2. Proof of proposition 1.4

Let $x_1,x_2$ attain the maximum in definition (1.5) of $h$, and define, for $j=1,2$, $\rho _j(s)$ as the proportion of points $x\in B_r(x_j)$ at which $m(x)$ lies in the semi-circle of direction ${\rm e}^{is}$, that is, for every $s\in {\mathbb {R}}/2\pi {\mathbb {Z}}$, we set

\[ \rho_j(s)=\frac{1}{|B_r|}\left| B_r(x_j)\cap\left\lbrace m\cdot {\rm e}^{is}>0\right\rbrace\right| = \frac1{|B_r|}\int_{B_r(x_j)}\mathbf{1}_{E_m} (x,s)\,{\rm d}x, \]

where

(2.1)\begin{equation} E_m=\left\lbrace (x,s)\in\Omega\times{\mathbb{R}}/2\pi{\mathbb{Z}}\colon m(x)\cdot {\rm e}^{is}>0\right\rbrace. \end{equation}

Note that $|\rho _j|\leq 1$ and, since for every $x \in \Omega$ it holds $|D_s\mathbf {1}_{E_m} (x,\cdot )| ( {\mathbb {R}}/2\pi {\mathbb {Z}}) =2$, then $\rho _j\in BV({\mathbb {R}}/2\pi {\mathbb {Z}})$ with $|D\rho _j|({\mathbb {R}}/2\pi {\mathbb {Z}})\leq 2$. Moreover, by Fubini theorem, these functions satisfy the identities

\[ \int_{\mathbb{R}/2\pi{\mathbb{Z}}}\,{\rm e}^{is}\rho_j(s)\,{\rm d}s = \int_{\mathbb{R}/2\pi{\mathbb{Z}}} {\int\hskip -1,05em -\,}_{B_r(x_j)}\mathbf{1}_{E_m} (x,s) {\rm e}^{is}\,{\rm d}x\,{\rm d}s = 2 {\int\hskip -1,05em -\,}_{B_r(x_j)}m(x)\,{\rm d}x. \]

For $s\in {\mathbb {R}}$ and $\rho >0$ we denote by $I_\rho (s)$ the segment

\[ I_\rho(s)=[s-\rho,s+\rho]. \]

For a small enough absolute constant $\delta \in (0,1)$, the subset $S\subset {\mathbb {R}}/2\pi {\mathbb {Z}}$ given by

\[ S=\left\lbrace s\in{\mathbb{R}}/2\pi{\mathbb{Z}}\colon (|D\rho_1|+|D\rho_2|)(I_{\delta h^2}(s))\geq \frac h {4\pi}\right\rbrace, \]

satisfies $|S|\leq h /2$ (as follows e.g. from a Besicovitch covering argument). Thus, we have

\begin{align*} h & =\frac 12\left|\int_{\mathbb{R}/2\pi{\mathbb{Z}}}{\rm e}^{is}\rho_1(s)\,{\rm d}s -\int_{{\mathbb{R}}/2\pi{\mathbb{Z}}}{\rm e}^{is}\rho_2(s)\,{\rm d}s \right|\leq\frac 12\int_{{\mathbb{R}}/2\pi{\mathbb{Z}}}|\rho_1(s)-\rho_2(s)|\,{\rm d}s \\ & \leq \frac 12 \int_{({\mathbb{R}}/2\pi{\mathbb{Z}}){\setminus} S} |\rho_1(s)-\rho_2(s)|\,{\rm d}s + \frac{h}{2}. \end{align*}

We may therefore find $s\in {\mathbb {R}}/2\pi {\mathbb {Z}}$ such that $s\notin S$ and $|\rho _1(s)-\rho _2(s)|\geq h/{2\pi }$. We assume without loss of generality that $\rho _1(s)-\rho _2(s)\geq h/2\pi$, and by definition of $S$ we deduce

\[ \inf_{I_{\delta h^2}(s) }\rho_1- \sup_{I_{\delta h^2}(s)}\rho_2\geq \frac h {4\pi} . \]

In particular, setting $s_0=s-\pi /2 -3\delta h^2/4$, we have

\begin{align*} & \inf_{I_{\delta h^2/4}(s_0+\pi/2) }\rho_1- \sup_{I_{\delta h^2/4}(s_0 +\pi/2 +\delta h^2)}\rho_2 \geq \frac h {4\pi},\\ & \inf_{I_{\delta h^2/4}(s_0+\pi/2 +\delta h^2) }\rho_1- \sup_{I_{\delta h^2/4}(s_0+\pi/2)}\rho_2 \geq \frac h {4\pi}. \end{align*}

As $\rho _j(s+\pi )=1-\rho _j(s)$ for a.e. $s\in {\mathbb {R}}/2\pi {\mathbb {Z}}$, this implies

(2.2)\begin{align} & \mathop {{\rm ess}\hspace{0.8pt}{\rm inf}}\limits_{I_{\delta h^2/4}(s_0+\pi/2) }\rho_1 + \mathop {{\rm ess}\hspace{0.8pt}{\rm inf}}\limits_{I_{\delta h^2/4}(s_0-\pi/2 +\delta h^2)}\rho_2 \geq 1 + \frac h {4\pi} , \end{align}

(2.3)\begin{align} & \mathop {{\rm ess}\hspace{0.8pt}{\rm inf}}\limits_{I_{\delta h^2/4}(s_0 +\pi/2 +\delta h^2) }\rho_1 + \mathop {{\rm ess}\hspace{0.8pt}{\rm inf}}\limits_{I_{\delta h^2/4}(s_0-\pi/2)}\rho_2 \geq 1+ \frac h {4\pi}. \end{align}

The relevance of (2.2)–(2.3) comes from the following geometric observation. Given two directions $s_1\in I_{\delta h^2/4}(s_0+\pi /2)$ and $s_2\in I_{\delta h^2/4}(s_0-\pi /2+\delta h^2)$ and two points $y_1\in B_r(x_1)\cap \lbrace m\cdot {\rm e}^{is_1}>0\rbrace$, $y_2\in B_r(x_2)\cap \lbrace m\cdot {\rm e}^{is_2}>0\rbrace$, we have $|s_1-s_2|\geq \delta h^2$, and the two lines $y_j +{\mathbb {R}} {\rm e}^{is_j}$ intersect at a point $z\in B_{8r/(\delta h^2)}$. In the absence of dissipation, one would have $m(z)\cdot {\rm e}^{is_j}>0$ for $j=1,2$, and therefore $m(z)\cdot {\rm e}^{is_0}\geq \cos (2\delta h^2)\geq 1/2$. The last lower bound is valid provided $\delta \leq \pi /24$, since $|h|\leq 2$. The same argument with $s_1\in I_{\delta h^2/4}(s_0+\pi /2+\delta h^2)$ and $s_2\in I_{\delta h^2/4}(s_0-\pi /2)$ implies instead $m(z)\cdot {\rm e}^{is_0}\leq -1/2$.

Thanks to the techniques in [Reference Marconi16], in the presence of dissipation this can be made quantitative. The main idea is that (1.3) provides an estimate on the difference between the ‘epigraph’ $E_m$ defined in (2.1) and its free transport ${\mathrm {FT}}(E_m,t)$, where the free transport operator ${\mathrm {FT}}(\cdot,t)$ is defined for $t\in {\mathbb {R}}$ by

\[ {\mathrm{FT}}(E,t)=\left\lbrace (x,s)\in\Omega\times{\mathbb{R}}/2\pi{\mathbb{Z}} \colon (x-t {\rm e}^{is},s)\in E\right\rbrace. \]

Lemma 2.1 Let $t\in {\mathbb {R}}$ and $\rho >0$ such that $B_{\rho + |t|}\subset \Omega$. For all $\phi \in C_c^1(B_\rho \times {\mathbb {R}}/2\pi {\mathbb {Z}})$ we have

\[ \int_{\Omega\times{\mathbb{R}}/2\pi{\mathbb{Z}}}\phi(x,s)\left( \mathbf 1_{{\mathrm{FT}}(E_m,t)}-\mathbf 1_{E_m}\right) \,{\rm d}x\,{\rm d}s \leq\left( |t|\, \|\partial_s\phi\|_\infty +t^2\|\nabla_x\phi\|_\infty\right)\nu(B_{\rho+|t|}). \]

Proof of lemma 2.1 Define $\chi,\chi ^{{\mathrm {FT}}}\colon [-|t|,|t|]\times \Omega \times {\mathbb {R}}/2\pi {\mathbb {Z}}\to {\mathbb {R}}$ by

\[ \chi(\tau,x,s)=\mathbf 1_{(x,s)\in E_m},\quad \chi^{{\mathrm{FT}}}(\tau,x,s)=\mathbf 1_{(x,s)\in {\mathrm{FT}}(E_m,\tau)} =\chi(x-\tau {\rm e}^{is},s), \]

so we have, in the sense of distributions,

\[ \partial_\tau\chi +{\rm e}^{is}\cdot\nabla_x \chi =\partial_s\sigma(x,s),\quad\partial_\tau\chi^{{\mathrm{FT}}} +{\rm e}^{is}\cdot\nabla_x \chi^{{\mathrm{FT}}} =0. \]

Setting $\hat \chi =\chi ^{{\mathrm {FT}}}-\chi$, and $\psi (\tau,x,s)=\phi (x+{\rm e}^{is}(t-\tau ),s)$ which satisfies $\partial _\tau \psi +{\rm e}^{is}\cdot \nabla _x\psi =0$, we deduce

\[ \partial_\tau \left[\psi\hat\chi \right] +{\rm e}^{is}\cdot\nabla_x \left[\psi\hat\chi \right] ={-}\psi \partial_s\sigma. \]

Integrating with respect to $(x,s)$ (this is formal but makes sense distributionally) we deduce

\[ \frac{{\rm d}}{{\rm d}\tau} \int_{\Omega\times{\mathbb{R}}/2\pi{\mathbb{Z}}} \psi\hat\chi\,{\rm d}x\,{\rm d}s =\int_{\Omega\times{\mathbb{R}}/2\pi{\mathbb{Z}}} \partial_s\psi\,{\rm d}\sigma(x,s). \]

Integrating this from $0$ to $t$ and recalling $\nu (A)=|\sigma |(A\times {\mathbb {R}}/2\pi {\mathbb {Z}})$ for $A\subset \Omega$, we obtain

\begin{align*} \int_{\Omega\times{\mathbb{R}}/2\pi{\mathbb{Z}}}\phi(x,s)\left( \mathbf 1_{{\mathrm{FT}}(E_m,t)}-\mathbf 1_{E_m}\right) \,{\rm d}x\,{\rm d}s & = \int_0^t \int_{\Omega\times {\mathbb{R}}/2\pi{\mathbb{Z}}} \partial_s\psi\,{\rm d}\sigma\,{\rm d}\tau \\ & \leq |t|\, \|\partial_s\psi\|_\infty |\nu|(B_{\rho+|t|}\times{\mathbb{R}}/2\pi{\mathbb{Z}}). \end{align*}

Noting that $\|\partial _s\psi \|_\infty \leq \|\partial _s\phi \|_\infty +|t| \,\|\nabla _x\phi \|_\infty$ completes the proof.

Equipped with lemma 2.1 we continue the proof of proposition 1.4. First, we make use of (2.2). We define $\hat z\in {\mathbb {R}}^2$ as the intersection of the lines $x_1+{\mathbb {R}}\,{\rm e}^{i(s_0+\pi /2)}$ and $x_2+{\mathbb {R}}\,{\rm e}^{i(s_0-\pi /2+\delta h^2)}$, that is,

\[ x_1+t_1\,{\rm e}^{i(s_0 +\pi/2)}=x_2+t_2\,{\rm e}^{i(s_0-\pi/2+\delta h^2)}=\hat z, \]

for some $t_1,t_2\in {\mathbb {R}}$. Since $|x_1-x_2|\leq 4r$, we have

(2.4)\begin{equation} |t_1|,|t_2|\leq \frac{4r}{\sin(\delta h^2)}\leq \frac{8r}{\delta h^2}, \end{equation}

and therefore $B_r(\hat z)\subset B_{R/2}$. We will use lemma 2.1 to compare $E_m$ with ${\mathrm {FT}}(E_m,t_1)$ and ${\mathrm {FT}}(E_m,t_2)$ on $B_r(\hat z)$. We define

\begin{align*} C_1 & = B_r(\hat z)\times I_{c}(s_0+\pi/2), \quad C_2 = B_r(\hat z)\times I_{c}(s_0-\pi/2+\delta h^2), \\ A_1 & =E_m\cap C_1, \quad A_2 = E_m \cap C_2 \end{align*}

with $c= \delta h^3/128\pi \le \delta h^2/4$, and their free transport counterparts

\[ A_1^{{\mathrm{FT}}} ={\mathrm{FT}}(E_m,t_1)\cap C_1, \quad A_2^{{\mathrm{FT}}} ={\mathrm{FT}}(E_m,t_2)\cap C_2. \]

We estimate

\begin{align*} |A_1^{{\mathrm{FT}}}| & = | E_m \cap {\mathrm{FT}}({\cdot}, t_1)^{{-}1}(C_1)| \\ & \ge |E_m \cap (B_r(x_1)\times I_c(s_0 + \pi/2))|\\ & - | {\mathrm{FT}}({\cdot},t_1)^{{-}1}(C_1) {\setminus} (B_r(x_1)\times I_c(s_0 +\pi/2))| \\ & = \int_{s_0 +\pi/2 -c}^{s_0 +\pi/2 +c}\rho_1(s)\,{\rm d}s - | {\mathrm{FT}}({\cdot},t_1)^{{-}1}(C_1) {\setminus} (B_r(x_1)\times I_c(s_0 +\pi/2))|. \end{align*}

Moreover,

\begin{align*} | {\mathrm{FT}}({\cdot},t_1)^{{-}1}(C_1) {\setminus} (B_r(x_1)\times I_c(s_0 +\pi/2))| & =\int_{s_0 +\pi/2 -c}^{s_0 +\pi/2 +c}|B_r(\hat z - t_1 {\rm e}^{is}) {\setminus} B_r(x_1) |\,{\rm d}s \\ & \le 2r \int_{s_0 +\pi/2 -c}^{s_0 +\pi/2 +c} | \hat z - t_1 {\rm e}^{is} - x_1|\,{\rm d}s \\ & \le 2r \int_{s_0 +\pi/2 -c}^{s_0 +\pi/2 +c} |t_1| | {\rm e}^{i (s_0 + \pi/2)}-{\rm e}^{is}|\,{\rm d}s \\ & \le 32 \frac{c^2 r^2}{\delta h^2}, \end{align*}

where in the last inequality we used (2.4). Therefore, we have

(2.5)\begin{equation} |A_1^{{\mathrm{FT}}}| \ge \int_{s_0 +\pi/2 -c}^{s_0 +\pi/2 +c}\rho_1(s)\,{\rm d}s -32 \frac{c^2 r^2}{\delta h^2}, \end{equation}

and similarly

(2.6)\begin{equation} |A_2^{{\mathrm{FT}}}| \ge\int_{ s_0 -\pi/2 +\delta h^2 -c}^{ s_0 - \pi/2 + \delta h^2 +c}\rho_2(s)\,{\rm d}s -32 \frac{c^2 r^2}{\delta h^2}. \end{equation}

From (2.2) we know that

\[ \rho_1\left(s_0+ \frac\pi 2 +s\right) + \rho_2\left(s_0-\frac\pi 2 +\delta h^2 +s\right)\geq 1+\frac h{4\pi}\quad\text{for all }|s|\leq \frac{\delta}{4}h^2. \]

Integrating this inequality in $s\in [-c,c]$, it follows from (2.5) and (2.6) that

(2.7)\begin{equation} |A_1^{{\mathrm{FT}}}| +|A_2^{{\mathrm{FT}}}|\geq 2c|B_r|\left(1+\frac h{4\pi}\right) -64 \frac{c^2 r^2}{\delta h^2}\ge 2c|B_r|\left(1+\frac h{8\pi}\right), \end{equation}

by the choice $c=\delta h^3/128\pi$. Next, we consider two cases, depending on whether $A_1$ and $A_2$ satisfy a similar inequality.

Case 1. Assume first that

\[ |A_1|+|A_2|\geq 2c|B_r|\left(1+\frac h{16\pi}\right), \]

then

\[ |\pi_x(A_1)|+ |\pi_x(A_2)|\geq |B_r|\left(1+ \frac{h}{16\pi}\right). \]

Moreover, since $\pi _x(A_1) \cup \pi _x(A_2) \subset B_r(\hat z)$, it follows that $A:= \pi _x(A_1) \cap \pi _x(A_2)$ satisfies $|A|\ge h|B_r|/16$. By construction, we have

\begin{align*} A & =\left\lbrace x\in B_r(\hat z)\colon \exists s_1\in I_{c}\left(s_0+\frac\pi2\right),\, s_2\in I_{c}\left(s_0-\frac\pi 2 +\delta h^2\right),\right.\\ & \qquad \left.m(x)\cdot {\rm e}^{is_1}>0\text{ and }m(x)\cdot {\rm e}^{is_2}>0 \right\rbrace \\ & \subset B_r(\hat z) \cap \lbrace m\cdot {\rm e}^{is_0} \geq \cos(2\delta h^2)\Big\rbrace, \end{align*}

so this implies

(2.8)\begin{equation} \left| B_{R/2}\cap \left\lbrace m\cdot m_0\geq \frac 12\right\rbrace\right|\gtrsim hr^2. \end{equation}

Case 2. Assume now that

\[ |A_1|+|A_2|< 2c |B_r|\left(1+\frac h{16\pi}\right). \]

Then using (2.7) we obtain

\[ |A_1^{{\mathrm{FT}}}|-|A_1| + |A_2^{{\mathrm{FT}}}|-|A_2| > 2c |B_r| \frac h{16\pi}, \]

so either $|A_1^{{\mathrm {FT}}}|-|A_1|$ or $|A_2^{{\mathrm {FT}}}|-|A_2|$ is larger than half the right-hand side. We consider without loss of generality only the first case:

\[ |A_1^{{\mathrm{FT}}}|-|A_1| > |B_r| \frac{ch}{16\pi}. \]

This implies a lower bound on the entropy dissipation $\nu (B_{R})$ thanks to lemma 2.1. Specifically, we apply lemma 2.1 to $t=t_1$ and $\phi \in C_c^\infty (B_{2r(\hat z)}\times I_{2c}(s_0+\pi /2))$ such that

\[ \mathbf 1_{x\in B_r(\hat z)}\mathbf 1_{s\in I_{c}(s_0+\pi/2)} \leq\phi(x,s)\leq\mathbf 1_{x\in B_{(1+\varepsilon)r}(\hat z)}\mathbf 1_{s\in I_{(1+\varepsilon)c}(s_0+\pi/2)}, \]

and $|\partial _s\phi |\leq 2/(\varepsilon c)$, $|\nabla _x\phi |\leq 2/(\varepsilon r)$. We choose $\varepsilon =h/192\pi$ to ensure

\[ \left|\left(B_{(1+\varepsilon)r}(\hat z)\times I_{(1+\varepsilon)c}\right) {\setminus}\left(B_{r}(\hat z)\times I_{c}\right)\right| \leq \frac{ch}{32\pi}|B_r|. \]

Since $|t_1|\leq 8r/(\delta h^2)$ and $B_{2r+|t_1|}(\hat z)\subset B_R$, we deduce $\nu (B_R)\gtrsim \delta ^3 h^{11} r \gtrsim h^{11}r$.

Similarly, using (2.3) we have two cases: either

(2.9)\begin{equation} \left| B_{R/2}\cap \left\lbrace m\cdot m_0\leq{-}\frac 12\right\rbrace \right|\gtrsim hr^2, \end{equation}

or $\nu (B_R)\gtrsim h^{11} r$. So gathering all cases, we see that either both (2.8) and (2.9) are satisfied, or $\nu (B_R)\gtrsim h^{11}r$, which is exactly the dichotomy of proposition 1.4.

3. Proof of proposition 1.5

To prove proposition 1.5, we briefly recall from [Reference Marconi14] the notion of Lagrangian representation of an entropy solution $m$ of the eikonal equation. In [Reference Marconi14, Reference Marconi15], the second author shows the existence of a finite non-negative Radon measure $\omega$ on the set of curves:

\begin{align*} \Gamma& = \left\{ (\gamma,t^-_\gamma,t^+_\gamma)\colon 0\le t^-_\gamma\le t^+_\gamma\le 1, \right.\\ & \qquad\gamma=(\gamma_x,\gamma_s)\in {\mathrm{BV}}((t^-_\gamma,t^+_\gamma);\Omega \times {\mathbb{R}}/2\pi {\mathbb{Z}}), \\ & \qquad\left.\gamma_x \text{is Lipschitz} \right\}, \end{align*}

with the following three properties:

• • for every $t\in (0,1)$, the pushforward of $\omega$, restricted to the section $\Gamma (t)=\lbrace (\gamma,t^-_\gamma,t^+_\gamma )\in \Gamma \colon t_\gamma ^- < t < t_\gamma ^+\rbrace$, by the evaluation map $e_t\colon \gamma \mapsto \gamma (t)$ (a right-continuous representative of $\gamma _s$ is always considered), is uniform on the ‘epigraph’ $E_m=\lbrace m(x)\cdot {\rm e}^{is}>0\rbrace$, that is,

  (3.1)\begin{equation} ( e_t)_\sharp \left[ \omega \lfloor \Gamma(t)\right]= \mathbf 1_{m(x)\cdot {\rm e}^{is}>0}\,{\rm d}x\,{\rm d}s ; \end{equation}

• • the measure $\omega$ is concentrated on curves $(\gamma,t^-_\gamma,t^+_\gamma )\in \Gamma$ solving the characteristic equation:

  (3.2)\begin{equation} \dot\gamma_x(t)= {\rm e}^{i \gamma_s(t)}\quad\text{for a.e. }t\in (t^-_\gamma,t^+_\gamma); \end{equation}

• • the entropy dissipation measure (1.4) disintegrates along the Lagrangian curves as

  (3.3)\begin{equation} \nu(A)= \int_\Gamma \mu_\gamma (\gamma_x^{{-}1}(A))\,{\rm d}\omega(\gamma)\quad\text{for all measurable }A\subset\Omega, \end{equation}
  where $\mu _\gamma =|D_t\gamma _s|$, with the convention that a jump of $\gamma _s$ from $s^-$ to $s^+$ at time $t_0\in (t_\gamma ^-,t_\gamma ^+)$ contributes $\operatorname {dist}_{{\mathbb {R}}/2\pi {\mathbb {Z}}}(s^-,s^+)\delta _{t=t_0}$ to the jump part of $\mu _\gamma$ (see [Reference Marconi14, proposition 2.5]).

Moreover, the Lagrangian property (3.1) implies that $\omega$ is concentrated on curves $\gamma$ such that $\gamma _x(t)$ is a Lebesgue point of $m$ with $m(\gamma _x(t))\cdot {\rm e}^{i\gamma _s(t^+)}>0$, for a.e. $t\in (0,1)$ [Reference Marconi14, lemma 2.7]. We denote by $\Gamma _g\subset \Gamma$ the full-measure subset of Lagrangian curves which satisfy that property together with the characteristic equation (3.2).

The proof of proposition 1.5 is based on two main tools. The first, lemma 3.1, is a dichotomy stating that either Lagrangian curves passing through a given set create a lot of dissipation, or one can find an almost-straight Lagrangian curve passing through that set. The second ([Reference Lamy and Marconi12, lemma 5.2], a slightly more precise version of [Reference Marconi14, lemma 3.1], itself adapted from [Reference Marconi15, lemma 22]) is another dichotomy: given an almost-straight Lagrangian curve, either the density of points at which $m$ lies in the semi-circle indicated by the $s$-component of that curve is high, or a lot of dissipation must be created. The succession of these two dichotomies is reflected in the three alternatives in the conclusion of proposition 1.5. We first state and prove the first tool, and then proceed to the proof of proposition 1.5.

Lemma 3.1 For any $R>0$ such that $B_R\subset \Omega$, any measurable set $A\subset B_{R}\times {\mathbb {R}}/2\pi {\mathbb {Z}}$, and any $\eta \in (0,1)$, we have either

(3.4)\begin{equation} \nu(B_R)\gtrsim \frac{\eta}{ R}\left|\lbrace (x,s)\in A\colon m(x)\cdot {\rm e}^{is}>0\rbrace\right|, \end{equation}

or there exists a curve $\gamma \in \Gamma _g$ and a connected component $J$ of $\gamma _x^{-1}(B_R)$ such that

\[ J\cap\gamma^{{-}1}(A)\neq \emptyset\quad\text{and}\quad \mu_\gamma(J)\leq \eta. \]

Proof of lemma 3.1 Assume that the second alternative of lemma 3.1 is not verified: for every curve $\gamma \in \Gamma _g$ and every connected component $J$ of $\gamma _x^{-1}(B_R)$ intersecting $\gamma ^{-1}(A)$, we have $\mu _\gamma (J) >\eta$. Then we claim that

(3.5)\begin{equation} \mu_\gamma(\gamma_x^{{-}1}(B_R)) \gtrsim \frac{\eta }{ R}T(\gamma),\quad T(\gamma) =\left| \lbrace t\in (t_\gamma^-,t_\gamma^+)\colon \gamma(t)\in A \rbrace\right|, \end{equation}

for all $\gamma \in \Gamma _g$. To prove (3.5), denote by $J_k=(t_k^-,t_k^+)$ the connected components of $\gamma _x^{-1}(B_R)$ which intersect $\gamma ^{-1}(A)$. We show next that $\mu _\gamma (J_k)\gtrsim \eta |J_k|/ R$ for all $k$. On the one hand, if $|J_k|\leq 4R$ then $\mu _\gamma (J_k)\gtrsim \eta |J_k|/ R$ because $\mu _\gamma (J_k)> \eta$ by assumption. On the other, from the characteristic equation (3.2) and the definition of $\mu _\gamma =|D_t\gamma _s|$, we have the inequality:

\[ |\gamma_x(t_2)-\gamma_x(t_1)-{\rm e}^{i\gamma_s(t_1)}(t_2-t_1)|\leq \mu_\gamma([t_1,t_2])|t_2-t_1|, \]

and we deduce that in any interval $J\subset (0,1)$ such that $\gamma _x(J)\subset B_R$ and $|J|\geq 4R$, we must have $\mu _\gamma (J) \geq 1/2$. Therefore, if $|J_k|\geq 4R$, cutting $J_k$ in disjoint subintervals of length between $4R$ and $8R$, we obtain that $\mu _\gamma (J_k) \gtrsim |J_k|/R \geq \eta |J_k|/ R$. So we have

\[ \mu_\gamma(\gamma_x^{{-}1}(B_R)) \geq \sum_k \mu_\gamma(J_k) \gtrsim \frac{\eta}{R}\sum_k |J_k|, \]

which implies (3.5) since $\gamma ^{-1}(A)\subset \bigcup _k J_k$. From (3.5) and the fact that $\omega (\Gamma {\setminus} \Gamma _g)=0$ we infer

\[ \nu(B_{R})= \int_{\Gamma} \mu_\gamma(\gamma_x^{{-}1}(B_R ))\,{\rm d}\omega(\gamma)\gtrsim \frac{\eta}{R}\int_{\Gamma} T(\gamma)\,{\rm d}\omega(\gamma), \]

where the first equality comes from disintegration (3.3). Making use of the Lagrangian property (3.1) to rewrite the last expression, we see that it is precisely equal to the right-hand side of (3.4), which concludes the proof of lemma 3.1.

Proof of proposition 1.5 We recall that $m_0={\rm e}^{is_0}$ and the sets $X_\pm$ are defined by

\[ X_\pm{=}B_{R/2}\cap \lbrace \pm m\cdot {\rm e}^{is_0}\geq 1/2\rbrace. \]

For any $\hat s\in [s_0-\pi /4,s_0+\pi /4]$, we apply lemma 3.1 to $A(\hat s)=B_{R/2}\times I_\eta (\hat s)$, where $I_\eta (\hat s)=[\hat s-\eta,\hat s+\eta ]$. If $\eta \in (0,\pi /12)$ then we have $m(x)\cdot {\rm e}^{is}>0$ for all $(x,s)\in X_+\times I_\eta (\hat s)$, and therefore,

\[ \left|\lbrace (x,s)\in A(\hat s)\colon m(x)\cdot {\rm e}^{is}>0\rbrace\right|\geq \eta |X_+|. \]

So, we have either $\nu (B_R)\gtrsim \eta ^2|X_+|/R$, or there exists a curve $\gamma \in \Gamma _g$ and a connected component $J$ of $\gamma _x^{-1}(B_R)$ intersecting $A(\hat s)$ such that $\mu _\gamma (J)<\eta$. In that second case, applying [Reference Lamy and Marconi12, lemma 5.2] we deduce that either $\nu (B_R)\gtrsim \eta ^3 R$ or $|B_R\cap \lbrace m\cdot {\rm e}^{i\hat s}\geq -2\eta \rbrace |\gtrsim \eta R^2$. We fix $\eta =1/4$ and summarize the preceding discussion: for all $\hat s\in [s_0-\pi /4,s_0+\pi /4]$, we have

\[ \nu(B_R)\gtrsim \frac{|X_+|}{R},\quad\text{ or }\nu(B_R)\gtrsim R,\quad\text{or }\left|B_R\cap \lbrace m\cdot {\rm e}^{i\hat s}\geq{-}1/2\rbrace\right|\gtrsim R^2. \]

Similarly, for all $\hat s\in [s_0+3\pi /4,s_0+5\pi /4]$, we have

\[ \nu(B_R)\gtrsim \frac{|X_-|}{R},\quad\text{or }\nu(B_R)\gtrsim R,\quad\text{or }\left|B_R\cap \lbrace m\cdot {\rm e}^{i\hat s}\geq{-}1/2\rbrace\right|\gtrsim R^2. \]

We conclude that we have either (1.10), or $\nu (B_R)\gtrsim R$, or

\[ \left|B_R\cap \lbrace m\cdot {\rm e}^{i s}\geq{-}1/2\rbrace\right|\gtrsim R^2, \]

for all $s\in [s_0-\pi /4,s_0+\pi /4]\cup [s_0+3\pi /4,s_0+5\pi /4]$. This corresponds exactly to the three alternatives in the statement of proposition 1.5.