded contraction hypothesis, we deduce from the Kœbe distortion theorem that there exist three constants $K'>0$ and $0<\beta '<\beta $ such that

(4.19) $$ \begin{align} \frac{1}{K'}\operatorname{\mbox{{dist}}}(x,\mathbb{X})^\beta\le e^{-n} \le K'\operatorname{\mbox{{dist}}}(x,\mathbb{X})^{\beta'}. \end{align} $$

It is important to note that $K'$ , $\beta '$ , and $\beta $ only depend on the constant $\boldsymbol {\eta }$ used to control the geometry of the NACIFS.

Let $x\in \mathbb {D}_X$ , with $X=a_1\cdots a_n$ and $|X|=n(x):=n$ . Let $X_0=\mathbb {X}$ and, for any $1\le k\le n$ , $X_k=a_1\cdots a_k$ and set $u_k:y\mapsto \omega (y, X_k,\mathbb {D}_{\gamma X_k}\setminus X_k)$ . The function $u_k$ is harmonic on $\mathbb {D}_{\gamma X_k}\setminus X_k$ and equal $1$ on $X_k$ . Our goal is to prove that

$$ \begin{align*} u_0(x)=\omega(x,\mathbb{X},\mathbb{D}_\gamma\setminus \mathbb{X})\ge 1-K\operatorname{\mbox{{dist}}}(x,\mathbb{X})^\alpha. \end{align*} $$

First, we note using Lemma 4.6, conformal invariance, and maximum principle that there exists $c>0$ such that for any $x\in \mathbb {D}$ ,

(4.20) $$ \begin{align} u_k(x)\ge c. \end{align} $$

For any $1\le k\le n$ , we have

Because of our geometric assumptions on the NACIFS and Harnack inequalities, there exists a constant $\unicode{x3bb}>0$ such that for any $Y\in \mathcal {C}$ , any continuous function $h>0$ , defined on $\mathbb {D}_Y$ and harmonic on $\mathbb {D}_Y\setminus Y$ , and any y and $y'$ in $\mathbb {D}_Y\setminus Y$ , $ \unicode{x3bb} h(y)\le h(y')$ .

From this, we deduce that

Let $v_k=1-u_k$ . We drop x to lighten notation.

Using (4.20), we get $\unicode{x3bb} c v_k\le \unicode{x3bb} u_kv_k\le v_k-v_{k-1}$ and

$$ \begin{align*} c\unicode{x3bb}\le \frac{v_k-v_{k-1}}{v_k}\le \int_{v_{k-1}}^{v_k}\frac{1}{t}\,dt=\log (v_k)-\log (v_{k-1}). \end{align*} $$

Summing for $1\le k\le n$ ,

$$ \begin{align*} c\unicode{x3bb} n\le \log v_n-\log v_0, \end{align*} $$

so that

$$ \begin{align*} 1-u_0=v_0\le e^{-c\unicode{x3bb} n}. \end{align*} $$

We use (4.19) to conclude that $ u_0(x)=\omega (x,\mathbb {X},\mathbb {D}_\gamma \setminus \mathbb {X})\ge 1-({1}/{K'}\operatorname {\mbox {{dist}}}(x,\mathbb {X})^\beta )^{c\unicode{x3bb} }$ . This is the desired result with and $\alpha =c\unicode{x3bb} \beta $ .

Proof. Let $\varepsilon>0$ .

With $K>0$ and $\alpha>0$ appearing in Lemma 4.10, let $\unicode{x3bb} $ be such that $K\unicode{x3bb} ^{\alpha }<\varepsilon $ .

Let n big enough so that for any $k\in \mathbb {N}$ and any $Y\in \mathcal {C}$ , with $|Y|=n$ , we have $\operatorname {\mathrm {diam}} \mathbb {D}_{\gamma Y} < \unicode{x3bb} $ . Let $X \in \mathcal {C}$ and , with $|Y|=n$ . We have

$$ \begin{align*}\omega (x,X,\mathbb{C}\setminus \mathbb{X})\ge \omega (x,X,\mathbb{D}_{\gamma X}\setminus \mathbb{X}).\end{align*} $$

By conformal invariance, with $k=|X|$ , we have

$$ \begin{align*}\omega (x,X,\mathbb{D}_{\gamma X}\setminus \mathbb{X})=\omega_k(\varphi_X(x),\mathbb{X}_k,\mathbb{D}_{\gamma}\setminus \mathbb{X}_k).\end{align*} $$

Since , we have , so that $\operatorname {\mbox {{dist}}} (\varphi _X(x),\mathbb {X}_k)\le \operatorname {\mathrm {diam}} \mathbb {D}_{\gamma Y}<\unicode{x3bb} $ and Lemma 4.10 gives $\omega _k(\varphi _X(x),\mathbb {X}_k,\mathbb {D}_\gamma \setminus \mathbb {X}_k)>1-\varepsilon $ .

We have proved that for any , we have

$$ \begin{align*} \omega (x,X,\mathbb{C}\setminus \mathbb{X})\ge \omega (x,X,\mathbb{D}_{\gamma XY}\setminus \mathbb{X})=\omega_k(\varphi_X(x),\mathbb{X}_k,\mathbb{D}_\gamma\setminus \mathbb{X}_k)\ge 1-\varepsilon. \end{align*} $$

This inequality extends to $\mathbb {D}_{XY}$ by maximum principle.

Proof of Theorem 4.4

Let $\varepsilon>0$ and $N_1\in \mathbb {N}$ be big enough so that $Kq^{N_1}$ in Corollary 4.8 is strictly less than $\varepsilon $ .

We use this first assumption to show that the proof of continuity may be restricted to cylinders of length less than or equal to $N_1$ . More precisely, let $X\in \mathcal {C}_n$ , with $n>N_1$ . One may write $X=YZ$ , with $|Y|=k>N_1$ and $|Z|=\ell < N_1$ . Let a, with $|a|=1$ , we have

$$ \begin{align*} \omega_{X}(a)= \frac{\omega(YZa)}{\omega(YZ)}=\frac{\omega_{k}(Za)}{\omega_{k}(Z)} \frac{\omega(YZa)\omega_{k}(Z)}{\omega(YZ)\omega_{k}(Za)}. \end{align*} $$

The choice made for $N_1$ , together with Proposition 4.8, allows us to conclude that

$$ \begin{align*} \bigg\vert\!\log\bigg(\frac{\omega(YZa)\omega_{k}(Z)}{\omega(YZ)\omega_{k}(Za)}\bigg)\bigg\vert\le \varepsilon. \end{align*} $$

The same being true for $\tilde \omega $ , we deduce that

(4.21) $$ \begin{align} \bigg\vert\!\log\bigg(\frac{\omega_{X}(a)}{\tilde\omega_{X}(a)}\bigg)\bigg\vert\le \bigg\vert\!\log\bigg(\frac{\omega_{k}(Za)}{\tilde\omega_{k}(Za)}\bigg)\bigg\vert+\bigg\vert\!\log\bigg(\frac{\omega_{k}(Z)}{\tilde\omega_{k}(Z)}\bigg)\bigg \vert+2\varepsilon. \end{align} $$

From Corollary 4.9, we know that $cq^{N_1}\le \omega _k(Z)$ for any $k\in \mathbb {N}$ , any $Z\in T^k(\mathcal {C})$ with $|Z|\le N_1$ . Let $\varepsilon ' =cq^{N_1}\varepsilon $ . For any $k\in \mathbb {N}$ , any $Z\in T^k(\mathcal {C})$ with $|Z|\le N_1$ , we have

(4.22) $$ \begin{align} \frac{\varepsilon'}{\omega(X)}\le \varepsilon. \end{align} $$

Let $N_2$ be given by Corollary 4.11 with $\varepsilon '>0$ so that for any $XY\in \mathcal {C}$ with $|Y|=N_2$ and any , we have $\omega (x,X,\mathbb {C}\setminus \mathbb {X})\ge 1-\varepsilon '$ .

Let now be as small as necessary to ensure that for any $k\in \mathbb {N}$ , any $Y\in T^k(\mathcal {C})$ with $|Y|\le N_1+N_2$ , we have $\tilde Y\subset \mathbb {D}_{Y}$ , where $\tilde Y$ is the geometric cylinder associated to with coding Y. This means that, geometrically speaking, cylinders of length less than or equal to $N_1+N_2$ are very close to one another.

By (4.21), we are led to prove that $\log ({\omega _k(X)}/{\tilde \omega _k(X)})$ is small for any cylinder X with $|X|\le N_1$ and any k.

To simplify the notation, we omit the index k. Recall that our estimates for the harmonic measure depend only on the geometric constants appearing in our assumptions, namely, the Bounded Contraction and Annulus conditions, and are therefore uniform in k.

Let $|X|=k\le N_1$ , $F=\bigcup _{|Z|=N_1+N_2}\mathbb {D}_{Z}$ , and $\Omega =\mathbb {C}\setminus F$ . Our choices imply that for any Z with $|Z|=N_1+N_2$ , we have $\tilde Z \subset \mathbb {D}_Z$ so that $\tilde {\mathbb {X}}\subset F$ .

The functions $u:x\mapsto \omega (x,X,\mathbb {C}\setminus \mathbb {X})$ and $v:x\mapsto \omega (x,\tilde X,\mathbb {C}\setminus \tilde {\mathbb {X}})$ are harmonic on $\Omega $ .

Let $|Z|=N_1+N_2$ with $Z\subset X$ , then from Corollary 4.11, we deduce, for any , that $u(x)\ge 1-\varepsilon $ as well as $v(x)\ge 1-\varepsilon '$ , which implies that $|u(x)-v(x)|\le \varepsilon '$ for any .

Let now $Z=YW$ , with $|Y|=|X|$ , $|Z|=N_1+N_2$ , and $Z\not \subset X$ . By Corollary 4.11, we have $\omega (x,Y,\mathbb {C}\setminus \mathbb {X})>1-\varepsilon '$ and also $\omega (x,\tilde Y,\mathbb {C}\setminus \tilde {\mathbb {X}})>1-\varepsilon '$ for any .

Since

$$ \begin{align*}1=\sum_{|V|=|X|}\omega(x,V,\mathbb{C}\setminus\mathbb{X})=\sum_{|V|=|X|}\omega(x,\tilde V,\mathbb{C}\setminus\tilde{\mathbb{X}}),\end{align*} $$

we deduce that $u(x)\le 1-\omega (x,Y,\mathbb {C}\setminus \mathbb {X})\le \varepsilon '$ and $v(x)\le 1-\omega (x,\tilde Y,\mathbb {C}\setminus \tilde {\mathbb {X}})\le \varepsilon '$ . We thus have, for any , $|u(x)-v(x)|\le 2\varepsilon '$ .

The functions u and v are both harmonic on $\Omega $ and for any , we have $|u(x)-v(x)|\le 2\varepsilon '$ . By maximum principle, we have $|u(x)-v(x)|\le 2\varepsilon '$ for any $x\in \Omega $ and in particular for $x=\infty $ , so that $|\omega (X)-\tilde \omega (X)|\le 2\varepsilon '$ .

We deduce from (4.22) that for any cylinder X with $|X|\le N_1$ , we have

$$ \begin{align*} \bigg|\frac{\tilde\omega(X)}{\omega(X)}-1\bigg|\le 2\varepsilon'\frac{1}{\omega(X)}\le 2\varepsilon.\end{align*} $$

From here, it is easy to deduce that we have, for such cylinders,

$$ \begin{align*} \bigg\vert\!\log\bigg(\frac{\omega_X(a)}{\tilde\omega_X(a)}\bigg)\bigg\vert=\bigg\vert\!\log\bigg(\frac{\omega(Xa)}{\tilde\omega(Xa)}\bigg) +\log\bigg(\frac{\tilde\omega(X)}{\omega(X)}\bigg)\bigg\vert\le 4\varepsilon. \end{align*} $$

With inequality (4.21), we can conclude that for any cylinder X of any length, we have

$$ \begin{align*} \bigg\vert\!\log\bigg(\frac{\omega_X(a)}{\tilde\omega_X(a)}\bigg)\bigg\vert\le 6\varepsilon, \end{align*} $$

which tells us that $\mathcal {D}(\omega ,\tilde \omega )\le 6\varepsilon $ and concludes the proof.