Proof. It suffices to prove the result for the case $\eta = \sigma \cup \{s\}$, for some $s \in[a,b] \setminus \sigma$. Indeed, if $\eta$ had more than one extra point than $\sigma$, as partitions are finite we can just iteratively apply the previous case until we include all points of $\eta$. As $\sigma = \{t_i\}_{i=0}^m$ is a partition of $[a,b]$ and $s \not \in \sigma$, there exists a unique $i \in \{0, \dots, m-1\}$ such that $s \in (t_i, t_{i+1})$. We then see that by the reverse triangular inequality of $\tau$ we get

\begin{align*} \max_{t \in [t_i, t_{i+1}]} \Omega(\gamma(t)) \, \tau(\gamma(t_i), \gamma(t_{i+1}))&\geq \max_{t \in [t_i, t_{i+1}]}\Omega(\gamma(t)) \,\tau(\gamma(t_i), \gamma(s)) \nonumber\\ &\quad + \max_{t \in [t_i, t_{i+1}]}\Omega(\gamma(t)) \,\tau(\gamma(s), \gamma(t_{i+1}))\\ &\geq \max_{t \in [t_i, s]}\Omega(\gamma(t)) \,\tau(\gamma(t_i), \gamma(s)) \nonumber\\ &\quad + \max_{t \in [s, t_{i+1}]}\Omega(\gamma(t)) \,\tau(\gamma(s), \gamma(t_{i+1})). \end{align*}

Since all of the other terms required to compute the variations $V_{\Omega, \sigma}$ and $V_{\Omega, \eta}$ are the same, we immediately get that $ V_{\Omega, \sigma}(\gamma) \geq V_{\Omega, \eta}(\gamma)$ which proves the first claim.

The second claim immediately follows from noticing that the inequality ‘ $\leq$’ holds by definition and the inequality ‘ $\geq $’ is implied by the fact that for any partition $\theta$, we can always consider the partition $\theta \cup \sigma$, which will contain $\sigma$ and will give a smaller variation by the first claim.

Proof. We prove each point separately:

1. (i) By definition, $L_\Omega^\tau $ is equal to the infimum over all partitions of $[a,b]$ of the variations $V_{\Omega, \sigma}$. In particular, this includes as well partitions which have modulus less than $\delta$, hence the inequality ‘ $\leq$’ is trivial. About the other inequality, for any $\varepsilon \gt 0$ we can consider a partition $\sigma = \{t_i\}_{i=0}^m$ such that $V_{\Omega,\sigma}(\gamma) \leq L_\Omega^\tau (\gamma) + \varepsilon$. We split each interval $[t_i, t_{i+1}]$ of the partition into subintervals which have modulus less than $\delta$ as follows: let $M\in\mathbb{N}$ be such that $\left\vert \sigma \right\vert/M \lt \delta$. For each $i \in \{0, \dots, m-1\}$ we define the points

   \begin{equation*} t_j^i := t_i + \frac{t_{i+1} - t_i}{M}\cdot j, \quad \text{for} \ j \in \{0, \dots, M\}. \end{equation*}

   The collection of points $\eta = \{t_j^i\}_{i=0,j=0}^{m-1,M}$ is again a partition of $[a,b]$. Moreover, $\left\vert \eta \right\vert \lt \delta$: indeed, it is clear that for any $i \in \{0, \dots,m-1\}$ and any $j \in \{0, \dots, M-1\} $ we have that

   \begin{equation*} \left\vert t^i_{j+1} - t_j^i \right\vert = \frac{t_{i+1}-t_i}{M} \leq \frac{\left\vert \sigma \right\vert}{M} \lt \delta. \end{equation*}

   If instead we look at any point of the form $t_M^i$, for $i \in \{0, \dots, m-1\}$, we know that $t_M^i = t_0^{i+1}$, which is either equal to $b$ so there is nothing to be proved, or we can apply the previous reasoning. The partition $\eta$ contains $\sigma$ by construction, so by Lemma 3.3 we immediately get that

   \begin{equation*} V_{\Omega, \eta}(\gamma) \leq V_{\Omega,\sigma}(\gamma) \leq L_\Omega^\tau (\gamma) + \varepsilon. \end{equation*}

   As we have proven that for any $\varepsilon \gt 0$ there is a partition $\eta$ with modulus less than $\delta$ such that the variation is less than $ L_\Omega^\tau (\gamma) + \varepsilon$, we have that the inequality ‘ $\geq$’ holds and thus the two infimum are equal.

2. (ii) Again, we prove that both the inequalities ‘ $\leq$’ and ‘ $\geq$’ hold. For ‘ $ \leq $’, take any two partitions $\sigma$ and $\eta$ of $[a,s]$ and $[s,b]$, respectively. Clearly, the set $\theta := \sigma \cup \eta$ is a partition of the full interval $[a,b]$. From the definition, it is immediate to see that

   \begin{equation*} L_\Omega^\tau (\gamma) \leq V_{\Omega, \theta}(\gamma) = V_{\Omega, \sigma}(\gamma\vert_{[a,s]}) + V_{\Omega, \eta}(\gamma\vert_{[s,b]}). \end{equation*}

   Taking the infimum over all $\sigma$ and $\eta$ yields the inequality ‘ $\leq$’. For the opposite one, we start with the partition $\sigma = \{a,s,b\}$ and consider any partition $\theta$ of the interval $[a,b]$ containing $\sigma$. We then define $\theta_{\leq s}$ and $\theta_{\geq s}$ to be the set of points in $\theta$ which are less than or equal to $s$ and greater or equal to $s$, respectively. The two sets $\theta_{\leq s}$ and $\theta_{\geq s}$ are then partitions of $[a,s]$ and $[s,b]$; from the definition, it is immediate to see that

   \begin{equation*} V_{\Omega, \theta}(\gamma) = V_{\Omega, \theta_{\leq s}}(\gamma\vert_{[a,s]}) + V_{\Omega, \theta_{\geq s}}(\gamma\vert_{[s,b]}) \geq L_\Omega^\tau (\gamma\vert_{[a,s]}) + L_\Omega^\tau (\gamma\vert_{[s,b]}). \end{equation*}

   Since this holds for any $\theta$ containing $\sigma$, by Lemma 3.3 taking the infimum over all such $\theta$ yields the inequality ‘ $\geq$’, which concludes the proof.

3. (iii) This last point is immediate, since if $\phi : [c,d] \to [a,b]$ is an increasing Lipschitz homeomorphism, any partition $\sigma$ of $[a,b]$ yields a partition $\phi^{-1}(\sigma)$ of $[c,d]$ with the same variation, and for any partition $\eta$ of $[c,d]$ we get the natural partition $\phi(\eta)$ of $[a,b]$, which again has the same variation.

Proof. If we denote by $m$ and $M$ the minimum and the maximum, respectively, of the function $\Omega\circ \gamma$, it is immediate to see from the definition that

\begin{equation*} m L^\tau(\gamma) \leq L_\Omega^\tau(\gamma) \leq M L^\tau(\gamma). \end{equation*}

From these previous inequalities we get that $ L_\Omega^\tau (\gamma) \gt 0$ if and only if $ L^\tau(\gamma) \gt 0$, as both $m$ and $M$ are positive, which proves the first item. If we now assume that $\gamma$ is length maximizing with respect to $L^\tau$ and of zero length, it must also be length maximizing with respect to $L_\Omega^\tau $; otherwise, there would be a causal curve $\alpha$ joining the endpoints of $\gamma$ with greater $L_\Omega^\tau $-length, and this would imply that $L_\Omega^\tau (\alpha) \gt 0$. By the previous item, we know then that $L^\tau(\alpha) \gt 0$, which then contradicts $\gamma$ being $L^\tau$-maximal.

Proof. Taking the supremum over all causal curves with endpoints $x$ and $y$ on item (i) of Proposition 3.6 yields that $\tau(x,y) \gt 0$ if and only if $\tau_{\Omega}(x,y) \gt 0$. This, together with the definitions of $\ll_\Omega ,\leq_{\Omega}$, immediately proves the second claim.

Proof. For any $\varepsilon \gt 0$, we can consider causal curves $\gamma$ from $x$ to $y$ and $\alpha$ from $y$ to $z$ such that

\begin{equation*} \tau_\Omega(x,y)\leq L_\Omega^\tau (\gamma) + \varepsilon \ \land \ \tau_\Omega(y,z)\leq L_\Omega^\tau (\alpha) + \varepsilon. \end{equation*}

Concatenating the two curves yields a causal curve $\gamma\alpha$ from $x$ to $z$. By item (ii) of Proposition 3.4, we get

\begin{equation*} \tau_\Omega(x,z) \geq L_\Omega^\tau (\gamma\alpha) = L_\Omega^\tau (\gamma) + L_\Omega^\tau (\alpha) \geq \tau_\Omega(x,y) + \tau_\Omega(y,z) - 2\varepsilon. \end{equation*}

Since this holds for any $\varepsilon \gt 0$, we get the desired result.

Proof. If $p \not \leq q$ there is nothing to do. Let us thus assume that $p \leq q$ and consider any causal curve $\gamma : [a,b] \to X$ joining them. For any partition of $[a,b]$ we see that

\begin{equation*} \sum_{i=0}^{m-1}L_{\Omega}^\tau(\gamma\vert_{[t_i, t_{i+1}]}) \leq \sum_{i=0}^{m-1} \tau_\Omega (\gamma(t_i), \gamma(t_{i+1})) \leq \tau_\Omega(p,q), \end{equation*}

where the left inequality follows from the definition of $\tau_\Omega$ as a supremum of $L_\Omega^\tau$-lengths and the right inequality from the reverse triangle inequality shown in Proposition 3.8. We apply additivity of the $L_{\Omega}^\tau$-length and get that

\begin{equation*} L_\Omega^\tau(\gamma) \leq \sum_{i=0}^{m-1} \tau_\Omega (\gamma(t_i), \gamma(t_{i+1})) \leq \tau_\Omega(p,q). \end{equation*}

Taking the infimum over all partitions of $[a,b]$ we have

\begin{equation*} L_\Omega^\tau(\gamma) \leq L^{\tau_\Omega}(\gamma) \leq \tau_\Omega(p,q), \end{equation*}

and then taking the supremum over all such curves $\gamma$ we finally get

\begin{equation*} \tau_\Omega(p,q) \leq \hat{\tau}_\Omega(p,q) \leq \tau_\Omega(p,q), \end{equation*}

which proves that $\tau_\Omega(p,q) = \hat{\tau}_\Omega(p,q)$.

Proof. We just consider the case where the sequence $\{p_n\}_n$ is contained in $J^+(p)$—the other case being completely analogous.

Using continuity of $\Omega$ at $p$, we can find an open ball $U$ around $p$ such that the oscillation of $\Omega$ is bounded by $\varepsilon$ on $U$. We then employ quasi-strong causality to find a smaller neighbourhood $V \subset U$ such that any causal curve with endpoints in $V$ stays in $U$. Then there exists an $N \in \mathbb{N}$ such that for every $n\geq N$ we have $p_n \in V$. If we now take a curve $\gamma_n$ from $p$ to $p_n$, we see that for any partition $\sigma$ of its domain, we have that

\begin{equation*} (\Omega(p) -\varepsilon)V_{\sigma}(\gamma_n) \leq V_{\Omega,\sigma}(\gamma_n) \leq (\Omega(p) +\varepsilon)V_{\sigma}(\gamma_n) , \end{equation*}

since the terms featuring $\Omega$ can be bounded above and below by $\Omega(x)$ up to an error $\varepsilon$, as $\gamma_n$ is entirely contained in $U$ by quasi-strong causality. Taking the infimum over all partitions $\sigma$ then yields (3). Now, for any $\varepsilon \gt 0$ eventually in $n$ we have that for any causal curve $\gamma$ between $p$ and $p_n$ it holds that

\begin{equation*} (\Omega(p) -\varepsilon)L^{\tau}(\gamma) \leq L_\Omega^\tau (\gamma) \leq (\Omega(p) +\varepsilon)L^{\tau}(\gamma). \end{equation*}

By intrinsicness of $\tau$, taking the supremum over all such curves $\gamma$ then yields (4).

If we now consider the additional assumption that $\tau(p,p_n)\in(0,\infty)$, we can divide the previous inequalities by $\tau(p,p_n)$ getting

\begin{equation*} (\Omega(p) -\varepsilon)\leq \frac{\tau_\Omega(p, p_n)}{\tau(p, p_n)} \leq (\Omega(p) +\varepsilon)\qquad\Longleftrightarrow\qquad \left\vert \frac{\tau_\Omega(p, p_n)}{\tau(p, p_n)} - \Omega(p) \right\vert \leq \varepsilon. \end{equation*}

Since this inequality holds for any $\varepsilon \gt 0$ and for $n$ sufficiently large, this proves the last claim.

Proof. Let $p,q \in X$ be any two points. If $\tau(p,q)=0$ (which happens if and only if $\tau_\Omega(p,q)=0$ by Proposition 3.7), then lower semicontinuity follows immediately. We therefore assume $\tau(p,q) \gt 0$, and consider two arbitrary sequences of points $p_n$ and $q_n$ converging to $p$ and $q$, respectively.

If $p = q$, the reverse triangle inequality for $\tau$ implies that $\tau(p,p) = +\infty$. Then Proposition 3.6 ensures that $\tau_{\Omega}(p,p)\neq0$, hence $\tau_\Omega(p,p) = +\infty $ by the reverse triangle inequality for $\tau_{\Omega}$. We now fix any $\varepsilon \in (0, \Omega(p))$ and find a neighbourhood $U$ of $p$ where the oscillation of $\Omega$ is bounded by $\varepsilon$, as well as the corresponding quasi-strong causality neighbourhood $V\subseteq U$ (cf. Definition 2.12).

Eventually in $n$, both sequences $p_n$ and $q_n$ are contained in $V$, and lower semicontinuity of $\tau$ yields

\begin{equation*} \liminf_{n \to +\infty} \tau(p_n,q_n) \geq \tau(p,p) = +\infty, \end{equation*}

which implies that eventually in $n$ we have $p_n \ll q_n$. Therefore, for sufficiently large $n$ we can find causal curves $\gamma_n$ between $p_n$ and $q_n$ which are contained in $U$ with $L^\tau(\gamma_n) \to +\infty$ as $n \to +\infty$. Estimating the $L^\tau_\Omega$-length as in Lemma 3.10 we see that

\begin{equation*} \tau_\Omega(p_n,q_n) \geq L_\Omega^\tau(\gamma_n) \geq (\Omega(p) - \varepsilon)L^\tau(\gamma_n). \end{equation*}

Taking the $\liminf$ as $n \to +\infty$ we see that

\begin{equation*} \liminf_{n \to +\infty} \tau_\Omega(p_n,q_n) \geq +\infty = \tau_\Omega(p,p), \end{equation*}

which proves lower semicontinuity.

Now, if $p \neq q$, as $\tau(p,q) \gt 0$ we get a causal curve $\gamma:[a,b]\rightarrow X$ from $p$ to $q$ with positive $\tau$-length.

We define

\begin{equation*} \begin{aligned} t_1 &:= \sup \{t \in [a,b] \ : \ p \not \ll \gamma(t)\}, \\ t_2 &:= \inf \{t \in [a,b] \ : \ \gamma(t) \not \ll q\}, \end{aligned} \end{equation*}

and set $\widetilde{p} := \gamma(t_1)$, $\widetilde{q} := \gamma(t_2)$; if the sets are empty, we set $t_1 = a$ and $t_2 = b$, so that $\widetilde{p} = p$ and $\widetilde{q}=q$, respectively. We now notice the following facts:

• • $L_\Omega^\tau(\gamma\vert_{[t_1,t_2]}) = L^\tau_\Omega(\gamma)$; indeed, this is trivially true if $t_1 =a$ and $t_2 = b$. If $t_1 \neq a$ (or $t_2 \neq b$), we see that by continuity of $\gamma$ and by the fact that the complement of $I^+(p)$ ( $I^-(q)$, respectively) is a closed set we must have $p \not \ll \widetilde{p}$ ( $\widetilde{q} \not \ll q$, respectively). In particular then the $L^\tau$-length of the curve $\gamma\vert_{[a, t_1]}$ ( $\gamma\vert_{[t_2,b]}$, respectively) vanishes and therefore so does the $L^\tau_\Omega$-length, by Proposition 3.6. By additivity of the length, we see then that the length of $\gamma\vert_{[t_1,t_2]}$ coincides with the length of $\gamma$.

• • $t_1 \lt t_2$; this is trivially true when $t_1 = a$ and $t_2 = b$ as $a \lt b$. If $t_1 \neq a$ ( $t_2 \neq b$, respectively), assuming by contradiction that $t_1 \geq t_2$ gives in particular that $t_2 \neq b$ ( $t_1 \neq a$, respectively): if $t_1 = b$ ( $t_2 = a$, respectively) then the sup (inf) exists and $p \not \ll \gamma(b) = q$ ( $p = \gamma(a) \not \ll q$), since the complement of $I^+(p)$ ( $I^-(q)$) is closed; on the other hand we assumed $p \ll q$, which is a contradiction. As we noticed before in such case we have $p \not \ll \widetilde{p}$ and $\widetilde{q} \not \ll q$. We then estimate the length of $\gamma$ as

  \begin{equation*} L^\tau(\gamma) \leq \tau(p, \widetilde{p}) + \tau(\widetilde{p},q) =0, \end{equation*}
  as the first term vanishes since $p \not \ll \widetilde{p}$ and the second one vanishes as the assumption that $t_1 \geq t_2$ implies that $\gamma(t_1) \not \ll q$ by the push-up property; this contradicts the fact that $\gamma$ has positive length.

For $h \in (0, \frac{t_2-t_1}{2})$ we define $p^+(h) := \gamma(t_1+h)$ and $q^-(h) := \gamma(t_2-h)$. Notice that $p^+(h) \in I^+(p)$ and $q^-(h) \in I^-(q)$ since $t_1+h \gt t_1$ and $t_2 -h \lt t_2$. In particular $I^-(p^+(h))$ and $I^+(q^-(h))$ are open neighbourhoods of $p$ and $q$, respectively. Therefore, for a fixed $h$ the two sequences are eventually contained in them, i.e. $p_n \ll p^+(h)$ and $q_n \ll q^-(h)$ for sufficiently large $n$. We fix $\varepsilon \gt 0$ with $\varepsilon \lt \min(\Omega(\widetilde{p}), \Omega(\widetilde{q}))$ and consider neighbourhoods at $\widetilde{p}$ and $\widetilde{q}$ such that the oscillation of $\Omega$ is bounded by $\varepsilon$. We then employ quasi-strong causality to find smaller neighbourhoods $V_1$ and $V_2$ which satisfy the quasi-strong causality condition. Lastly, we take $h \leq h(\varepsilon)$ small enough so that $p^+(h)$ and $q^-(h)$ are in $V_1$ and $V_2$, respectively.

As $p^+(h) \leq q^-(h)$, we can apply the reverse triangle inequality to see that

\begin{equation*} \tau_{\Omega}(p_n,q_n) \geq \tau_{\Omega}(p_n, p^+(h)) + \tau_{\Omega}(p^+(h),q^-(h)) + \tau_{\Omega}(q^-(h), q_n) \geq \tau_\Omega(p^+(h), q^-(h)). \end{equation*}

Notice that such estimate holds whenever $n$ is big enough, hence taking the $\liminf$ yields

\begin{equation*} \liminf_{n \to +\infty} \tau_{\Omega}(p_n,q_n) \geq \tau_\Omega(p^+(h), q^-(h)). \end{equation*}

As $V_1$ and $V_2$ are open neighbourhoods of $p^+(h(\varepsilon))$ and $q^-(h(\varepsilon))$, respectively, we can find two points $p_1$ and $p_2$ along $\gamma$ so that $p_i \in V_i$ for $i=1,2$ and for every $h \leq h(\varepsilon)$ it holds $p^+(h) \leq p_1 \leq p_2 \leq q^-(h)$. Applying the triangle inequality then yields

\begin{equation*} \liminf_{n \to +\infty} \tau_{\Omega}(p_n,q_n) \geq \tau_\Omega(p^+(h), q^-(h)) \geq \tau_\Omega(p^+(h), p_1) + \tau_\Omega(p_1, p_2) + \tau_\Omega(p_2, q^-(h)). \end{equation*}

As $p^+(h), p_1 \in V_1$ and $p_2,q^-(h) \in V_2$ every causal curve between each pair is contained in a neighbourhood where the oscillation of $\Omega$ is bounded by $\varepsilon$. Analogously to what is done in Lemma 3.10, we can estimate $\tau_\Omega$ by

\begin{align*} &\tau_\Omega(p^+(h), p_1) + \tau_\Omega(p_1, p_2) + \tau_\Omega(p_2, q^-(h)) \nonumber\\ &\quad \geq (\Omega(\widetilde{p}) - \varepsilon) \tau(p^+(h), p_1) + \tau_\Omega(p_1, p_2) + (\Omega(\widetilde{q}) - \varepsilon)\tau(p_2, q^-(h)). \end{align*}

Notice that this estimate holds for any $h \leq h(\varepsilon)$. We can then employ lower semicontinuity of $\tau$ sending $h \to 0$ to see that

\begin{align*} & \liminf_{n \to +\infty}\tau_\Omega(p_n,q_n) \geq\liminf_{h \to 0^+} \tau_\Omega(p^+(h), q^-(h)) \nonumber\\ &\quad \geq (\Omega(\widetilde{p}) - \varepsilon) \tau(\widetilde{p}, p_1) + \tau_\Omega(p_1, p_2) + (\Omega(\widetilde{q}) - \varepsilon) \tau(p_2, \widetilde{q}). \end{align*}

We can once again estimate as in Lemma 3.10 to get

\begin{equation*} \begin{aligned} \liminf_{n \to +\infty}\tau_\Omega(p_n,q_n) &\geq \frac{\Omega(\widetilde{p}) - \varepsilon}{\Omega(\widetilde{p}) + \varepsilon} \tau_\Omega(\widetilde{p}, p_1) + \tau_\Omega(p_1, p_2) + \frac{\Omega(\widetilde{q}) - \varepsilon}{\Omega (\widetilde{q}) + \varepsilon}\tau_\Omega(p_2, \widetilde{q})\\ &\geq \min \left( \frac{\Omega(\widetilde{p}) - \varepsilon}{\Omega(\widetilde{p}) + \varepsilon}, \frac{\Omega(\widetilde{q}) - \varepsilon}{\Omega (\widetilde{q}) + \varepsilon}, 1 \right) (\tau_\Omega(\widetilde{p},p_1) + \tau_\Omega(p_1,p_2) + \tau_\Omega(p_2, \widetilde{q})) \end{aligned} \end{equation*}

Each term of the sum on the right-hand side can be estimated from below by the $L_\Omega^\tau$-length of $\gamma$ restricted to the corresponding interval. Using then additivity of the length, we get

\begin{equation*} \liminf_{n \to +\infty}\tau_\Omega(p_n,q_n) \geq \min \left( \frac{\Omega(\widetilde{p}) - \varepsilon}{\Omega(\widetilde{p}) + \varepsilon}, \frac{\Omega(\widetilde{q}) - \varepsilon}{\Omega (\widetilde{q}) + \varepsilon}, 1 \right) L_\Omega^\tau(\gamma\vert_{[t_1,t_2]}) \end{equation*}

Sending $\varepsilon \to 0^+$ yields

\begin{equation*} \liminf_{n \to +\infty}\tau_\Omega(p_n,q_n) \geq L_\Omega^\tau(\gamma\vert_{[t_1,t_2]}) = L_\Omega^\tau(\gamma). \end{equation*}

Notice that this final estimate holds trivially if $\gamma$ has zero length, so it holds for any causal curve between $p$ and $q$. Taking the supremum over all such curves then gives

\begin{equation*} \liminf_{n \to +\infty}\tau_\Omega(p_n,q_n) \geq \tau_\Omega(p,q), \end{equation*}

which proves lower semicontinuity.

Proof. We first prove that $L_\Omega^\tau (\gamma)\leq L_{\tau_{\Omega}}(\gamma)$. For any partition $\sigma$ of $[a,b]$ we see that

\begin{equation*} \begin{aligned} \sum_{i=1}^{m-1}\tau_{\Omega}\big(\gamma(t_i),\gamma(t_{i+1})\big) &\geq \sum_{i=1}^{m-1}L_\Omega^\tau \left(\gamma\vert_{[t_i,t_{i+1}]}\right) =L_\Omega^\tau (\gamma), \end{aligned} \end{equation*}

as $\tau_\Omega(\gamma(t_i), \gamma(t_{i+1}))$ is defined as the supremum of the $L_\Omega^\tau$-length of curves between the two points and thus can be estimated from below by the $L_\Omega^\tau$-length of $\gamma\vert_{[t_i,t_{i+1}]}$. Taking the infimum over all partitions $\sigma$ immediately yields $L^{\tau_{\Omega}}(\gamma)\geq L_\Omega^\tau (\gamma)$.

We now prove $L^{\tau_{\Omega}}(\gamma)\leq L_\Omega^\tau (\gamma)$. Consider any positive real number $\varepsilon$. Since $\Omega$ is continuous, for all $t\in[a,b]$ there exists an open ball $U_t = B(\gamma(t),\delta_t)$ centred at $\gamma(t)$ with radius $\delta_t \gt 0$ such that, for all $x\in U_t$, we have $\big\vert\Omega\big(\gamma(t)\big)-\Omega\big(x\big)\big\vert\leq \varepsilon$. By quasi-strong causality, there exists an open set $V_t\subset U_t$ containing $\gamma(t)$, such that any causal curve with endpoints in $V_t$ is contained in $U_t$. This is true for any $t$, so $\gamma$ is entirely contained in $\bigcup_{t\in[a,b]} V_t\subset X$. Moreover, since $\gamma([a,b])$ is compact, there exists a finite collection $\{V_{t_j}\}_{j=1}^{k}, k\in\mathbb{N}$ whose union contains $\gamma$. Lebesgue’s number lemma then ensures that there is $\delta \gt 0$ such that

\begin{align*} \vert t-t'\vert \lt \delta\quad\Longrightarrow\quad \gamma\vert_{[t,t']}\subset V_{t_j}\textrm{for some }t_j. \end{align*}

Now, take any partition $\sigma$ of $\gamma$ with modulus $\vert\sigma\vert \lt \delta$, and fix some $\widetilde{\varepsilon} \gt 0$. Then for all $i \in \{1,\dots,m-1\}$ there exists a causal curve $\eta_i$ from $\gamma(t_i)$ to $\gamma(t_{i+1})$ such that

\begin{equation*} L_\Omega^\tau (\eta_i)+\widetilde{\varepsilon}\geq \tau_{\Omega}\left(\gamma(t_i),\gamma(t_{i+1})\right). \end{equation*}

The endpoints of $\eta_i$ and $\gamma\vert_{[t_i,t_{i+1}]}$ are the same and are contained in $V_{t_j}$ for some $t_j\in[a,b]$, which (by quasi-strong causality) means $\eta_i\subset U_{t_j}$. Then,

\begin{equation*} \begin{aligned} V_{\Omega,\sigma}(\gamma) & = \sum_{i=0}^{m-1} \max_{t \in [t_i, t_{i+1}]} \Omega\left(\gamma(t)\right) \, \tau\big(\gamma(t_i), \gamma(t_{i+1})\big) \\ & \geq \sum_{i=0}^{m-1} \max_{t \in [t_i, t_{i+1}]} \Omega\big(\gamma(t)\big) \, L^{\tau}\big(\eta_i\big) \geq \sum_{i=0}^{m-1} \frac{\max_{t \in [t_i, t_{i+1}]}\Omega\big(\gamma(t)\big) }{\max_{s\in [t_i,t_{i+1}]} \Omega(\eta_i(s))}L_\Omega^\tau (\eta_i), \end{aligned} \end{equation*}

where in the last inequality we used the same estimate as in the proof of Proposition 3.6 to recover $L_\Omega^\tau $ from $L^\tau$. We now recall that for each $i \in \{1, \dots, m-1\}$ the $L^\tau_\Omega$-length of $\eta_i$ and the value of $\tau_\Omega$ at its endpoints differ by at most $\widetilde{\varepsilon}$, so that

\begin{equation*} V_{\Omega,\sigma}(\gamma) \geq \sum_{i=0}^{m-1} \frac{\max_{t \in [t_i, t_{i+1}]}\Omega\big(\gamma(t)\big) }{\max_{s\in [t_i,t_{i+1}]} \Omega(\eta_i(s))}\Big(\tau_{\Omega}\big(\gamma(t_i),\gamma(t_{i+1})\big)-\widetilde{\varepsilon}\Big). \end{equation*}

We then use the fact that for each summand, the curves are contained in $U_{t_j}$ and thus the oscillation of $\Omega$ is bounded by $2\varepsilon$; this allows us to replace the maximum term in the denominator with the same one in the numerator up to a $2\varepsilon$ contribution:

\begin{equation*} \begin{aligned} V_{\Omega,\sigma}(\gamma) &\geq \sum_{i=0}^{m-1} \frac{\max_{t \in [t_i, t_{i+1}]} \Omega \big(\gamma(t)\big) }{\max_{t \in [t_i, t_{i+1}]} \Omega\big(\gamma(t)\big)+2\varepsilon}\Big(\tau_{\Omega}\big(\gamma(t_i),\gamma(t_{i+1})\big)-\widetilde{\varepsilon}\Big)\\ &= \sum_{i=0}^{m-1} \left( 1 -\frac{2\varepsilon}{\max_{t \in [t_i, t_{i+1}]} \Omega \big(\gamma(t)\big)+2\varepsilon} \right) \Big(\tau_{\Omega}\big(\gamma(t_i),\gamma(t_{i+1})\big)-\widetilde{\varepsilon}\Big). \end{aligned} \end{equation*}

We now define $m:=\min_{t\in[a,b]}\Omega\big(\gamma(t)\big)$. Since nothing depends on $\eta_i$ in the inequality above, we can let $\widetilde{\varepsilon} \to0$ and find

\begin{equation*} \begin{aligned} V_{\Omega,\sigma}(\gamma) &\geq \sum_{i=0}^{m-1} \left( 1-\frac{2\varepsilon}{\max_{t \in [t_i, t_{i+1}]}\Omega\big(\gamma(t)\big)+2\varepsilon}\right) \tau_{\Omega}\big(\gamma(t_i),\gamma(t_{i+1})\big) \\ &\geq \left( 1-\frac{2\varepsilon}{m +2\varepsilon}\right) \sum_{i=0}^{m-1} \tau_{\Omega}\big(\gamma(t_i),\gamma(t_{i+1})\big) \geq \left( 1-\frac{2\varepsilon}{m +2\varepsilon}\right) L^{\tau_{\Omega}}\big(\gamma\big). \end{aligned} \end{equation*}

Taking the infimum over all partitions $\sigma$ yields $L_\Omega^\tau (\gamma)\geq \left( 1-\frac{2\varepsilon}{m +2\varepsilon}\right) L^{\tau_{\Omega}} (\gamma)$, which upon letting $\varepsilon\to0$ becomes $L_\Omega^\tau (\gamma)\geq L^{\tau_{\Omega}} (\gamma)$.

The intrinsicness of $\tau_\Omega$ is now immediate, as we proved that the length functionals $L^\tau_\Omega$ and $L^{\tau_\Omega}$ agree: since the causal relations $\leq$ and $\leq_\Omega$ coincide (as we discussed at the beginning of this section), the causal curves between any two given points are the same and thus taking the supremum with respect to either length will give the same value.

Proof. Let $\gamma:[a,b]\to X$ be any causal curve. We take $\varepsilon \gt 0$ and consider $\delta \gt 0$ small enough so that $\Omega\circ\gamma$, $\Omega'\circ \gamma$, and $\left(\Omega\cdot\Omega'\right)\circ \gamma$ are uniformly continuous with modulus of continuity $(\varepsilon, \delta)$. We also take $\varepsilon$ small enough so that $2\varepsilon$ is less than the minimum value of all the conformal factors along the curve. We take a partition $\sigma$ with modulus $\left\vert \sigma \right\vert \lt \delta$ and for each interval $[t_i, t_{i+1}]$ of the partition we define

\begin{align*} M_{i} &:= \max_{[t_i, t_{i+1}]} \Omega \circ \gamma, & M_{i}' &:= \max_{[t_i, t_{i+1}]} \Omega' \circ \gamma, & M_{i}^{\prime\prime} &:= \max_{[t_i, t_{i+1}]} (\Omega\cdot\Omega') \circ \gamma. & \end{align*}

Notice that $M_i^{\prime\prime}$ is not necessarily equal to the product of $M_i$ with $M_i'$, but just less than or equal; however, we can bound it from below by the product of the minimum of $\Omega$ over $[t_i, t_{i+1}]$ with $M_i'$, which in turn is bounded from below by $(M_i - \varepsilon)M_i'$ by uniform continuity. Therefore

(5)\begin{equation} M_iM_i' - \varepsilon M \leq (M_i - \varepsilon)M_i' \leq M_i^{\prime\prime} \leq M_i M_i' \implies \left\vert M_i^{\prime\prime} - M_i'M_i \right\vert \leq \varepsilon M, \end{equation}

where $M$ is the maximum of all of the conformal factors along the curve.

We now consider any partition $\theta=\{s_j\}_{j=0}^r$ of the interval $[t_i, t_{i+1}]$. It is then immediate to see, as done in Proposition 3.6, that

\begin{align*} (M_i - \varepsilon) \sum_{j=0}^{r-1} \tau(\gamma(s_j), \gamma(s_{j+1})) & \leq \sum_{j=0}^{r-1} \max_{s\in [s_j, s_{j+1}]} \Omega\left(\gamma(s)\right)\tau(\gamma(s_j), \gamma(s_{j+1})) \nonumber\\ & \quad \leq (M_i + \varepsilon) \sum_{j=0}^{r-1} \tau(\gamma(s_j), \gamma(s_{j+1})), \end{align*}

which means that

\begin{equation*} (M_i - \varepsilon)V_\theta({\gamma}\vert_{[t_i, t_{i+1}]}) \leq V_{\Omega, \theta}({\gamma}\vert_{[t_i, t_{i+1}]}) \leq (M_i + \varepsilon)V_\theta({\gamma}\vert_{[t_i, t_{i+1}]}), \end{equation*}

so that taking the infimum over all partitions $\theta$ yields

(6)\begin{equation} (M_i - \varepsilon) L^\tau({\gamma}\vert_{[t_i, t_{i+1}]}) \leq L_\Omega^\tau ({\gamma}\vert_{[t_i, t_{i+1}]}) \leq (M_i + \varepsilon) L^\tau({\gamma}\vert_{[t_i, t_{i+1}]}). \end{equation}

In a completely analogous way, we also get the inequalities

(7)\begin{equation} (M_i' - \varepsilon) L_\Omega^\tau ({\gamma}\vert_{[t_i, t_{i+1}]}) \leq L_{\Omega'}^{\tau_\Omega} ({\gamma}\vert_{[t_i, t_{i+1}]}) \leq (M_i' + \varepsilon) L_\Omega^\tau ({\gamma}\vert_{[t_i, t_{i+1}]}), \end{equation}

(8)\begin{equation} (M_i^{\prime\prime} - \varepsilon) L^\tau({\gamma}\vert_{[t_i, t_{i+1}]}) \leq L_{\Omega\cdot\Omega'}^\tau ({\gamma}\vert_{[t_i, t_{i+1}]}) \leq (M_i^{\prime\prime} + \varepsilon) L^{\tau}({\gamma}\vert_{[t_i, t_{i+1}]}), \end{equation}

where in (7) we have used that $L_\Omega^\tau =L^{\tau_{\Omega}}$ (by Lemma 3.12). Equations (6)-(7) can be rewritten as

\begin{equation*} \begin{aligned} \frac{L_\Omega^\tau ({\gamma}\vert_{[t_i, t_{i+1}]})}{M_{i} + \varepsilon} &\leq L^\tau({\gamma}\vert_{[t_i, t_{i+1}]}) \leq \frac{L_\Omega^\tau ({\gamma}\vert_{[t_i, t_{i+1}]})}{M_{i} - \varepsilon},\\ \frac{L_{\Omega'}^{\tau_\Omega}({\gamma}\vert_{[t_i, t_{i+1}]})}{M_{i}' + \varepsilon} &\leq L_\Omega^\tau ({\gamma}\vert_{[t_i, t_{i+1}]}) \leq \frac{L_{\Omega'}^{\tau_\Omega}({\gamma}\vert_{[t_i, t_{i+1}]})}{M_{i}' - \varepsilon}, \end{aligned} \end{equation*}

which can then be used in (8) to give the estimate

\begin{align*} \frac{M_i^{\prime\prime} - \varepsilon}{(M_i' + \varepsilon)(M_i + \varepsilon)} L_{\Omega'}^{\tau_\Omega}({\gamma}\vert_{[t_i, t_{i+1}]}) &\leq L_{\Omega\cdot\Omega'}^\tau({\gamma}\vert_{[t_i, t_{i+1}]}) \nonumber\\ & \quad\leq \frac{M_i^{\prime\prime} + \varepsilon}{(M_i' - \varepsilon)(M_i -\varepsilon)} L_{\Omega'}^{\tau_\Omega}({\gamma}\vert_{[t_i, t_{i+1}]}). \end{align*}

Thanks to (5) we can replace the denominators by $M_i^{\prime\prime} \pm O(\varepsilon)$, where $O(\varepsilon)$ does not depend on $i$. The fractions can then be simplified to

\begin{equation*} (1 - O(\varepsilon)) L_{\Omega'}^{\tau_\Omega}({\gamma}\vert_{[t_i, t_{i+1}]})\leq L_{\Omega\cdot\Omega'}^\tau ({\gamma}\vert_{[t_i, t_{i+1}]}) \leq (1+ O(\varepsilon))L_{\Omega'}^{\tau_\Omega}({\gamma}\vert_{[t_i, t_{i+1}]}). \end{equation*}

Summing over $i$ then gives

\begin{equation*} (1 - O(\varepsilon)) L_{\Omega'}^{\tau_\Omega}({\gamma})\leq L_{\Omega\cdot\Omega'}^{\tau} ({\gamma}) \leq (1+ O(\varepsilon))L_{\Omega'}^{\tau_\Omega}({\gamma}), \end{equation*}

which implies, letting $\varepsilon \to 0^+$, that $L_{\Omega\cdot\Omega'}(\gamma) = L_{\Omega'}^{\tau_\Omega}(\gamma)$, thus concluding the proof. The special case $L^{\tau_\Omega}_{1/\Omega}=L^\tau$ then follows immediately.

Proof. From Lemma 3.10 it follows

\begin{equation*} \Omega(p) = \lim_{n \to +\infty} \frac{\tau_{\Omega}(p,p_n)}{\tau(p,p_n)} = \lim_{n \to +\infty} \frac{\tau_{\Omega'}(p,p_n)}{\tau(p,p_n)} = \Omega'(p). \end{equation*}

Proof. Reflexivity immediately follows by taking $\iota= \operatorname{id}_X$ and $\Omega\equiv 1$. For symmetry, if $X \sim_{\iota, \Omega} \widetilde{X}$ we can set $j = \iota^{-1}$ and $\Omega' := \frac{1}{\Omega \circ j}$. It is then clear that $j$ is a homeomorphism which preserves the causal and timelike relations. Moreover, using $\iota^*\widetilde{\tau}=\tau_{\Omega}$ together with Proposition 3.13, we get $\tau = (\iota^*\widetilde{\tau})_{1/\Omega}$ and from the previous remark we then get $\tau = \iota^*(\widetilde{\tau}_{1/\Omega \circ j}) = \iota^*\widetilde{\tau}_{\Omega'}$. Then we see easily that

\begin{equation*} j^*\tau = j^*\big(\iota^*\widetilde{\tau}_{\Omega'}\big) = \widetilde{\tau}_{\Omega'}, \end{equation*}

which proves symmetry. Regarding transitivity, if $X \sim_{\iota, \Omega}X'$ and $X' \sim_{\iota', \Omega'}X^{\prime\prime}$ we define the map $\iota^{\prime\prime} := \iota' \circ \iota$ and the conformal factor $\Omega^{\prime\prime} := \Omega \cdot (\Omega' \circ \iota)$. Again, it is immediate to see that $\iota^{\prime\prime}$ is a homeomorphism that preserves the causal relations. Using Remark 3.19, we rewrite the identity $\iota^{\prime*}\tau^{\prime\prime} = \tau'_{\Omega'}$ as $\tau' = \iota^{\prime*}(\tau^{\prime\prime}_{1/\Omega' \circ {\iota^{\prime-1}}})$. This can be plugged in the identity $\iota^*\tau' = \tau_\Omega$ to obtain $\iota^*\iota^{\prime*} \big(\tau^{\prime\prime}_{1/\Omega' \circ \iota^{\prime-1}}\big) = \tau_\Omega$. On noticing that $i^* \circ i^{\prime*} = i^{\prime\prime*}$ we get

\begin{equation*} \tau_\Omega = i^{\prime\prime*}(\tau^{\prime\prime}_{1/\Omega^{\prime} \circ \iota^{\prime-1}}) = (\iota^{\prime\prime*} \tau^{\prime\prime})_{1/\Omega^{\prime} \circ \iota^{\prime-1} \circ \iota^{\prime\prime}} = (\iota^{\prime\prime*}\tau^{\prime\prime})_{1/\Omega^{\prime} \circ \iota}. \end{equation*}

Applying Proposition 3.13 we see that $ \iota^{\prime\prime*}\tau'' = (\tau_\Omega)_{\Omega' \circ \iota} = \tau_{\Omega \cdot \Omega' \circ \iota} = \tau_{\Omega^{\prime\prime}}, $ which concludes the proof.

Proof. Consider the neighbourhood $U$ of $p$ given by the locally-causally-closed condition. Without loss of generality, since both $\alpha$ and $\beta$ are continuous, we can assume that the image of both of the curves is contained in $U$; otherwise, we just work with the restriction of the curves to an interval where both of them are in $U$.

Fix any $k \in \mathbb{N}$ and consider the point $q_k=\beta(b/k)$. Since the curve $\beta$ is timelike, we have that $p \in I^-(q_k)$, which is an open set. By continuity of $\alpha$ then there exists some time $s_k \in (0,b/k)$ such that $a_k := \alpha(s_k) \in I^-(q_k)$. Equivalently, we have that $q_k \in I^+(a_k)$. Since $p \ll a_k$, we have that $p \not \in I^+(a_k)$ as the space is chronological. We can then define

\begin{equation*} t_k := \sup \left\{t \in [0,b/k] \ : \ \beta(t) \not \in I^+(a_k) \right\}\,. \end{equation*}

Notice that $t_k \in (0,b/k)$: firstly, since $p \neq a_k$, we can take a small neighbourhood of $p$ which is disjoint from $a_k$ and contained in $I^-(a_k)$. Using continuity of $\beta$ at 0, the curve stays in the neighbourhood for a positive time and thus as the space is chronological none of these points belong to $I^+(a_k)$, hence $t_k \gt 0$. Analogously, since $\beta(b/k) = q_k \in I^+(a_k)$, which is an open set, by continuity the curve $\beta$ must stay in such neighbourhood for all times sufficiently close to $b/k$ and thus $t_k \lt b/k$. By the same continuity argument, along with the fact that $t_k$ is defined as a supremum, shows that $b_k := \beta(t_k) \not \in I^+(a_k) $. By local causal closedness, since for every $t \in (t_k, b/k)$ we have $a_k \ll \beta(t)$, we get that $a_k \leq b_k$. Since $a_k \not \ll b_k$, we have that $\tau(a(s_k), b(t_k)) = 0$.

To conclude, notice that both $t_k$ and $s_k$ are contained in $(0,b/k)$, hence the sequences $\{t_k\}_k$ and $\{s_k\}_k$ converge to 0 as $k \to +\infty$; up to extracting subsequences, we may assume that both sequences are decreasing.

Proof. First, note that the timelike and causal relations and the time orientation of the curves are unchanged. Without loss of generality, we assume $\alpha, \beta$ to be future directed (the past directed case works completely analogous), hence the sign $\sigma$ of the angle is $-1$.

By assumption, we have

(9)\begin{equation} \measuredangle_p(\alpha,\beta) = \lim_{(s,t)\in A_0;\,s,t\searrow 0} \tilde\measuredangle_p(\alpha(s),\beta(t))\,, \end{equation}

(10)\begin{equation} \measuredangle_p^\Omega(\alpha,\beta) = \lim_{(s,t)\in A_0;\,s,t\searrow 0} \tilde\measuredangle_p^\Omega(\alpha(s),\beta(t))\,, \end{equation}

where $A_0:=\{(s,t)\in(0,b)^2: \alpha(s)\leq\beta(t)$ or $\beta(t)\leq\alpha(s)\}$ and $\tilde\measuredangle_p^\Omega(\alpha(s),\beta(t))$ is the comparison angle of the triangle with side lengths $\tau_\Omega(p,\alpha(s))$, $\tau_\Omega(p,\beta(t))$ and $\max(\tau_\Omega(\alpha(s),\beta(t)),\tau_\Omega(\beta(t),\alpha(s)))$. By finiteness of $\tau$ and the hyperbolic law of cosines (see e.g. [Reference Beran and Sämann6, Lem. 2.4]) we have for $(s,t)\in A_0$ that

\begin{align*} -\sigma\cosh(\tilde\measuredangle_p(\alpha(s),\beta(t)))= \frac{a^2 + b^2 - c^2}{2 a b} &= \frac{1}{2} \frac{a}{b} + \frac{1}{2} \frac{b-c}{a} \left(1+\frac{c}{b}\right) \,, \end{align*}

where $a,b,c$ are the side lengths with respect to $\tau$ of the timelike triangle $(p,\alpha(s),\beta(t))$.

Let $(s_k, t_k)_k$ be a sequence given by Lemma 3.21, hence

\begin{align*} \measuredangle_p(\alpha,\beta) = \lim_{k\to\infty} \tilde\measuredangle_p(\alpha(s_k),\beta(t_k))\,. \end{align*}

We write $a_k:=\tau(p,\alpha(s_k))$, $b_k:=\tau(p,\beta(t_k))$, $x_k:= \frac{a_k}{b_k}$, $y_k:= \frac{b_k}{a_k}=\frac{1}{x_k}$ and notice that $c_k := \tau(a_k,b_k)$, which vanishes by Lemma 3.21. Hence, $\cosh(\measuredangle_p(\alpha,\beta))=\lim_{k\to\infty}\Bigl(\frac{1}{2}x_k + \frac{1}{2} y_k\Bigr)$ with $x_k,y_k\geq 0$. We also denote by $\tilde a_k, \tilde b_k, \tilde c_k, \tilde x_k, \tilde y_k$ the corresponding side-lengths and their fractions with respect to $\tau_\Omega$; notice that $\tilde{c}_k = \tau_\Omega(\tilde a_k, \tilde b_k) = 0$ by Lemma 3.10.

By construction we have for all $k\in\mathbb{N}$ that $a_k\leq b_k$, hence $x_k = \frac{a_k}{b_k} \leq 1$. By going over to further subsequences, this allows us to assume without loss of generality that $x_k\to x\in[0,1]$ and hence $y_k = \frac{1}{x_k} \to \frac{1}{x}=:y$, with the understanding that $\frac{1}{0}=+\infty$. At this point we observe that since everything is finite and positive we can multiply and divide to get

(11)\begin{equation} \tilde x_k = \frac{\tilde a_k}{\tilde b_k}= \frac{a_k \frac{\tilde a_k}{a_k}}{b_k\frac{\tilde b_k}{b_k}} = x_k \frac{\tilde a_k}{a_k}\frac{b_k}{\tilde b_k}\,, \end{equation}

and use that $\frac{\tilde a_k}{a_k}\to \Omega(p)$ and $\frac{b_k}{\tilde b_k}\to \frac{1}{\Omega(p)}$ by Lemma 3.10. Thus, we have $\tilde x_k\to x$ and analogously that $\tilde y_k \to y$.

Finally, this allows us to conclude that

\begin{align*} \measuredangle_p^\Omega(\alpha,\beta) = \lim_{k\to\infty} \Bigl(\frac{1}{2} \tilde x_k + \frac{1}{2} \tilde y_k \Bigr) = \measuredangle_p(\alpha,\beta)\,. \end{align*}

Proof. The proof is analogous to the one in [Reference Kunzinger and Sämann38, Prop. 2.32]. Since we are assuming (quasi-)strong causality, the notions of causal curve induced by the Lorentzian pre-length structure and the one induced by the smooth structure coincide [Reference Kunzinger and Sämann38, Lemma 2.21]. For any partition $\sigma = \{t_i\}_{i=0}^m$ of the interval $[a,b]$, by additivity of $L_{\widetilde{g}}$ we have (recall (1))

\begin{align*} L_{\widetilde{g}}(\gamma) &= \sum_{i=0}^{m-1} \int_{t_i}^{t_{i+1}} \Omega(\gamma(t)) \sqrt{-g(\dot{\gamma}(t), \dot{\gamma}(t))} \, dt \nonumber\\ &\quad \leq \sum_{i=0}^{m-1} \max_{t \in [t_i, t_{i+1}]} \Omega(\gamma(t)) \int_{t_i}^{t_{i+1}} \ \sqrt{-g(\dot{\gamma}(t), \dot{\gamma}(t))} \, dt \\ &= \sum_{i=0}^{m-1} \max_{t \in [t_i, t_{i+1}]} \Omega(\gamma(t)) L_g(\gamma\vert_{[t_i,t_{i+1}]}) \nonumber\\ &\quad\leq \sum_{i=0}^{m-1} \max_{t \in [t_i, t_{i+1}]} \Omega(\gamma(t)) \, \tau(\gamma(t_i), \gamma(t_{i+1})) = V_{\Omega,\sigma}(\gamma), \end{align*}

where for the last inequality we have used that $L_g(\gamma\vert_{[t_i, t_{i+1}]}) \leq \tau(\gamma(t_i), \gamma(t_{i+1}))$ for any $i \in \{0, \dots, m-1\}$ (cf. (2)). Taking then the infimum over all partitions $\sigma$ immediately yields that $L_{\widetilde{g}}(\gamma) \leq L_\Omega^\tau (\gamma)$. Now, at any point of differentiability of $\gamma$, it holds [Reference Kunzinger and Sämann38, Prop. 2.32]

\begin{equation*} \lim_{h \to0^+} \frac{\tau(\gamma(t), \gamma(t+h))}{h} = \left\vert \dot{\gamma}(t) \right\vert_{g}. \end{equation*}

By continuity of $\Omega \circ \gamma$ it then follows that

\begin{equation*} \lim_{h \to0^+} \max_{s \in[t,t+h]} \Omega(\gamma(s)) \,\frac{\tau(\gamma(t), \gamma(t+h))}{h} = \Omega(\gamma(t)) \, \left\vert \dot{\gamma}(t) \right\vert_{g} = \left\vert \dot{\gamma} \right\vert_{\widetilde{g}}. \end{equation*}

We now notice that $ \max_{s \in[t,t+h]} \Omega(\gamma(s)) \, \tau(\gamma(t), \gamma(t+h)) \geq L_\Omega^\tau (\gamma\vert_{[t,t+h]}), $ since the left hand side is just $V_{\Omega,\sigma}(\gamma\vert_{[t,t+h]})$ for $\sigma = \{t,t+h\}$. This, together with the previous estimate $L_\Omega^\tau (\gamma) \geq L_{\widetilde{g}}(\gamma)$, gives

\begin{align*} &\max_{s \in[t,t+h]} \Omega(\gamma(s)) \,\frac{\tau(\gamma(t), \gamma(t+h))}{h} \geq \frac{L^\tau_{\Omega}(\gamma\vert_{[t,t+h]})}{h} \geq \frac{L_{\widetilde{g}}(\gamma\vert_{[t,t+h]})}{h} \nonumber\\ &\quad = \frac{1}{h}\int_t^{t+h} \left\vert \dot{\gamma}(s) \right\vert_{\widetilde{g}} \, ds. \end{align*}

Both the initial and the last term of this chain of inequalities converge to $\left\vert \dot{\gamma}(t) \right\vert_{\widetilde{g}}$ as $h \to 0^+$, therefore

\begin{equation*} \lim_{h\to 0^+} \frac{L_\Omega^\tau (\gamma\vert_{[t,t+h]})}{h} = \left\vert \dot{\gamma}(t) \right\vert_{\widetilde{g}}. \end{equation*}

Thus, the function $f(t) := L_\Omega^\tau (\gamma\vert_{[a,t]})$ admits right derivatives almost everywhere. Indeed, for almost every $t \in [a,b]$ we have

\begin{equation*} \lim_{h \to 0^+} \frac{L_\Omega^\tau (\gamma\vert_{[a,t+h]}) - L_\Omega^\tau (\gamma\vert_{[a,t]})}{h} = \lim_{h \to 0^+} \frac{L_\Omega^\tau (\gamma\vert_{[t,t+h]})}{h} = \left\vert \dot{\gamma}(t) \right\vert_{\widetilde{g}}, \end{equation*}

where in the first equality we used additivity of the length. Notice that the function $q(t) := L_{\widetilde{g}}(\gamma\vert_{[a,t]})$ is absolutely continuous (as the primitive of a continuous function) and satisfies $f(0)=q(0)=0$. To conclude, we prove that $f$ is absolutely continuous, so that we may apply the Fundamental Theorem of Calculus. First we note that, for any subinterval $[a_i,b_i]$ in $[a,b]$, $L_\Omega^\tau (\gamma\vert_{[a_i,b_i]}) \leq M \cdot L_{g}(\gamma\vert_{[a_i,b_i]})$, where $M \gt 0$ is the maximum of $\Omega \circ \gamma$ over the whole interval $[a,b]$. Indeed, for any partition $\sigma$ of the interval $[a_i,b_i]$, one gets

\begin{align*} V_{\Omega,\sigma}(\gamma\vert_{[a_i,b_i]}) & = \sum_{j=0}^{m-1} \max_{t \in [t_j, t_{j+1}]} \Omega(\gamma(t)) \, \tau(\gamma(t_j), \gamma(t_{j+1})) \leq M \sum_{j=0}^{m-1} \tau(\gamma(t_j), \gamma(t_{j+1})) \nonumber\\ &= M\cdot V_\sigma(\gamma\vert_{[a_i,b_i]}), \end{align*}

so that taking the infimum over $\sigma$ yields $L_\Omega^\tau (\gamma\vert_{[a_i,b_i]}) \leq M L_{g}(\gamma\vert_{[a_i,b_i]})$. On the other hand, the function $h(t) := L_g(\gamma\vert_{[a,t]})$ is absolutely continuous, hence for any $\varepsilon \gt 0$ there exists $\delta \gt 0$ such that, for any disjoint family $\{[a_i,b_i]\}_{i=0}^m$ of intervals in $[a,b]$, it holds

\begin{equation*} \sum_{i=0}^m \left\vert b_i-a_i \right\vert \lt \delta \implies \sum_{i=0}^m \left\vert h(b_i)-h(a_i) \right\vert \lt \frac{\varepsilon}{M}. \end{equation*}

For any such family $\{[a_i,b_i]\}_{i=0}^m$, we get

\begin{equation*} \begin{aligned} \sum_{i=0}^m \left\vert f(b_i)-f(a_i) \right\vert &= \sum_{i=0}^m \left\vert L_\Omega^\tau (\gamma\vert_{[a,b_i]}) - L_{\Omega}(\gamma\vert_{[a,a_i]}) \right\vert = \sum_{i=0}^m L_\Omega^\tau (\gamma\vert_{[a_i,b_i]}) \\ &\leq \sum_{i=0}^m M \, L_{g}({\gamma}\vert_{[a_i,b_i]}) =M \sum_{i=0}^m \left\vert h(b_i)-h(a_i) \right\vert \lt \varepsilon. \end{aligned} \end{equation*}

Therefore, $f$ is absolutely continuous; in particular, it is differentiable almost everywhere, so its derivative is equal to its right-sided derivative, i.e. $\left\vert \dot{\gamma}(t) \right\vert_{\widetilde{g}}$. Applying the Fundamental Theorem of Calculus then gives

\begin{equation*} L_\Omega^\tau (\gamma) = f(b) = \int_a^b \frac{d}{dt} f(t) \, dt = \int_a^b \left\vert \dot{\gamma}(t) \right\vert_{\widetilde{g}} \, dt = \int_a^b \frac{d}{dt} q(t) \, dt = q(b) = L_{\widetilde{g}}(\gamma), \end{equation*}

which proves the theorem.

Proof. The proof is straightforward as the conformally transformed space has the same causal and timelike relations as the starting space by Proposition 3.7. Moreover, the conditions depend only on the topology of the space, the metric and on the causal structure. Since conformal changes do not alter any of these, it immediately follows that they also hold for the final space.

Proof. The proof of points (i), (ii), (iv), (vi)–(viii), (*) is the same as in the proof of Proposition 4.4. For strong causality we observe that $\iota(x)\,\widetilde\ll\,\iota(y)$ if and only if $x\ll y$, hence $I(\iota(x),\iota(y)) = \iota(I(x,y))$ for all $x,y\in X$. Moreover, as $\iota$ is a homeomorphism, it is immediate to see that it maps subbases of the starting space to subbases of the target. Hence, if we assume that $X$ is strongly causal, we immediately have that the chronological diamonds of $\widetilde{X}$, being the image of the chronological diamonds of $X$, make a subbase of the topology of $\widetilde{X}$ and thus $\widetilde{X}$ is strongly causal.

Finally, for Lorentzian length spaces global hyperbolicity does not depend on the metric inducing the topology, hence is conformally invariant, see [Reference Minguzzi47, Thm. 3.7, Cor. 3.8].

Proof. We begin by showing ‘ $\implies$’. Let us take any pair of points $p \leq q$ and consider the diamond $J(p,q)$. As the space is globally hyperbolic, this diamond is compact. We consider any conformal factor $\Omega$ and denote by $M$ the maximum value it attains on the diamond, which exists by compactness along with continuity of $\Omega$. For any causal curve $\gamma$ joining $p$ to $q$, which will then be contained in $J(p,q)$, we can estimate its $L_\Omega^\tau $-length analogously to what we did in Proposition 3.6 by

\begin{equation*} L_\Omega^\tau (\gamma) \leq M L^\tau(\gamma) \leq M \tau(p,q). \end{equation*}

Notice that here $M$ is independent of the curve, so by taking the supremum over all such curves $\gamma$ we see that

\begin{equation*} \tau_\Omega(p,q) \leq M \tau(p,q) \lt +\infty, \end{equation*}

as we assumed our initial time separation function to be finite.

For ‘ $\impliedby$’, we proceed by contradiction. Suppose that in the initial space there exist two points $p \leq q$ such that their diamond $J(p,q)$ is not compact. Then there must exist a sequence of points $\{p_n\}_n$ in $J(p,q)$ such that none of its subsequences converges; without loss of generality, we can assume that the sequence is injective, so that $n \neq m \implies p_n \neq p_m$. First, we take a point $p^- \in I^-(p)$, so that by the push-up property of Lorentzian pre-length spaces we have $p^- \ll p_n$ for all $n \in \mathbb{N}$. We see that for every $n \in \mathbb{N}$, we must have

\begin{equation*} r(n) := \inf_{m \neq n} d(p_n,p_m) \gt 0, \end{equation*}

as otherwise we would be able to extract a subsequence converging to $p_n$. We now see that the balls defined as

\begin{equation*} B_n := B_{d}\left( p_n, \frac{r(n)}{2} \right) \end{equation*}

must be mutually disjoint; indeed, if $\exists z \in B_n \cap B_m$ for $n \neq m$ by the triangle inequality we would have

\begin{equation*} d(p_n,p_m) \leq d(p_n,z) + d(z,p_m) \lt \frac{r(n)}{2} + \frac{r(m)}{2} \leq \max(r(n),r(m)), \end{equation*}

but we also know that $d(p_n,p_m)$ is greater or equal than both $r(n)$ and $r(m)$, therefore we would get $d(p_n,p_m) \geq \max(r(n),r(m))$, which contradicts the previous chain of inequalities as one of them is a strict inequality.

The aim now is to build a conformal factor which takes high values inside each $B_n$ but is constant outside of them. To this end, for any $B_n$ we look at a smaller ball $U_n := B(p_n, r(n)/4)$ and employ quasi-strong causality to get the existence of a smaller neighbourhood $V_n \subset U_n$ such that any causal curve with endpoints in $V_n$ stays in $U_n$. By causal path connectedness, as $p^- \ll p_n$, there exists a timelike curve $\gamma_n$ joining $p^-$ to $p_n$. By continuity, we can consider a point $s_n$ on this curve which is contained in $V_n$; as $\gamma_n$ is timelike, $s_n \ll p_n$, so in particular $\tau(s_n,p_n) \gt 0$. Since $\tau$ is intrinsic, we can find a causal curve $\sigma_n$ joining $s_n$ to $p_n$ with positive length $L^\tau(\sigma_n)$, and by quasi-strong causality this curve is entirely contained in $U_n$. The concatenation of the curve $\gamma_n$ until the point $s_n$ and the curve $\sigma_n$ produces a curve $\eta_n$ from $p^-$ to $p_n$ which has positive length in a neighbourhood of the point $p_n$.

We define the map $\Omega_n : X \to (0, +\infty)$ by

\begin{equation*} \Omega_n(x) := \frac{2n}{r(n)L^\tau(\sigma_n)}\operatorname{dist}(x, (B_n)^c) + 1 \end{equation*}

and notice that this map satisfies the following properties:

• • $\Omega_n$ is continuous and in fact Lipschitz with constant equal to $2n/(r(n)L^\tau(\sigma_n))$;

• • $\Omega_n \equiv 1$ outside of $B_n$;

• • $\Omega_n(x) \geq \frac{n}{L^n(\sigma_n)} +1$ for every $x \in U_n$.

We then define the map $\Omega = \prod_{n \in \mathbb{N}} \Omega_n$ and see that such product is well defined since at every point at most only one of the factors is not equal to 1; in fact, such a product is continuous. Moreover, for any $n\in \mathbb{N}$ it holds that $\Omega \geq \Omega_n$, so that measuring the length of the curve $\eta_n$ yields

\begin{equation*} \tau_\Omega(p^-,p_n) \geq L_\Omega^\tau (\eta_n) \geq L_\Omega^\tau (\sigma_n) \geq L^\tau_{\Omega_n}(\sigma_n). \end{equation*}

As $\sigma_n \subset U_n$, we estimate the last term in the same way as in the proof of Proposition 3.6 by bounding $\Omega_n$ from below with its minimum over $U_n$, which we know to be greater than $\frac{n}{L(\sigma_n)} +1$:

\begin{equation*} L^\tau_{\Omega_n}(\sigma_n) \geq L^\tau(\sigma_n) \left( \frac{n}{L(\sigma_n)} +1 \right) \geq n. \end{equation*}

This shows that

\begin{equation*} \tau_\Omega(p^-, p_n) \geq n. \end{equation*}

As $p_n \ll q$, we see that

\begin{equation*} \tau_\Omega(p^-,q) \geq \tau(p^-,p_n) + \tau(p_n, q) \geq n, \end{equation*}

and since this inequality holds for every $n \in \mathbb{N}$, we must have $\tau_\Omega(p^-,q) = + \infty$, which shows that for this choice of conformal factor the time separation function is not finite and therefore contradicts our assumption.

Proof. Suppose that $\tau_{\Omega}$ is finite for any choice of conformal factor. Then by Theorem 4.6 all diamonds in $(X,d,\ll,\leq,\tau) $ are compact. Since $(X,d,\ll,\leq,\tau) $ is by assumption non-totally imprisoning, it is globally hyperbolic. On the other hand, if $(X,d,\ll,\leq,\tau) $ is globally hyperbolic, then all its diamonds are compact, and again by Theorem 4.6 we have that $\tau_{\Omega}$ is finite for any choice of conformal factor.