Proof. That $d_{\mathbb{K}}$ is non-negative and symmetric is obvious. To prove that it is also finite, for any choice of $(K,\pi ^K,X^K,L^K)$ , $(K',\pi ^{K'},X^{K'},L^{K'})$ consider the disjoint union $Z=K\sqcup K'$ of $K$ and $K'$ . Then, set $d_{Z}(x,x')\;:\!=\;\textrm{diam}_{K}(K)+\textrm{diam}_{K'}(K')$ , for any $x\in K$ , $x'\in K'$ , where

\begin{equation*} \textrm {diam}_{K}(K)=\sup _{y,z\in K} d_K(y,z) \end{equation*}

denotes the diameter of $K$ with respect to the metric $d_K$ . Since $K$ and $K'$ are compact their diameters are finite. Therefore, $d_Z$ is finite for any $x\in K$ , $x'\in K'$ . To conclude that $d_{\mathbb{K}}$ is finite, simply suppose that $\mathcal{C}=K\times K'$ .

Next, we show that $d_{\mathbb{K}}$ is positive-definite. Let $(K,\pi ^K,X^K,L^K), (K',\pi ^{K'},X^{K'},L^{K'})$ be in $\mathbb{K}$ such that $d_{\mathbb{K}}((K,\pi ^K,X^K,L^K),(K',\pi ^{K'},X^{K'},L^{K'}))=0$ . Then, for every $\varepsilon \gt 0$ there exist $Z,\phi,\phi ',\mathcal{C}$ such that the sum of the quantities inside the infimum in the definition of $d_{\mathbb{K}}$ is bounded above by $\varepsilon$ . Furthermore, there exists $\lambda _{\varepsilon }\in \Lambda$ such that the sum of the quantities inside the infimum in the definition of $d_{J_1}^Z$ is bounded above by $2 \varepsilon$ . Recall that for every $t\in [0,T]$ , $L^K_t\;:\;K\to \mathbb{R}_{+}$ is a continuous function and since $K$ is a compact metric space, then it is also uniformly continuous. Therefore, there exists a $\delta \in (0,\varepsilon ]$ such that

(2.1) \begin{equation} \sup _{\substack{x_1,x_2\in K:\\ d_{K}(x_1,x_2)\lt \delta }} \sup _{t\in [0,T]} |L_t^K(x_1)-L_t^K(x_2)|\le \varepsilon . \end{equation}

Now, let $(x_i)_{i\ge 1}$ be a dense sequence of disjoint elements in $K$ . Since $K$ is compact, there exists an integer $N_{\varepsilon }$ such that the collection of open balls $(B_{K}(x_i,\delta ))_{i=1}^{N_{\varepsilon }}$ covers $K$ . Defining $A_1=B_{K}(x_1,\delta )$ and $A_i=B_{K}(x_i,\delta )\setminus \cup _{j=1}^{i-1} B_{K}(x_j,\delta )$ , for $i=2,\ldots,N_{\varepsilon }$ , we have that $(A_i)_{i=1}^{N_{\varepsilon }}$ is a disjoint cover of $K$ . Consider a function $f_{\varepsilon }\;:\;K\rightarrow K'$ by setting

\begin{equation*} f_{\varepsilon }(x)\;:\!=\;x'_{\!\!i} \end{equation*}

on $A_i$ , where $x'_{\!\!i}$ is chosen such that $(x_i,x'_{\!\!i})\in \mathcal{C}$ , for $i=1,\ldots,N_{\varepsilon }$ . Note that by definition $f_{\varepsilon }$ is a measurable function defined on $K$ . For any $x\in K$ , such that $x\in A_i$ for some $i=1,\ldots,N_{\varepsilon }$ , we have that

(2.2) \begin{align} d_{Z}(\phi (x),\phi '(f_{\varepsilon }(x)))&=d_{Z}(\phi (x),\phi '(x'_{\!\!i})) \nonumber \\[5pt] &\le d_{Z}(\phi (x),\phi (x_i))+d_{Z}(\phi (x_i),\phi '(x'_{\!\!i}))\le \delta +\varepsilon \le 2 \varepsilon . \end{align}

From (2.2), it follows that for any $x\in K$ and $y\in K$

\begin{align*} |d_{Z}(\phi (x),\phi (y))-d_{Z}(\phi '(f_{\varepsilon }(x)),\phi '(f_{\varepsilon }(y))|&\le d_{Z}(\phi (y),\phi '(f_{\varepsilon }(y)))+d_{Z}(\phi (x),\phi '(f_{\varepsilon }(x))) \\[5pt] &\le 2 \varepsilon +2 \varepsilon =4 \varepsilon . \end{align*}

This immediately yields

(2.3) \begin{equation} \sup _{x,y\in K} |d_{K}(x,y)-d_{K'}(f_{\varepsilon }(x),f_{\varepsilon }(y))|\le 4 \varepsilon . \end{equation}

From (2.3), we deduce the bound

(2.4) \begin{equation} d_{P}^{K'}(\pi ^{K}\circ f_{\varepsilon }^{-1},\pi ^{K'})\le 5 \varepsilon \end{equation}

for the Prokhorov distance between $\pi ^K\circ f_{\varepsilon }^{-1}$ and $\pi ^{K'}$ in $K'$ . Using (2.1) and the fact that the last quantity inside the infimum in the definition of $d_{\mathbb{K}}$ is bounded above by $\varepsilon$ , we deduce

(2.5) \begin{equation} \sup _{x\in K,t\in [0,T]} |L_t^K(x)-L_t^{K'}(f_{\varepsilon }(x))|\le 2 \varepsilon . \end{equation}

Using (2.2) and the fact that the second quantity in the infimum is bounded above by $\varepsilon$ , we deduce that for any $t\in [0,T]$

\begin{align*} d_{Z}(\phi '(f_{\varepsilon }(X^K_{\lambda _{\varepsilon }(t)})),\phi '(X_t^{K'}))&\le d_{Z}(\phi '(f_{\varepsilon }(X^K_{\lambda _{\varepsilon }(t)})),\phi (X^K_{\lambda _{\varepsilon }(t)}))+d_{Z}(\phi (X^K_{\lambda _{\varepsilon }(t)}),\phi '(X_t^{K'})) \\[5pt] &\le 2 \varepsilon +2 \varepsilon =4 \varepsilon . \end{align*}

Therefore,

(2.6) \begin{equation} \sup _{t\in [0,T]} d_{K'}(f_{\varepsilon }(X_{\lambda _{\varepsilon }(t)}^K),X_t^{K'})\le 4 \varepsilon . \end{equation}

Using a diagonalization argument we can find a sequence $(\varepsilon _n)_{n\ge 1}$ such that $f_{\varepsilon _n}(x_i)$ converges to some limit $f(x_i)\in K'$ , for every $i\ge 1$ . From (2.3) we immediately get that $d_{K}(x_i,x_j)=d_{K'}(f(x_i),f(x_j))$ , for every $i,j\ge 1$ . By [Reference Burago, Burago and Ivanov12, Proposition 1.5.9], this map can be extended continuously to the whole $K$ . This shows that $f$ is distance-preserving. Reversing the roles of $K$ and $K'$ , we are able to find also a distance-preserving map from $K'$ to $K$ . Hence $f$ is an isometry. We are now able to check that $\pi ^K\circ f^{-1}=\pi ^{K'}$ , $L_t^{K'}\circ f=L_t^K$ , for all $t\in [0,T]$ , and $f\circ X^K=X^{K'}$ . Since $f_{\varepsilon _n}(x_i)$ converges to $f(x_i)$ in $K'$ , we can find $\varepsilon '\in (0,\varepsilon ]$ such that $d_{K'}(f_{\varepsilon '}(x_i),f(x_i))\le \varepsilon$ , for $i=1,\ldots,N_{\varepsilon }$ . Recall that $(x_i)_{i=1}^{N_{\varepsilon }}$ is an $\varepsilon$ -net in $K$ . Then, for $i=1,\ldots,N_{\varepsilon }$ , such that $x\in A_i$ , using (2.3) and the fact that $f$ is an isometry, we deduce

(2.7) \begin{equation} d_{K'}(f_{\varepsilon '}(x),f(x))\le d_{K'}(f_{\varepsilon '}(x),f_{\varepsilon '}(x_i))+d_{K'}(f_{\varepsilon '}(x_i),f(x_i))+d_{K'}(f(x_i),f(x))\le 7 \varepsilon . \end{equation}

This, combined with (2.4) implies

\begin{equation*} d_{P}^{K'}(\pi ^K\circ f^{-1},\pi ^{K'})\le d_{P}^{K'}(\pi ^K\circ f^{-1},\pi ^{K}\circ f_{\varepsilon '}^{-1})+d_{P}^{K'}(\pi ^K\circ f_{\varepsilon '}^{-1},\pi ^{K'})\le 12 \varepsilon . \end{equation*}

Since $\varepsilon \gt 0$ was arbitrary, $\pi ^{K}\circ f^{-1}=\pi ^{K'}$ . Moreover, from (2.5) and (2.7) we have that

\begin{align*} &\sup _{x\in K,t\in [0,T]} |L_t^K(x)-L_t^{K'}(f(x))| \\[5pt] &\le \sup _{x\in K,t\in [0,T]} |L_t^{K}(x)-L_t^{K'}(f_{\varepsilon '}(x))|+\sup _{x\in K,t\in [0,T]} |L_t^{K'}(f_{\varepsilon '}(x))-L_t^{K'}(f(x))| \\[5pt] &\le 2 \varepsilon +\sup _{\substack{x_1',x_2'\in K':\\ d_{K'}(x_1',x_2')\le 7\varepsilon }} \sup _{t\in [0,T]} |L_t^{K'}(x_1')-L_t^{K'}(x_2')|. \end{align*}

Now, this and the uniform continuity of $L^{K'}$ (replace $L^K$ by $L^{K'}$ in (2.1)) gives $L_t^{K'}\circ f=L_t^K$ , for all $t\in [0,T]$ . Finally, we verify that $f\circ X^K=X^{K'}$ . For any $t\in [0,T]$

\begin{equation*} d_{K'}(f(X_{\lambda _{\varepsilon }(t)}^K),X_t^{K'})\le d_{K'}(f(X_{\lambda _{\varepsilon }(t)}^K),f_{\varepsilon '}(X_{\lambda _{\varepsilon }(t)}^K))+d_{K'}(f_{\varepsilon '}(X_{\lambda _{\varepsilon }(t)}^K),X_t^{K'})\le 7 \varepsilon +4 \varepsilon =11 \varepsilon, \end{equation*}

where we used (2.6) and (2.7). Therefore,

(2.8) \begin{equation} \sup _{t\in [0,T]} d_{K'}(f(X_{\lambda _{\varepsilon }(t)}^K),X_t^{K'})\le 11 \varepsilon . \end{equation}

Recall that $\sup _{t\in [0,T]} |\lambda _{\varepsilon }(t)-t|\le 2 \varepsilon$ . From this and (2.8), it follows that for every $t\in [0,T]$ , there exists a sequence $(z_n)_{n\ge 1}$ , such that $z_n\rightarrow t$ and $d_{K'}(f(X_{z_n}^{K}),X_t^{K'})\rightarrow 0$ , as $n\rightarrow \infty$ . If $t$ is a continuity point of $f\circ X^K$ , then $d_{K'}(f(X_{z_n}^{K}),f(X_t^K))\rightarrow 0$ , as $n\rightarrow \infty$ . Thus, $f(X_t^K)=X_t^{K'}$ . If $f\circ X^K$ has a jump at $t$ and $(z_n)_{n\ge 1}$ has a subsequence $(z_{n_k})_{k\ge 1}$ , such that $z_{n_k}\ge t$ , for any $k\ge 1$ , then $d_{K'}(f(X_{z_{n_k}}^{K}),X_t^{K'})\rightarrow 0$ , as $n\rightarrow \infty$ , and $d_{K'}(f(X_{z_{n_k}}^{K}),f(X_t^K))\rightarrow 0$ , as $n\rightarrow \infty$ . Therefore, $f(X_t^K)=X_t^{K'}$ . Otherwise, $z_n\lt t$ , for $n$ large enough and $d_{K'}(f(X_{z_n}^{K}),f(X_{t-}^{K}))\rightarrow 0$ , as $n\rightarrow \infty$ , which implies $f(X_{t-}^{K})=X_t^{K'}$ . Essentially, what we have proved is that if $f\circ X^K$ has a jump, then either $f(X^K_t)=X_t^{K'}$ or $f(X^K_{t-})=X_t^{K'}$ . But, since $X^{K'}$ is càdlàg, $f\circ X^K=X^{K'}$ . This completes the proof that the quadruples $(K,\pi ^K,X^K,L^K)$ and $(K',\pi ^{K'},X^{K'},L^{K'})$ are equivalent in $(\mathbb{K},d_{\mathbb{K}})$ , and consequently that $d_{\mathbb{K}}$ is positive-definite.

For the triangle inequality we follow the proof of [Reference Burago, Burago and Ivanov12, Proposition 7.3.16], which proves the triangle inequality for the standard Gromov-Hausdorff distance. Let $\mathcal{K}^i=(K^i,\pi ^i,X^i,L^i)$ be an element of $(\mathbb{K},d_{\mathbb{K}})$ for $i=1,2,3$ . Suppose that

\begin{equation*} d_{\mathbb {K}}(\mathcal {K}^1,\mathcal {K}^2)\lt \delta _1. \end{equation*}

Thus, there exists a metric space $Z_1$ , isometric embeddings $\phi _{1,1}\;:\;K^1\rightarrow Z_1$ , $\phi _{2,1}\;:\;K^2\rightarrow Z_1$ and a correspondence $\mathcal{C}_1$ between $K^1$ and $K^2$ such that the sum of the quantities inside the infimum that defines $d_{\mathbb{K}}$ is bounded above by $\delta _1$ . Similarly, if

\begin{equation*} d_{\mathbb {K}}(\mathcal {K}^2,\mathcal {K}^3)\lt \delta _2, \end{equation*}

there exists a metric space $Z_2$ , isometric embeddings $\phi _{2,2}\;:\;K^2\rightarrow Z_2$ , $\phi _{2,3}\;:\;K^3\rightarrow Z_2$ and a correspondence $\mathcal{C}_2$ between $K^2$ and $K^3$ such that the sum of the quantities inside the infimum that defines $d_{\mathbb{K}}$ is bounded above by $\delta _2$ . Next, we set $Z=Z_1\sqcup Z_2$ to be the disjoint union of $Z_1$ and $Z_2$ and we define a distance on $Z$ in the following way. Let $d_{Z|Z_i\times Z_i}=d_{Z_i}$ , for $i=1,2$ , and for $x\in Z_1$ , $y\in Z_2$ set

\begin{equation*} d_{Z}(x,y)\;:\!=\;\inf _{z\in K^2} \{d_{Z_1}(x,\phi _{2,1}(z))+d_{Z_2}(\phi _{2,2}(z),y)\}. \end{equation*}

It is obvious that $d_Z$ is symmetric and non-negative. It is also easy to check that $d_Z$ satisfies the triangle inequality. Identifying points that are separated by zero distance and slightly abusing notation, we turn $(Z,d_{Z})$ into a metric space, which comes with isometric embeddings $\phi _i$ of $Z_i$ for $i=1,2$ . Using the triangle inequality of the Prokhorov metric on $Z$ , gives us that

\begin{equation*} d_P^Z(\pi ^1\circ (\phi _1\circ \phi _{1,1})^{-1},\pi ^3\circ (\phi _2\circ \phi _{3,2})^{-1}) \end{equation*}

\begin{equation*} \le d_P^Z(\pi ^1\circ (\phi _1\circ \phi _{1,1})^{-1},\pi ^2\circ (\phi _1\circ \phi _{2,1})^{-1})+d_P^Z(\pi ^2\circ (\phi _1\circ \phi _{2,1})^{-1},\pi ^3\circ (\phi _2\circ \phi _{3,2})^{-1}). \end{equation*}

Now, since $\phi _1(\phi _{2,1}(y))=\phi _2(\phi _{2,2}(y))$ , for all $y\in K^2$ , we deduce

(2.9) \begin{equation} d_P^Z(\pi ^1\circ (\phi _1\circ \phi _{1,1})^{-1},\pi ^3\circ (\phi _2\circ \phi _{3,2})^{-1}) \le d_P^{Z_1}(\pi ^1\circ \phi _{1,1}^{-1},\pi ^2\circ \phi _{2,1}^{-1})+d_P^{Z_2}(\pi ^2\circ \phi _{2,2}^{-1},\pi ^3\circ \phi _{3,2}^{-1}). \end{equation}

A similar bound also applies for the embedded càdlàg paths. Namely, using the same methods as above, we deduce

(2.10) \begin{equation} d_{J_1}^Z((\phi _1\circ \phi _{1,1})(X^1),(\phi _2\circ \phi _{3,2})(X^3))\le d_{J_1}^Z(\phi _{1,1}(X^1),\phi _{2,1}(X^2))+d_{J_1}^Z(\phi _{2,2}(X^2),\phi _{3,2}(X^3)). \end{equation}

Now, let

\begin{equation*} \mathcal {C}\;:\!=\;\{(x,z)\in K^1\times K^3\;:\;(x,y)\in \mathcal {C}_1,(y,z)\in \mathcal {C}_2,\textrm { for some }y\in K^2\}. \end{equation*}

Observe that $\mathcal{C}$ is a correspondence between $K^1$ and $K^3$ . Then, if $(x,z)\in \mathcal{C}$ , there exists $y\in K^2$ such that $(x,y)\in \mathcal{C}_1$ and $(y,z)\in \mathcal{C}_2$ , and noting again that $\phi _1(\phi _{2,1}(y))=\phi _2(\phi _{2,2}(y))$ , for all $y\in K^2$ , we deduce

(2.11) \begin{equation} d_Z(\phi _1(\phi _{1,1}(x)),\phi _2(\phi _{3,2}(z)))\le d_{Z_1}(\phi _{1,1}(x),\phi _{2,1}(y))+d_{Z_2}(\phi _{2,2}(y),\phi _{3,2}(z)). \end{equation}

Using the same arguments one can prove a corresponding bound involving $L^i$ , $i=1,2,3$ . Namely, if $(x,z)\in \mathcal{C}$ , there exists $y\in K^2$ such that $(x,y)\in \mathcal{C}_1$ and $(y,z)\in \mathcal{C}_2$ , and moreover

(2.12) \begin{equation} \sup _{t\in [0,T]}|L_t^1(x)-L_t^3(z)|\le \sup _{t\in [0,T]} |L_t^1(x)-L_t^2(y)|+\sup _{t\in [0,T]} |L_t^2(y)-L_t^3(z)|. \end{equation}

Putting (2.9), (2.10), (2.11) and (2.12) together gives

\begin{equation*} d_{\mathbb {K}}(\mathcal {K}^1,\mathcal {K}^3)\le \delta _1+\delta _2, \end{equation*}

and the triangle inequality follows. Thus, $(\mathbb{K},d_{\mathbb{K}})$ forms a metric space.

To finish the proof, we need to show that $(\mathbb{K},d_{\mathbb{K}})$ is separable. Let $(K,\pi,X,L)$ be an element of $\mathbb{K}$ . First, let $K^n$ be a finite $n^{-1}$ -net of $K$ , which exists since $K$ is compact. Furthermore, we can endow $K^n$ with a metric $d_{K^n}$ , such that $d_{K^n}(x,y)\in \mathbb{Q}$ , and moreover $|d_{K^n}(x,y)-d_K(x,y)|\le n^{-1}$ , for every $x,y \in K^n$ . Since, $K^n$ is a finite $n^{-1}$ -net of $K$ we can choose a partition for $K$ , $(A_x)_{x\in K^n}$ , such that $x\in A_x$ , and $\textrm{diam}_{K}(A_x)\le 2 n^{-1}$ . We can even choose the partition in such a way that $A_x$ is measurable for all $x\in K^n$ (see for example the definition of $(A_i)_{i=1}^{N_{\varepsilon }}$ after (2.1)). Next, we construct a Borel probability measure $\pi ^n$ in $K^n$ that takes rational mass at each point, i.e. $\pi ^n(\{x\})\in \mathbb{Q}$ , and $|\pi ^n(\{x\})-\pi (A_x)|\le n^{-1}$ . Define $\varepsilon _n$ by

\begin{equation*} \varepsilon _n\;:\!=\;\sup _{\substack {s,t\in [0,T]:\\ |s-t|\le n^{-1}}} \sup _{\substack {x,x'\in K:\\ d_{K}(x,x')\le n^{-1}}} |L_s(x)-L_t(x')|. \end{equation*}

By the joint continuity of $L$ , $\varepsilon _n\rightarrow 0$ , as $n\rightarrow \infty$ . Let $0=s_0\lt s_1\lt \cdot \cdot \cdot \lt s_r=T$ be a set of rational times such that $|s_{i+1}-s_i|\le n^{-1}$ , for $i=0,\ldots,r-1$ . Choose $L_{s_i}^n(x)\in \mathbb{Q}$ with $|L_{s_i}^n(x)-L_{s_i}(x)|\le n^{-1}$ , for every $x\in K^n$ . We interpolate linearly between the finite collection of rational time points in order to define $L^n$ to the whole domain $K^n\times [0,T]$ . Let $\mathcal{C}^n\;:\!=\;\{(x,x')\in K\times K^n\;:\;d_K(x,x')\le n^{-1}\}$ . Clearly $\mathcal{C}^n$ defines a correspondence between $K$ and $K^n$ . Let $(x,x')\in \mathcal{C}^n$ and $s\in [s_i,s_{i+1}]$ , for some $i=0,\ldots,r-1$ . Then, using the triangle inequality we observe that

(2.13) \begin{equation} |L_s^n(x)-L_s(x')|\le |L_s^n(x)-L_s(x)|+|L_s(x)-L_s(x')|\le |L_s^n(x)-L_s(x)|+\varepsilon _n. \end{equation}

Since we interpolated linearly to define $L^n$ beyond rational time points on the whole space $K^n\times [0,T]$ we have that

(2.14) \begin{equation} |L_s^n(x)-L_s(x)|\le |L_{s_{i+1}}^n(x)-L_s(x)|+|L_{s_i}^n(x)-L_s(x)|. \end{equation}

Applying the triangle inequality again yields

\begin{align*} |L_{s_i}^n(x)-L_s(x)|&\le |L_{s_i}^n(x)-L_{s_i}(x)|+|L_{s_i}(x)-L_s(x)| \\[5pt] &\le n^{-1}+\varepsilon _n. \end{align*}

The same upper bound applies for $|L_{s_{i+1}}^n(x)-L_s(x)|$ , and from (2.13) and (2.14) we conclude that for $(x,x')\in \mathcal{C}^n$ and $s\in [s_i,s_{i+1}]$ , for some $i=0,\ldots,r-1$ ,

\begin{equation*} |L_s^n(x)-L_s(x')|\le 2 n^{-1}+3 \varepsilon _n. \end{equation*}

For $X\in D([0,T],K)$ and $A\subseteq [0,T]$ put

\begin{equation*} w(X;\;A)\;:\!=\;\sup _{s,t\in A} d_K(X_t,X_s). \end{equation*}

Now, for $\delta \in (0,1)$ , define the càdlàg modulus to be

\begin{equation*} w'(X;\;\delta )\;:\!=\;\inf _{\Sigma } \max _{1\le i\le k} w(X;\;[t_{i-1},t_i)), \end{equation*}

where the infimum is taken over all partitions $\Sigma =\{0=t_0\lt t_1\lt \cdot \cdot \cdot \lt t_k=T\}$ , $k\in \mathbb{N}$ , with $\min _{1\le i\le k}(t_i-t_{i-1})\gt \delta$ . For a function to lie in $D([0,T],K)$ , it is necessary and sufficient to satisfy $w'(X;\;\delta )\to 0$ , as $\delta \to 0$ . Let $B_n$ be the set of functions having a constant value in $K^n$ over each interval $[(u-1)T/n,uT/n)$ , for some $n\in \mathbb{N}$ and also a value in $K^n$ at time $T$ . Take $B=\cup _{n\ge 1} B_n$ , and observe that is countable. Clearly, putting $z=(z_u)_{u=0}^{n}$ , with $z_u=uT/n$ , for every $u=0,\ldots,n$ satisfies $0=z_0\lt z_1\lt \cdot \cdot \cdot \lt z_n=T$ . Let $T_z\;:\;D([0,T],K)\to D([0,T],K)$ be the map that is defined in the following way. For $X\in D([0,T],K)$ take $T_zX$ to have a constant value $X(z_{u-1})$ over the interval $[z_{u-1},z_u)$ for $1\le u\le n$ and the value $X(T)$ at $t=T$ . From an adaptation of [Reference Billingsley11, Lemma 3, p.127], considering càdlàg paths that take values on metric spaces, we have that

(2.15) \begin{equation} d_{J_1}(T_zX,X)\le T n^{-1}+w'(X;\;T n^{-1}). \end{equation}

Also, there exists $X^n\in B_n$ , for which

(2.16) \begin{equation} d_{J_1}(T_zX,X^n)\le T n^{-1}. \end{equation}

Combining (2.15) and (2.16), we have that

\begin{equation*} d_{J_1}(X^n,X)\le d_{J_1}(X^n,T_zX)+d_{J_1}(T_zX,X)\le 2 T n^{-1}+w'(X;\;T n^{-1}). \end{equation*}

With the choice of the sequence $(K^n,\pi ^n,X^n,L^n)$ , we find that

\begin{equation*} d_{\mathbb {K}}((K^n,\pi ^n,X^n,L^n),(K,\pi,X,L))\le (4 +2 T) n^{-1}+3 \varepsilon _n+w'(X;\;T n^{-1}). \end{equation*}

Recalling that $w'(X;\;T n^{-1})\to 0$ , as $n\to \infty$ , and noting that our sequence was drawn from a countable subset of $\mathbb{K}$ completes the proof of the proposition.

Proof. Since Assumption 1 holds, for each $n\ge 1$ we can find metric spaces $(F_n,d_n)$ , isometric embeddings $\phi _n\;:\;V(G^n)\rightarrow F_n$ , $\phi _n'\;:\;K\rightarrow F_n$ and correspondences $\mathcal{C}^n$ (that contain $(\rho ^n,\rho )$ ) between $V(G^n)$ and $K$ such that (identifying the relevant objects with their embeddings)

(2.21) \begin{align} d_{P}^{F_n}(\pi ^n,\pi )+d_{J_1}^{F_n}(\alpha (n) X^n,X)+\sup _{(x,x')\in \mathcal{C}^n}\bigg (d_n(x,x')+\sup _{t\in [0,T]} |\alpha (n) L^n_{\beta (n) t}(x)-L_t(x')|\bigg )\le \varepsilon _n, \end{align}

where $\varepsilon _n\rightarrow 0$ , as $n\rightarrow \infty$ . Now, let $F=\sqcup _{n\ge 1} F_n$ , be the disjoint union of $F_n$ , and define the distance $d_{F|F_n\times F_n}=d_{n}$ , for $n\ge 1$ , and for $x\in F_n$ , $x'\in F_{n'}$ , $n\neq n'$

\begin{equation*} d_F(x,x')\;:\!=\;\inf _{y\in K}\{d_n(x,y)+d_{n'}(y,x')\}. \end{equation*}

This distance, as the distance that was defined in order to prove the triangle inequality in Proposition 2.1, is symmetric and non-negative, so identifying points that are separated by a zero distance, we turn $(F,d_F)$ into a metric space, which comes with natural isometric embeddings of $(V(G^n),\alpha (n) d_{G^n})_{n\ge 1}$ and $(K,d_K)$ . In this setting, under the appropriate isometric embeddings (2.17), (2.18) and (2.19) readily hold from (2.21). Thus, it only remains to prove (2.20). For every $x\in V(G^n)$ , since $\mathcal{C}^n$ is a correspondence in $V(G^n)\times K$ , there exists an $x'\in K$ such that $(x,x')\in \mathcal{C}^n$ . Then, (2.21) implies that $d_F(x,x')\le \varepsilon _n$ . Now, let $(y,y')\in \mathcal{C}^n$ , $(z,z')\in \mathcal{C}^n$ and note that

\begin{align*} &\sup _{t\in [0,T]} \alpha (n) |L_{\beta (n) t}^n(y)-L_{\beta (n) t}^n(z)|\\[5pt] &\le \sup _{t\in [0,T]} |\alpha (n) L_{\beta (n) t}^n(y)-L_t(y')|+\sup _{t\in [0,T]} |\alpha (n) L_{\beta (n) t}^n(z)-L_t(z')|+\sup _{t\in [0,T]} |L_t(y')-L_t(z')|\\[5pt] &\le 2 \varepsilon _n+\sup _{t\in [0,T]} |L_t(y')-L_t(z')|. \end{align*}

For any $\delta \gt 0$ and $y, z\in V(G^n)$ , such that $\alpha (n) d_{G^n}(y,z)\lt \delta$ , we have that

\begin{equation*} d_K(y',z')\le d_F(y,y')+d_F(z,z')+\alpha (n) d_{G^n}(y,z)\lt 2 \varepsilon _n+\delta . \end{equation*}

Therefore,

(2.22) \begin{align} &\sup _{\substack{y,z\in V(G^n): \\ \alpha (n) d_{G^n}(y,z)\lt \delta }} \sup _{t\in [0,T]} \alpha (n) |L_{\beta (n) t}^n(y)-L_{\beta (n) t}^n(z)|\nonumber \\[5pt] &\le 2 \varepsilon _n+\sup _{\substack{y,z\in K: \\ d_K(y,z)\lt 2 \varepsilon _n+\delta }} \sup _{t\in [0,T]} |L_t(y)-L_t(z)|. \end{align}

Also, for every $x\in K$ there exists an $x'\in V(G^n)$ such that $(x',x)\in \mathcal{C}^n$ , and furthermore $d_F(x',x)\le \varepsilon _n$ . Let $x^n\in V(G^n)$ such that $d_F(x^n,x)\lt \delta$ . Then,

\begin{equation*} d_F(x^n,x')\le d_F(x^n,x)+d_F(x',x)\lt 2 \varepsilon _n+\delta . \end{equation*}

More generally, if we denote by $B_F(x,r)$ , the open balls of radius $r$ , centred in $x$ , we have the following inclusion

\begin{equation*} B_F(x,\delta )\cap V(G^n)\subseteq B_F(x',2 \varepsilon _n+\delta )\cap V(G^n). \end{equation*}

For $x\in K$ , and $x'\in V(G^n)$ with $d_F(x',x)\le \varepsilon _n$ , using (2.21), we deduce

\begin{align*} &\sup _{t\in [0,T]} |\alpha (n) L_{\beta (n) t}^n(x^n)-L_t(x)|\\[5pt] &\le \sup _{t\in [0,T]} \alpha (n) |L_{\beta (n) t}^n(x^n)-L_{\beta (n) t}^n(x')|+\sup _{t\in [0,T]} |\alpha (n) L_{\beta (n) t}^n(x')-L_t(x)|\\[5pt] &\le \varepsilon _n+\sup _{t\in [0,T]} \alpha (n) |L_{\beta (n) t}^n(x^n)-L_{\beta (n) t}^n(x')|. \end{align*}

Since $x^n \in B_F(x',2 \varepsilon _n+\delta )\cap V(G^n)$ , taking the supremum over all $x^n\in V(G^n)$ and $x\in K$ , for which $d_F(x^n,x)\lt \delta$ and using (2.22), we deduce

\begin{align*} &\sup _{\substack{x^n\in V(G^n),x\in K: \\ d_F(x^n,x)\lt \delta }} \sup _{t\in [0,T]} |\alpha (n) L_{\beta (n) t}^n(x^n)-L_t(x)|\\[5pt] &\le \varepsilon _n+\sup _{\substack{y,z\in V(G^n): \\ \alpha (n) d_{G^n}(y,z)\lt 2 \varepsilon _n+\delta }} \sup _{t\in [0,T]} \alpha (n) |L_{\beta (n) t}^n(y)-L_{\beta (n) t}^n(z)|\\[5pt] &\le 3 \varepsilon _n+\sup _{\substack{y,z\in K: \\ d_K(y,z)\lt 4 \varepsilon _n+\delta }} \sup _{t\in [0,T]} |L_t(y)-L_t(z)|. \end{align*}

Using the continuity of $L$ , as $n\rightarrow \infty$

(2.23) \begin{equation} \limsup _{n\rightarrow \infty } \sup _{\substack{x^n\in V(G^n),x\in K: \\ d_F(x^n,x)\lt \delta }} \sup _{t\in [0,T]} |\alpha (n) L_{\beta (n) t}^n(x^n)-L_t(x)|\le \sup _{\substack{y,z\in K: \\ d_K(y,z)\le \delta }} \sup _{t\in [0,T]} |L_t(y)-L_t(z)|. \end{equation}

Again appealing to the continuity of $L$ , the right-hand side converges to 0, as $\delta \rightarrow 0$ . Thus, we showed that (2.20) holds, and this finishes the proof of Lemma 2.2.

Proof. As we hinted upon when deriving (2.23), using the continuity of $L$ ,

\begin{equation*} \limsup _{n\rightarrow \infty } \sup _{\substack {y,z\in V(G^n): \\ \alpha (n) d_{G^n}(y,z)\lt \delta }} \sup _{t\in [0,T]} \alpha (n) |L_{\beta (n) t}^n(y)-L_{\beta (n) t}^n(z)|\le \sup _{\substack {y,z\in K: \\ d_K(y,z)\le \delta }} \sup _{t\in [0,T]} |L_t(y)-L_t(z)|. \end{equation*}

Sending $\delta \rightarrow 0$ gives the desired result.

Proof. There exist isometric embeddings of $(V(G^n),\alpha (n) d_{G^n})_{n\ge 1}$ and $(K,d_K)$ into a common metric space $(F,d_F)$ , under which the assumptions (2.17)-(2.20) hold. Since (2.17) gives the convergence of spaces under the Hausdorff metric, (2.18) gives the convergence of measures under the Prokhorov metric and (2.19) gives the convergence of paths under $d_{J_1}$ , it only remains to check the uniform convergence of local times. Let $\mathcal{C}^n$ be the set of all pairs $(x,x')\in K\times V(G^n)$ , for which $d_F(x,x')\le n^{-1}$ . Since (2.17) holds, $\mathcal{C}^n$ are correspondences for $n\ge 1$ . Then, for $(x,x')\in \mathcal{C}^n$

\begin{equation*} \sup _{t\in [0,T]} |\alpha (n) L_{\beta (n) t}^n(x')-L_t(x)|\le \sup _{\substack {x^n\in V(G^n),x\in K: \\ d_F(x^n,x)\lt n^{-1}}} \sup _{t\in [0,T]} |\alpha (n) L^n_{\beta (n) t}(x^n)-L_t(x)|, \end{equation*}

and using (2.20) completes the proof.

Proof. Fix $x\neq y$ . There is a strictly positive $\mathbb{P}_x$ -probability that $L_t(x)\gt 0$ for $t$ large enough, which is a consequence of [Reference Marcus and Rosen36, Lemma 3.6]. From the joint continuity of $(L_t(z))_{z\in K,t\ge 0}$ , there exist $r\equiv r(x)$ , $\delta \equiv \delta (x)\gt 0$ and $t_{*}\equiv t_{*}(x)\lt \infty$ such that

(4.1) \begin{equation} \mathbb{P}_x\left (\inf _{z\in B(x,r)} L_{t_{*}}(z)\gt \delta \right )\gt 0. \end{equation}

Now, let $\tau _{x,y}(t_{*})\;:\!=\;\inf \{t\gt t_{*}+\tau _x\;:\;X_t=y\}$ , where $\tau _x\;:\!=\;\inf \{t\gt 0\;:\;X_t=x\}$ is the hitting time of $x$ . In other words, $\tau _{x,y}(t_{*})$ is the first hitting of $y$ by $X$ after time $t_{*}+\tau _x$ . The commute time identity for a resistance derived in the proof of [Reference Croydon, Hambly and Kumagai20, Lemma 2.9], lets it be deduced that

\begin{equation*} \mathbb {E}_z \tau _w\le \mathbb {E}_z \tau _w+\mathbb {E}_w \tau _z=R(z,w) \pi (K), \end{equation*}

for every $z,w\in K$ , which in turn deduces that $\mathbb{E}_x \tau _y\lt \infty$ . Here $\pi (K)=1$ . Applying this observation about the finite first moments of hitting times, it is easy to check that $\tau _{x,y}(t_{*})\lt \infty$ , $\mathbb{P}_y$ -a.s., and also that

\begin{equation*} \mathbb {P}_y\left (\inf _{z\in B(x,r)} L_{\tau _{x,y}(t_{*})}(z)\gt \delta \right )\gt 0. \end{equation*}

The latter simply follows from an application of (4.1) and the Strong Markov property. We only need to show that $\mathbb{E}_y \tau _{x,y}(t_{*})\lt \infty$ in order to conclude that the former holds. This is easy to show since we have for instance:

(4.2) \begin{align} \mathbb{E}_y \tau _{x,y}(t_{*})&\le t_*+\mathbb{E}_y \tau _x+\sup _{z\in K} \mathbb{E}_z \tau _y \nonumber \\[5pt] &\le t_*+\pi (K) R(x,y)+\pi (K) \sup _{z,w\in K} R(z,w) \le t_*+2 \sup _{z,w\in K} R(z,w)\lt \infty . \end{align}

The additivity of local times and the Strong Markov property implies that

(4.3) \begin{equation} \liminf _{t\to \infty } \inf _{z\in B(x,r)} \frac{L_t(z)}{t}\ge \left (\sum _{i=1}^{\infty }\xi _i^1\right ) \left (\sum _{i=1}^{\infty } \xi _i^2\right )^{-1}, \end{equation}

where $(\xi _i^1)_{i\ge 1}$ are independent random variables distributed as $\inf _{z\in B(x,r)} L_{\tau _{x,y}(t_{*})}(z)$ and $(\xi _i^2)_{i\ge 1}$ are independent copies of $\tau _{x,y}(t_{*})$ . The strong law of large numbers along with (4.1) yields that the right-hand side of the inequality above converges to

\begin{equation*} \mathbb {E}_y \left [\inf _{z\in B(x,r)} L_{\tau _{x,y}(t_{*})}(z)\right ] (\mathbb {E}_y \tau _{x,y}(t_{*}))^{-1}, \end{equation*}

$\mathbb{P}_x$ -a.s. Using basic properties of the resolvents of killed processes of resistance forms (e.g. [Reference Croydon18, (6)-(8), p. 1945]), the way the stopping times are defined, and the upper bound in (4.2),

\begin{align*} \mathbb{E}_y \left [L_{\tau _{x,y}(t_{*})}(x)\right ] (\mathbb{E}_y \tau _{x,y}(t_{*}))^{-1} &\ge \mathbb{E}_x L_{\tau _y}(x) (\mathbb{E}_y \tau _{x,y}(t_{*}))^{-1} \\[5pt] &=\frac{R(x,y)}{t_*+R(x,y)+\sup _{z,w\in K} R(z,w)}, \end{align*}

and therefore, the joint continuity of local times lets it to be deduced that the right-hand side of (4.3) satisfies

(4.4) \begin{equation} \liminf _{t\to \infty } \inf _{z\in B(x,r)} \frac{L_t(z)}{t}\ge \varepsilon _{*}(x), \end{equation}

$\mathbb{P}_y$ -a.s., for some $\varepsilon _*(x)\in (\varepsilon,1)$ . To extend this statement that holds uniformly over $B(x,r)$ , $\mathbb{P}_y$ -a.s., we use the compactness of $K$ . Consider the open cover $(B(x,r))_{x_i\in K}$ the open cover for $K$ , which say admits a finite subcover $(B(x_i,r_i))_{i=1}^{N}$ . Since the right-hand of (4.4) is greater than $\varepsilon _{*}$ , $\mathbb{P}_y$ -a.s., the result clearly follows as

\begin{equation*} \liminf _{t\to \infty } \frac {L_t(x)}{t}\ge \min _{1\le i\le N} \liminf _{t\to \infty } \inf _{x\in B(x_i,r_i)} \frac {L_t(x)}{t}, \end{equation*}

which implies that there exists $t_0\equiv t_0(x)\lt \infty$ , such that $L_{t_0}(x)\ge \varepsilon t_0$ , for every $x\in K$ , and recalling the definition of the $\varepsilon$ -blanket time variable of $K$ in (1.5), we deduce that $\tau _{\textrm{bl}}(\varepsilon )\le t_0\lt \infty$ .