}g\text{ begins in }D\\}=\\{f\rfloor_{z}\in\Pi^{\downarrow}\,;\,z\in D\text{
and }f\in\mathcal{F}(z)\\}$ (5.7)

To see (5.7), first consider if $g$ does not cross $\mathcal{A}$ and begins in
$D$. By applying Theorem 4.3.9 (in reverse time) we obtain
$f\in\Pi^{\updownarrow}$ such that $g\subseteq f$ and $f$ does not cross
$\mathcal{A}$, which implies that
$f\in\operatorname{flow}(\mathcal{A})=\mathcal{F}$. We have $g=f\rfloor_{z}$,
so the left hand side of (5.7) is contained within the right hand side. For
the reverse inclusion, consider if $z\in D$ and $f\in\mathcal{F}(z)$. Clearly
$g=f\rfloor_{z}$ begins at $z\in D$ and does not cross $\mathcal{F}$, which
implies $g$ does not cross $\mathcal{A}$. We have thus established (5.7).

Let

$\displaystyle\widehat{\mathcal{W}}_{D}$
$\displaystyle=\overline{\\{g\in\Pi^{\downarrow}\,;\,g\text{ does not cross
}\mathcal{A}\text{ and }g\text{ begins in }D\\}_{\downarrow}}$ (5.8)
$\displaystyle\widehat{\mathcal{W}}^{\prime}_{D}$
$\displaystyle=\overline{\\{f\rfloor_{z}\in\Pi^{\downarrow}\,;\,z\in D\text{
and }f\in\mathcal{F}(z)\\}_{\downarrow}}$

Note that (5.8) is (2.16), repeated here for convenience. By Proposition
A.2.1, compactness of $\mathcal{F}$ implies relative compactness of (the right
hand side of) equation (5.7). With this in hand Lemma 4.5.1, applied in
reverse time, gives that
$\widehat{\mathcal{W}}_{D}=\widehat{\mathcal{W}}^{\prime}_{D}$. From (2.12) we
have that

$\widehat{\mathcal{W}}^{\prime}_{D}=(\operatorname{web}_{D^{\circlearrowright}}(\mathcal{F}^{\circlearrowright}))^{\circlearrowright}$
(5.9)

from which Theorem 2.4.3 gives that $\widehat{\mathcal{W}}^{\prime}_{D}$ is
both a dual web and (from Remark 4.5.4) does not depend on the choice of dense
and almost surely non-ramified subset $D\subseteq\mathbb{R}^{2}_{\rm c}$. From
this point on let us write
$\widehat{\mathcal{W}}=\widehat{\mathcal{W}}_{D}=\widehat{\mathcal{W}}^{\prime}_{D}$.
From (5.9) we have that
$(\widehat{\mathcal{W}}^{\prime}_{D})^{\circlearrowright}=\operatorname{web}_{D^{\circlearrowright}}(\mathcal{F}^{\circlearrowright})$,
which in words says that
$(\widehat{\mathcal{W}}^{\prime}_{D})^{\circlearrowright}$ is the web
associated to $\mathcal{F}^{\circlearrowright}$. Theorem 2.4.3 implies that
$(\widehat{\mathcal{W}}^{\prime}_{D})^{\circlearrowright}$ does not cross
$\mathcal{F}^{\circlearrowright}$, thus $\widehat{\mathcal{W}}^{\prime}_{D}$
does not cross $\mathcal{F}$, which by Lemma 4.1.3 implies it also does not
cross $\mathcal{A}$. Therefore $(\mathcal{W},\widehat{\mathcal{W}})$ is a
double web.

The same argument that led to (5.7), but now used forwards in time, gives that
$\\{f\in\Pi^{\uparrow}\,;\,f\text{ does not cross }\mathcal{A}\text{ and
}f\text{ begins in }D\\}=\\{f\lceil_{z}\in D\text{ and
}f\in\mathcal{F}(z)\\}$. It follows from (2.12) and Theorem 2.4.3 that

$\mathcal{W}=\overline{\\{f\in\Pi^{\uparrow}\,;\,f\text{ does not cross
}\mathcal{A}\text{ and }f\text{ begins in }D\\}_{\uparrow}}.$ (5.10)

We now turn our attention to (2.17). Let

$\mathcal{F}^{\prime}=\overline{\\{g_{\hookrightarrow}f\in\Pi^{\updownarrow}\,;\,g\in\widehat{\mathcal{W}}\text{
ends and }f\in\mathcal{W}\text{ begins at the same point of }D\\}}.$ (5.11)

We must show that $\mathcal{F}^{\prime}\stackrel{{\scriptstyle
a.s}}{{=}}\mathcal{F}$. Let $h\in\mathcal{F}^{\prime}$. Then there exists
$z_{n}\in D$ such that $f_{n}\in\mathcal{W}$ begins and
$g_{n}\in\widehat{\mathcal{W}}$ ends at $z_{n}$, and $h_{n}\rightarrow h$
where $h_{n}=(g_{n})_{\hookrightarrow}(f_{n})$. Using that $z_{n}$ is non-
ramified, let $h^{\prime}_{n}$ be the unique element of $\mathcal{F}(z_{n})$.
Note that $f_{n}$ does not cross $\mathcal{F}$. Hence
$((h^{\prime}_{n})\rfloor_{z_{n}})_{\hookrightarrow}(f_{n})$ is a path passing
through $z_{n}$ that does not cross $\mathcal{F}$, which is therefore an
element of $\mathcal{F}$, and therefore equal to $h_{n}$. Thus
$f_{n}=(h^{\prime}_{n})\lceil_{z_{n}}$. By a symmetrical argument,
$g_{n}=(h^{\prime}_{n})\rfloor_{z_{n}}$ which implies that
$h^{\prime}_{n}=(g_{n})_{\hookrightarrow}(f_{n})=h_{n}$. Hence
$h_{n}\in\mathcal{F}$, which implies that $h\in\mathcal{F}$.

To see the reverse inclusion, let $f\in\mathcal{F}$. By Lemma 4.2.8 we have
$\mathcal{F}=\overline{\mathcal{F}(D)}$, hence there exists $z_{n}\in D$ and
$h_{n}\in\mathcal{F}(z_{n})$ such that $h_{n}\rightarrow f$. Let
$f_{n}=(h_{n})\lceil_{z_{n}}$ and $g_{n}=(h_{n})\rfloor_{z_{n}}$. We have that
$h_{n}$ does not cross $\mathcal{F}$, hence $f_{n}$ and $g_{n}$ do not cross
$\mathcal{A}$. From (5.8) and (5.10) we have that $f_{n}\in\mathcal{W}$ and
$g_{n}\in\widehat{\mathcal{W}}$. Hence $h\in\mathcal{F}^{\prime}$. We thus
have $\mathcal{F}=\mathcal{F}^{\prime}$ which establishes (2.17).

It remains only to show the uniqueness claim. Let $\widehat{\mathcal{U}}$ be a
dual web and suppose that $(\mathcal{W},\widehat{\mathcal{U}})$ is a double
web. Then $\widehat{\mathcal{U}}^{\circlearrowright}$ is a web that almost
surely does not cross $\mathcal{F}^{\circlearrowright}$, and the same is true
of $\widehat{\mathcal{W}}^{\circlearrowright}$. It is straightforward to check
that $\mathcal{F}^{\circlearrowright}$ is a $\Pi^{\updownarrow}$ valued random
variable that inherits closedness, pervasiveness and the non-crossing property
from $\mathcal{F}$. Proposition A.2.1 implies that relative compactness is
also inherited through $\cdot^{\circlearrowright}$, so
$\mathcal{F}^{\circlearrowright}$ is a weave. Lemma 4.1.4 now implies that
$\widehat{\mathcal{U}}^{\circlearrowright}\cup\widehat{\mathcal{W}}^{\circlearrowright}$
is almost surely non-crossing, from which Lemma 4.5.5 implies that
$\operatorname{web}(\widehat{\mathcal{U}}^{\circlearrowright})\stackrel{{\scriptstyle\rm
a.s.}}{{=}}\operatorname{web}(\widehat{\mathcal{W}}^{\circlearrowright})$. By
Theorem 2.4.3 we thus have
$\widehat{\mathcal{U}}^{\circlearrowright}\stackrel{{\scriptstyle\rm
a.s.}}{{=}}\widehat{\mathcal{W}}^{\circlearrowright}$, which implies
$\widehat{\mathcal{U}}\stackrel{{\scriptstyle\rm
a.s.}}{{=}}\widehat{\mathcal{W}}$ as required.

###  5.7 Proof of Theorem 2.4.9

Recall that $\Pi^{\uparrow}_{c},\Pi^{\downarrow}_{c}$ and
$\Pi^{\updownarrow}_{c}$ respectively denote the sets of continuous forwards
half-infinite, backwards half-infinite, and bi-infinite càdlàg paths. The
proof of Theorem 2.4.9 is based on the following lemma.

###### Lemma 5.7.1

The following hold.

  1. 1.

Let $\mathcal{A}\subseteq\Pi^{\uparrow}_{c}$ be a deterministic weave. If
$h\in\Pi^{\updownarrow}$ does not cross $\mathcal{A}$ then
$h\in\Pi^{\updownarrow}_{c}$.

  2. 2.

Let $\mathcal{F}\subseteq\Pi^{\updownarrow}_{c}$ be a deterministic flow. If
$f\in\Pi^{\uparrow}$ does not cross $\mathcal{F}$ then
$f\in\Pi^{\uparrow}_{c}$.

###### Proof.

We will prove each claim in turn, starting with the first. We argue by
contradiction. Let $\mathcal{A}\subseteq\Pi_{c}$. Suppose that that
$h\in\Pi^{\updownarrow}$ does not cross $f$ and that $h$ is discontinuous at
$t\in\mathbb{R}$. Without loss of generality (or consider space reflected
about the origin) we may assume that $h(t-)<h(t+)$. By Lemma 4.1.1 there
exists $f\in\mathcal{A}$ such that $f(t-)\leq h(t-)$ and $h\vartriangleleft
f$. Hence $h(t+)\leq f(t+)$. We thus have $f(t-)\leq h(t-)<h(t+)\leq f(t+)$,
which means that $f\in\mathcal{A}$ is discontinuous at $t$. This is a
contradiction, which completes the proof.

It remains to establish the second claim. If $f\in\Pi^{\uparrow}$ does not
cross $\mathcal{F}$ then by Theorem 4.3.9 there exists
$f^{\prime}\in\Pi^{\updownarrow}$ such that $f\subseteq f^{\prime}$ and
$f^{\prime}$ does not cross $\mathcal{F}$. From what we have already proved we
have $f^{\prime}\in\mathcal{F}$, thus $f^{\prime}\in\Pi_{c}$ which implies
$f\in\Pi_{c}$. $\blacksquare$

We now give the proof of Theorem 2.4.9. Note that it suffices to prove the
results for deterministic weaves. We will prove the two claims in turn,
starting with the first. Let $\mathcal{A},\mathcal{B}$ be deterministic weaves
such that $\mathcal{A}\sim\mathcal{B}$. We seek to show that if
$\mathcal{A}\subseteq\Pi^{\uparrow}_{c}$ then
$\mathcal{B}\subseteq\Pi^{\uparrow}_{c}$. It then follows by symmetry that
$\mathcal{A}\subseteq\Pi^{\uparrow}_{c}$ if and only if
$\mathcal{B}\subseteq\Pi^{\uparrow}_{c}$.

By Theorem 2.4.4, without loss of generality we may assume that $\mathcal{A}$
are $\mathcal{B}$ coupled such that $\mathcal{A}\cup\mathcal{B}$ is almost
surely non-crossing. On that event, by Lemma 4.1.4, a path $f\in\Pi$ crosses
$\mathcal{A}$ if and only if it crosses $\mathcal{B}$. From (2.13) we thus
obtain $\operatorname{flow}(\mathcal{A})\stackrel{{\scriptstyle\rm
a.s.}}{{=}}\operatorname{flow}(\mathcal{B})$, which we henceforth refer to as
$\mathcal{F}$. By part 1 of Lemma 5.7.1, as
$\mathcal{A}\subseteq\Pi^{\uparrow}_{c}$ we have also that
$\mathcal{F}\subseteq\Pi^{\updownarrow}_{c}$. Since
$\mathcal{B}\preceq\mathcal{F}$ we have
$\mathcal{B}\subseteq\mathcal{F}_{\uparrow}$, thus
$\mathcal{B}\subseteq\Pi_{c}$. The same applies by symmetry with the roles of
$\mathcal{A}$ and $\mathcal{B}$ swapped. This proves the first claim of
Theorem 2.4.9.

It remains to prove the second claim. Let $\mathcal{A}$ be a deterministic
weave. Let $\mathcal{W},\mathcal{F}$ denote the corresponding web and flow. It
remains to show that $\hat{\mathcal{W}}\subseteq\Pi^{\uparrow}_{c}$ if and
only if $\mathcal{W}\subseteq\Pi^{\downarrow}_{c}$, where $\hat{\mathcal{W}}$
is given by (2.16) in Theorem 2.4.8. By symmetry (or consider reversing the
direction of time) it suffices to prove that if
$\mathcal{W}\subseteq\Pi^{\uparrow}_{c}$ then
$\hat{\mathcal{W}}\subseteq\Pi^{\downarrow}_{c}$. To this end, suppose that
$\mathcal{W}\subseteq\Pi_{c}$.

From what we have already proved, $\mathcal{F}\subseteq\Pi_{c}$. From (2.16)
we have

$\hat{\mathcal{W}}=\overline{\\{g\in\Pi^{\downarrow}\,;\,g\text{ does not
cross }\mathcal{F}\text{ and }(g(\tau_{g}),\tau_{g})\in D\\}_{\downarrow}}.$
(5.12)

Let $\hat{f}\in\hat{\mathcal{W}}$. Then there exists
$\hat{f}_{n}\in\Pi^{\downarrow}$ such that $\hat{f}_{n}$ does not cross
$\mathcal{F}$ and $\hat{f}_{n}\rightarrow\hat{f}$. By Theorem 4.3.9 (applied
in reverse time) there exists $h_{n}\in\mathcal{F}$ such that
$\hat{f}_{n}\subseteq h_{n}$. By compactness of $\mathcal{F}$ we may pass to a
convergent subsequence $h_{n}\rightarrow h\in\mathcal{F}$. By Lemma A.2.2 we
have $\hat{f}\subseteq h$. Since $h$ does not cross $\mathcal{F}$, also
$\hat{f}$ does not cross $\mathcal{F}$. By part 2 of Lemma 5.7.1 (applied in
reverse time) we thus have $\hat{f}\in\Pi^{\downarrow}_{c}$. Thus
$\mathcal{W}\subseteq\Pi^{\downarrow}_{c}$.

##  Appendix A Appendices

###  A.1 On the Hausdorff metric

Let $(M,d_{M})$ be a metric space and let $\mathcal{K}(M)$ denote the set of
compact subsets of $M$, including the empty set. We write
$\operatorname{dist}_{M}(x,A)=\inf_{a\in A}d_{M}(x,a)$ for the infimum
distance from the point $x\in M$ to $A\subseteq M$. We now state some well
known facts relating to $\mathcal{K}(M)$. The function

$d_{\mathcal{K}(M)}(A_{1},A_{2}):=\sup_{x_{1}\in
A_{1}}\operatorname{dist}_{M}(x_{1},A_{2})\vee\sup_{x_{2}\in
A_{2}}\operatorname{dist}_{M}(x_{2},A_{1}),$ (A.1)

defines a metric on $\mathcal{K}(M)$ known as the Hausdorff metric, or more
precisely the Hausdorff metric with respect to $d_{M}$. If $d$ and
$d^{\prime}$ are two metrics generating the same topology on $M$, then their
corresponding Hausdorff metrics generate the same topology on $M$. This
topology is known as the Hausdorff topology. Note the subtle difference
between ‘the Hausdorff topology’, which refers to this particular topology,
and ‘a Hausdorff topology’, which refers to any topology having the Hausdorff
property i.e. that distinct points possess disjoint neighbourhoods. (All
topological spaces mentioned within the present article have the Hausdorff
property.)

Completeness of $M$ implies completeness of $\mathcal{K}(M)$. The same
extension from $M$ to $\mathcal{K}(M)$ also holds for separability, and for
compactness. We now establish some more detailed connections.

###### Lemma A.1.1

Let $(M,d_{M})$ be a complete metric space.

  1. 1.

Let $\mathcal{X}\subseteq\mathcal{K}(M)$. Then $\mathcal{X}$ is relatively
compact if and only if $\cup_{X\in\mathcal{X}}X$ is a relatively compact
subset of $M$.

  2. 2.

Let $X\subseteq M$. Then $X$ is relatively compact if and only if
$\\{A\in\mathcal{K}(M)\,;\,A\subseteq X\\}$ is a relatively compact subset of
$\mathcal{K}(M)$.

###### Proof.

The two claims are readily seen to be equivalent: take
$\mathcal{X}=\\{A\in\mathcal{K}(M)\,;\,A\subseteq X\\}$ to see that the
(1)$\Rightarrow$(2) and take $X=\cup_{A\in\mathcal{X}}A$ to see that
(2)$\Rightarrow$(1). We will give proof of (1). As completeness of $M$ implies
completeness of $\mathcal{K}(M)$, in both $M$ and $\mathcal{K}(M)$ we have
that relative compactness is equivalent to total boundedness.

Suppose that $\mathcal{X}\subseteq\mathcal{K}(M)$ is totally bounded. Then,
for each $\epsilon>0$ there is a finite set $X_{1},\ldots,X_{n}$ of elements
of $\mathcal{K}(M)$ such that, for any $X\in\mathcal{X}$ there is some $X_{i}$
such that $d_{\mathcal{K}(M)}(X,X_{i})<\epsilon$. Let
$Y=\bigcup_{i=1}^{n}X_{i}$ and note that $\bigcup_{X\in\mathcal{X}}X\subseteq
Y^{(\epsilon)}$. Since each $X_{i}$ is compact in $M$, $Y$ is also compact in
$M$, and in particular $Y$ is totally bounded. Hence also
$\bigcup_{X\in\mathcal{X}}X$ is totally bounded.

Conversely, suppose that $\mathcal{X}\subseteq\mathcal{K}(M)$ is such that
$\bigcup_{X\in\mathcal{X}}X\subseteq M$ is totally bounded. Let $\epsilon>0$.
There exists a finite set $\\{x_{1},\ldots,x_{n}\\}\subseteq M$ and a map
$f:\bigcup_{X\in\mathcal{X}}X\rightarrow\\{x_{1},\ldots,x_{n}\\}$, such that
for any $x\in\bigcup_{X\in\mathcal{X}}X$ we have $d_{M}(x,f(x))<\epsilon$. For
any $X\in\mathcal{X}$, the set $f(X)=\\{f(x)\,;\,x\in X\\}$ is finite and
therefore compact, meaning that $f(X)\in\mathcal{K}(M)$. By construction we
have $d_{\mathcal{K}(M)}(X,f(X))<\epsilon$. Let $\mathscr{X}$ be the set of
subsets of $\\{x_{1},\ldots,x_{n}\\}$, hence $\mathscr{X}$ is a finite subset
of $\mathcal{K}(M)$. We have shown that for any $X\in\mathcal{X}$ there is
some $X^{\prime}\in\mathscr{X}$ such that
$d_{\mathcal{K}(M)}(X,X^{\prime})<\epsilon$. Thus $\mathcal{X}$ is totally
bounded. $\blacksquare$

The next Lemma is a key ingredient of the proof of Lemma A.3.1. It provides a
supply of closed (and consequently measurable) subsets of $\mathcal{K}(M)$. We
define

$\displaystyle A^{(\epsilon)}$ $\displaystyle=\\{x\in
M\,;\,\operatorname{dist}_{M}(x,A)<\epsilon\\}\qquad\text{ for }\epsilon>0,$
$\displaystyle A^{[\epsilon]}$ $\displaystyle=\\{x\in
M\,;\,\operatorname{dist}_{M}(x,A)\leq\epsilon\\}\qquad\text{ for
}\epsilon\geq 0,$

as the (respectively) open and closed $\epsilon$-expansions of $A\subseteq M$.
Note that $A^{[0]}=A$.

###### Lemma A.1.2

Let $X,Y\subseteq M$, $\epsilon\geq 0$ and
$\mathscr{C}\subseteq\mathcal{K}(M)$. Then:

  1. 1.

the set $\\{A\in\mathcal{K}(M)\,;\,X\subseteq A^{[\epsilon]}\\}$ is closed;

  2. 2.

the set $\\{A\in\mathcal{K}(M)\,;\,A\cap(Y\setminus X)=\emptyset\\}$ is closed
if $X$ is closed and $Y$ is open;

  3. 3.

the set $\\{A\in\mathcal{K}(M)\,;\,A\cap(X\setminus Y)\neq\emptyset\\}$ is
closed if $X$ is closed and $Y$ is open;

  4. 4.

the set $\\{A\in\mathcal{K}(M)\,;\,\exists C\in\mathscr{C},C\subseteq A\\}$ is
closed if $\mathscr{C}$ is closed.

###### Proof.

We prove the claims independently. For the first claim, assume that
$A_{n}\rightarrow A$ in $\mathcal{K}(M)$ and $X\subseteq A_{n}^{[\epsilon]}$
for all $n$, where $\epsilon\geq 0$. Let $f\in X$, so
$d_{\mathcal{K}(M)}(\\{f\\},A_{n})\leq\epsilon$. Letting $n\rightarrow\infty$
we obtain $d_{\mathcal{K}(M)}(\\{f\\},A)\leq\epsilon$. Since $f\in X$ was
arbitrary $X\subseteq A^{[\epsilon]}$, as required.

For the second claim, assume that $A_{n}\rightarrow A$ in $\mathcal{K}(M)$ and
$A_{n}\cap(Y\setminus X)=\emptyset$ for all $n$, where $Y$ is open and $X$ is
closed. We must show that $A\cap(Y\setminus X)$ is empty. We will argue by
contradiction. Suppose there exists $f\in A\cap(Y\setminus X)$, which means
that $f\in A\cap Y$ and $f\notin X$. Since $A_{n}\rightarrow A$ there exists
$f_{n}\in A_{n}$ such that $f_{n}\rightarrow f$. For each $n$, noting that
$A_{n}\cap(Y\setminus X)$ is empty we have (a) $f_{n}\notin Y$ or (b)
$f_{n}\in X$; thus one of these two alternatives must hold for an infinite
subsequence of $n\in\mathbb{N}$. If (a) holds for infinitely many $n$ then
along that subsequence we have $f_{n}\rightarrow f$ with $f_{n}\notin Y$ and
$f\in Y$, which contradicts the fact that $Y$ is open. If (b) holds for
infinitely many $n$ then along that subsequence we have $f_{n}\rightarrow f$
with $f_{n}\in X$ and $f\notin X$, which contradicts the fact that $X$ is
closed. We thus reach the desired contradiction.

For the third claim, suppose that $A_{n}\rightarrow A$ in $\mathcal{K}(M)$ and
$A_{n}\cap(X\setminus Y)\neq\emptyset$ for all $n$, where $X\subseteq M$ is
closed and $Y\subseteq M$ is open. Take $f_{n}\in A_{n}\cap(X\setminus Y)$, so
$f_{n}\in A_{n}\cap X$ and $f_{n}\in M\setminus Y$. By Lemma A.1.1 the set
$\cup_{n}A_{n}$ is a relatively compact $M$ and contains the sequence
$(f_{n})$, so we may pass to a convergence subsequence $f_{n}\rightarrow f$.
Since $A_{n}\rightarrow A$ we have $f\in A$. We have $f_{n}\in X$ and
$f_{n}\in M\setminus Y$, and since both $X$ and $M\setminus Y$ are closed we
thus have $f\in X$ and $f\in M\setminus Y$. Thus $f\in A\cap(X\setminus Y)$,
which completes the proof.

For the final claim, suppose that $A_{n}\rightarrow A$ in $\mathcal{K}(M)$ and
that $C_{n}\in\mathscr{C}$ with $C_{n}\subseteq A_{n}$. Using Lemma A.1.1 the
set $A\cup(\cup_{n}A_{n})$ is compact, which implies that $\cup_{n}C_{n}$ is
relatively compact, so we may pass to a subsequence and assume
$C_{n}\rightarrow C\in\mathcal{K}(M)$. As $C_{n}\subseteq A_{n}$ we have
$C\subseteq A$ and since $\mathscr{C}$ is closed we have $C\in\mathscr{C}$.
$\blacksquare$

###### Lemma A.1.3

If $F:M\rightarrow M$ is continuous then the map from $\mathcal{K}(M)$ to
itself given by $A\mapsto\\{F(f)\,;\,f\in A\\}$ is continuous.

###### Proof.

Note that if $F$ is uniformly continuous then the conclusion is clear from the
definition of the Hausdorff metric, see (A.1). For general continuous $F$,
note by Lemma A.1.1 that if $A_{n}\rightarrow A$ in $\mathcal{K}(M)$ then the
set $(\cup_{n}A_{n})\cup A$ is compact and thus the restriction of $F$ to this
set is uniformly continuous. The result follows. $\blacksquare$

###### Lemma A.1.4

Let $A_{n},B_{n},A,B\in\mathcal{K}(M)$ with $A_{n}\subseteq B_{n}$ for all
$n$. If $A_{n}\rightarrow A$ and $B_{n}\rightarrow B$ then $A\subseteq B$.

###### Proof.

Let $a\in A$. Since $A_{n}\rightarrow A$ there exists $a_{n}\in A_{n}$ such
that $a_{n}\rightarrow a$. Thus $a_{n}\in B_{n}$. Since $B_{n}\rightarrow B$
we thus have $a\in B$. $\blacksquare$

###  A.2 On the M1 topology of $\Pi$ and $\mathcal{K}(\Pi)$

The book of Whitt (2002) details relative compactness and weak convergence for
real valued stochastic processes (i.e. single càdlàg paths) in all four
Skorohod topologies. In Freeman and Swart (2023) we introduce a unified
framework for these four topologies, suitable for random sets of càdlàg paths.
We recall some properties of the M1 version of this framework here.

Compact subsets of $\Pi$ play a key role in our main results and, for this
reason, we require corresponding criteria for relative compactness (of subsets
of $\Pi$) and tightness (relating to convergence in law of $\mathcal{K}(\Pi)$
valued random variables). Fix a metric $d_{\overline{\mathbb{R}}}$ generating
the topology on $\overline{\mathbb{R}}$ and for
$A\subseteq\overline{\mathbb{R}}$ let us write
$\operatorname{dist}_{\overline{\mathbb{R}}}(x,A)=\inf\\{d_{\overline{\mathbb{R}}}(x,y)\,;\,y\in
A\\}$. Relative compactness for sets of continuous paths is often
characterised using the modulus of continuity, see for example Theorem 7.2 of
Billingsley (1995). The analogous object for the M1 topology is

$\displaystyle
w_{T,\delta}(f)=\sup\Big{\\{}\operatorname{dist}_{\overline{\mathbb{R}}}\big{(}f(t_{2}\star_{2}),[f(t_{1}\star_{1}),f(t_{3}\star_{3})]\big{)}\,;\,\;$
$\displaystyle t_{1}\star_{1},t_{2}\star_{2},t_{3}\star_{3}\in
I(f)_{\mathfrak{s}},$ $\displaystyle-T\leq t_{1}<t_{2}<t_{3}\leq T,\
t_{3}-t_{1}<\delta\Big{\\}},$ (A.2)

with the conventions that the supremum over the empty set is zero, and
$[a,b]=[a\wedge b,a\vee b]$.

###### Proposition A.2.1

The following hold:

  1. 1.

A subset ${\cal A}\subseteq\Pi$ is relatively compact if and only if for all
$0<T<\infty$

$\lim_{\delta\rightarrow 0}\sup_{f\in{\cal A}}w_{T,\delta}(f)=0.$

  2. 2.

A subset $\mathscr{A}\subseteq\mathcal{K}(\Pi)$ is relatively compact if and
only if for all $0<T<\infty$

$\lim_{\delta\rightarrow 0}\sup_{{\cal A}\in\mathscr{A}}\sup_{f\in{\cal
A}}w_{T,\delta}(f)=0.$

  3. 3.

A sequence of $\Pi$ valued random variables $(f_{n})$ is tight, in the sense
that their laws comprise a relatively compact sequence of probability measures
on $\Pi$, if and only if for all $0<T<\infty$ and $\epsilon>0$ we have

$\lim_{\delta\rightarrow
0}\limsup_{n\rightarrow\infty}\mathbb{P}\left[w_{T,\delta}(f_{n})\geq\epsilon\right]=0.$

  4. 4.

A sequence of $\mathcal{K}(\Pi)$ valued random variables $(\mathcal{A}_{n})$
is tight, in the sense that their laws comprise a relatively compact sequence
of probability measures on $\mathcal{K}(\Pi)$, if and only if for all
$0<T<\infty$ and $\epsilon>0$ we have

$\lim_{\delta\rightarrow
0}\limsup_{n\rightarrow\infty}\mathbb{P}\left[\sup_{f\in\mathcal{A}_{n}}w_{T,\delta}(f)\geq\epsilon\right]=0.$
(A.3)

###### Proof.

Part 1 follows from the relative compactness criteria given in Theorem 3.7 of
Freeman and Swart (2023). In the language of that theorem, compact containment
is automatic as $\overline{\mathbb{R}}$ is compact, and the Skorohod-
equiconinuity requirement thus becomes part 1 above. Note that, by Lemma
A.1.1, parts 1 and 2 of Proposition A.2.1 are in fact equivalent to each
other. Note also that part 3 follows from part 4 by taking
$\mathcal{A}_{n}=\\{f_{n}\\}$. Therefore, to complete the proof of Proposition
A.2.1 it suffices to deduce part 4 as a consequence of part 2. (In fact, a
similar argument deduces part 3 from part 1.)

Suppose first that $(\mathcal{A}_{n})$ is tight. That is, for each $\kappa>0$
there exists a compact set $\mathscr{B}\subseteq\mathcal{K}(\Pi)$ such that
$\liminf_{n\rightarrow\infty}\mathbb{P}[\mathcal{A}_{n}\in\mathscr{B}]\geq
1-\kappa$. By part 2, for any $\epsilon>0$ and $T\in(0,\infty)$ there exists
$\delta_{0}>0$ such that for all $\delta\in(0,\delta_{0})$ we have
$\sup_{B\in\mathscr{B}}\sup_{f\in B}w_{T,\delta}(f)\leq\epsilon$. Thus
$\liminf_{n\rightarrow\infty}\mathbb{P}[\sup_{f\in\mathcal{A}_{n}}w_{T,\delta}(f)\leq\epsilon]\geq
1-\kappa$.

It remains to show the reverse implication. Let $(\mathcal{A}_{n})$ satisfy
(A.3) and let $\kappa>0$. Note that $w_{T,\delta}(f)$ is an increasing
function of both $\delta$ and $T$. By (A.3), for each $k\in\mathbb{N}$ set
$\epsilon=1/k$ and choose $\delta_{k}>0$ and $T_{k}\in(0,\infty)$ such that

$\limsup_{n\rightarrow\infty}\mathbb{P}[\mathcal{A}_{n}\setminus
B_{k}\neq\emptyset]\leq\epsilon 2^{-k}$ (A.4)

where $B_{k}=\\{f\in\Pi\,;\,w_{T,\delta}(f)\leq k^{-1}\\}$. Note that
$B_{k+1}\subseteq B_{k}$ and let $B=\cap_{k}B_{k}$. Then
$\lim_{\delta\downarrow 0}\sup_{f\in B}w_{T,\delta}(f)=0$, so part 2 gives
that $\mathscr{B}=\\{X\in\mathcal{K}(\Pi)\,;\,X\subseteq B\\}$ is a compact
subset of $\mathcal{K}(\Pi)$. We have that

$\displaystyle\textstyle\limsup_{n}\mathbb{P}[\mathcal{A}_{n}\setminus
B\neq\emptyset]$
$\displaystyle=\textstyle\limsup_{n}\mathbb{P}[\cup_{k}(\mathcal{A}_{n}\setminus
B_{k})\neq\emptyset]$
$\displaystyle\leq\textstyle\limsup_{n}\sum_{k}\mathbb{P}[\mathcal{A}_{n}\setminus
B_{k}\neq\emptyset]$
$\displaystyle\leq\textstyle\sum_{k}\limsup_{n}\mathbb{P}[\mathcal{A}_{n}\setminus
B_{k}\neq\emptyset]$ $\displaystyle\leq\epsilon.$

In the above, the third line uses the reverse Fatou lemma and the final line
uses (A.4). Noting that $\\{\mathcal{A}_{n}\setminus
B=\emptyset\\}=\\{\mathcal{A}_{n}\subseteq
B\\}=\\{\mathcal{A}_{n}\in\mathscr{B}\\}$ we thus obtain
$\limsup_{n}\mathbb{P}[\mathcal{A}_{n}\in\mathscr{B}]\geq 1-\epsilon$. Thus
$(\mathcal{A}_{n})$ is tight, which completes the proof. $\blacksquare$

The following three lemmas are consequences of Propositions 2.2.1 and (part 1
of) A.2.1. They cover most of our interaction with the M1 topology within the
present article. Recall from Section 2.2 that for $f\in\Pi$ there is a natural
total order $\sqsubseteq$ on the interpolated graph $H(f)$. We write the
associated strict order relation as $\sqsubset$. We slightly extend the
terminology introduced in (2.8): if $w,z\in H(f)$ with $w\sqsubseteq z$ in the
total order on $H(f)$, we write $f|_{[w,z]}$ for the unique $g\subseteq f$
such that $g\in\Pi$ begins at $w$ and ends at $z$.

###### Lemma A.2.2

Let $f_{n},f\in\Pi$. Suppose that $f_{n}\rightarrow f$, and that
$w_{n},z_{n}\in H(f_{n})$ with $z_{n}\rightarrow z$ and $w_{n}\rightarrow w$,
and suppose that $w_{n}\sqsubseteq z_{n}$ in the induced order from
$H(f_{n})$. Then $w,z\in H(f)$, with $w\sqsubseteq z$ in the induced order
from $H(f)$, and $f_{n}|_{[w_{n},z_{n}]}\rightarrow f|_{[w,z]}$.

In particular, if $z_{n}\in H(f_{n})$ with $f_{n}\rightarrow f$ and
$z_{n}\rightarrow z$ then $z\in H(f)$.

###### Proof.

The second claim follows immediately from the first, so we will prove the
first. Let $g_{n}=f_{n}|_{[w_{n},z_{n}]}$ and $g=f_{[w,z]}$. From the
hypothesis of the lemma $(f_{n})$ is a relatively compact sequence in
$\mathcal{K}(\Pi)$. Since $g_{n}\subseteq f_{n}$ it follows from part 1 of
Proposition A.2.1 that $(g_{n})$ is also relatively compact. To establish the
present lemma it therefore suffices to show that any limit point of $(g_{n})$
is equal to $g$.

Let $g^{\prime}$ be a limit point of $(g_{n})$ and, with slight abuse of
notation, let us pass to a subsequence and assume that $g_{n}\rightarrow
g^{\prime}$, in the M1 topology. From Proposition 2.2.1 we thus have
$d_{\mathcal{K}((\mathbb{R}^{2}_{\rm
c})^{2})}(H^{(2)}(g_{n}),H^{(2)}(g^{\prime}))\rightarrow 0$. As we noted in
comments above Proposition 2.2.1, it is trivial to see that this implies
$d_{\mathcal{K}(\mathbb{R}^{2}_{\rm c})}(H(g_{n}),H(g^{\prime}))\rightarrow
0$. Moreover $(h,h^{\prime})\mapsto d_{\mathcal{K}(\mathbb{R}^{2}_{\rm
c})}(H(h),H(h^{\prime}))$ is the Hausdorff metric on $\Pi$, generating a
coarser topology than the M1 topology; as discussed in Section 3.4 of Freeman
and Swart (2023) the topology generated is Skorohod’s M2 topology. It follows
immediately by uniqueness of limits that if
$d_{\mathcal{K}(\mathbb{R}^{2}_{\rm c})}(H(g_{n}),H(g))\rightarrow 0$ then
$d_{\mathcal{K}((\mathbb{R}^{2}_{\rm
c})^{2})}(H^{(2)}(g_{n}),H^{(2)}(g))\rightarrow 0$. Therefore, to establish
the present lemma we need only show that $H(g^{\prime})=H(g)$.

Consider $v\in H(g^{\prime})$. As $H(g_{n})\rightarrow H(g^{\prime})$ there
exists $v_{n}\in H(g_{n})$ such that $v_{n}\rightarrow v$. Since
$g_{n}\subseteq f_{n}$ we thus have $v_{n}\in H(f_{n})$ with $w_{n}\sqsubseteq
v_{n}\sqsubseteq z_{n}$. Thus $(w_{n},v_{n}),(v_{n},z_{n})\in H^{(2)}(f_{n})$.
Since $H^{(2)}(f_{n})\rightarrow H^{(2)}(f)$ we thus obtain $(w,v),(v,z)\in
H^{(2)}(f)$, which implies that $v\in H(f)$ with $w\sqsubseteq v\sqsubseteq
z$. Hence $v\in H(g)$. We thus obtain $H(g^{\prime})\subseteq H(g)$.

It remains to show the reverse inclusion. Consider $v\in H(g)$. Thus $v\in
H(f)$ with $w\sqsubseteq v\sqsubseteq z$, which means that $(w,v),(v,z)\in
H^{(2)}(f)$. Hence there exists
$(w^{\prime}_{n},v_{n}),(v_{n},z^{\prime}_{n})\in H^{(2)}(f_{n})$ such that
$(w^{\prime}_{n},v_{n})\rightarrow(w,v)$ and
$(v_{n},z^{\prime}_{n})\rightarrow(v,z)$. If $v=w$ then $w_{n}\rightarrow v$
and as $w_{n}\in H(g_{n})$ we obtain $v\in H(g^{\prime})$. Similarly, if $v=z$
then $z\in H^{\prime}(g)$. Without loss of generality we may therefore assume
that $w\sqsubset v\sqsubset z$.

Note that $w_{n}\vee w^{\prime}_{n}$ and $z_{n}\wedge z^{\prime}_{n}$, where
$\wedge$ and $\vee$ respectively denote $\min$ and $\max$ under $\sqsubseteq$,
are elements of $H(f_{n})$. Moreover $w_{n}\vee w^{\prime}_{n}\rightarrow w$
and $z_{n}\wedge z^{\prime}_{n}\rightarrow z$. If $v_{n}\sqsubseteq w_{n}\vee
w^{\prime}_{n}$ for infinitely many $n\in\mathbb{N}$ then $(v_{n},w_{n}\vee
w^{\prime}_{n})\in H^{(2)}(f_{n})$ for such $n$ and we would have $(v,w)\in
H^{(2)}(f_{n})$, meaning that $v\sqsubseteq w$, which is a contradiction.
Hence $w_{n}\vee w^{\prime}_{n}\sqsubseteq v_{n}$ for all but finitely many
$n$. Similarly $v_{n}\sqsubseteq z_{n}\wedge z^{\prime}_{n}$. Thus
$w_{n}\sqsubseteq v_{n}\sqsubseteq z_{n}$ for all but finitely many $n$. For
such $n$ we have $v_{n}\in H(g_{n})$, which implies that $v\in H(g)$. We
therefore obtain $H(g^{\prime})\subseteq H(g)$, so in fact
$H(g^{\prime})=H(g)$, which completes the proof. $\blacksquare$

###### Lemma A.2.3

Let $f_{n},f\in\Pi$. Suppose that $f_{n}\rightarrow f$ and $t_{n}\in I(f_{n})$
with $t_{n}\rightarrow t\in\mathbb{R}$. For each $n$ let
$\star_{n}\in\\{-,+\\}$. Then $f\in I(f)$ and $f(t-)\wedge
f(t+)\leq\liminf_{n}f_{n}(t_{n}\star_{n})\leq\limsup_{n}f_{n}(t_{n}\star_{n})\leq
f(t-)\vee f(t+).$

###### Proof.

Let $x$ be any subsequential limit point of
$(f_{n}(t_{n}\star_{n}))_{n\in\mathbb{N}}$, which means that $(x,t)$ is a
subsequential limit point of $(f_{n}(t_{n}\star_{n}),t_{n})$. Note that
$(f_{n}(t_{n}\star_{n}),t_{n})\in H(f_{n})$. By Lemma A.2.2 we thus have
$(x,t)\in H(f)$, which implies that $x\in[f(t-)\wedge f(t+),f(t-)\vee f(t+)]$.
The result follows. $\blacksquare$

###### Lemma A.2.4

Let $f_{n},f\in\Pi$, with $f_{n}\rightarrow f$ and $t\star\in
I(f)_{\mathfrak{s}}$. There exists $t_{n}\in\mathbb{R}$ such that
$t_{n}\rightarrow t$ and $f_{n}(t_{n}\star)\rightarrow f(t\star)$.

###### Proof.

Noting that $H(f_{n})\rightarrow H(f)$, take $(x_{n},t_{n})\in H(f_{n})$ such
that $(x_{n},t_{n})\rightarrow(f(t\star),t)$. Then
$(f_{n}(t_{n}-),t_{n})\sqsubseteq(x_{n},t_{n})\sqsubseteq(f_{n}(t_{n}+),t_{n})$.
By compactness of $\overline{\mathbb{R}}$ we may, without loss of generality,
pass to a subsequence along which both $f_{n}(t_{n}-)$ and $f_{n}(t_{n}+)$
converge, say $f_{n}(t_{n}-)\rightarrow a$ and $f_{n}(t_{n}+)\rightarrow b$.
Hence, by Lemma A.2.2 we have $(a,t)\in H(f)$ and $(b,t)\in H(f)$ with
$(f(t-),t)\sqsubseteq(a,t)\sqsubseteq(f(t\star),t)\sqsubseteq(b,t)\sqsubseteq(f(t+),t)$.
If $\star=-$ then $(a,t)=(f(t\star),t)$ and $f_{n}(t_{n}-)\rightarrow f(t-)$.
If $\star=+$ then $(b,t)=(f(t\star),t)$ and $f_{n}(t_{n}+)\rightarrow f(t+)$.
$\blacksquare$

###  A.3 On measurability in $\mathcal{K}(\Pi)$

In this section we establish that various basic maps involving
$\mathcal{K}(\Pi)$ are measurable. Such things are required to work with
$\mathcal{K}(\Pi)$ valued random variables in Section 5. The vast majority of
the work involved in this section is Lemma A.3.1, which is also used in the
proof of Lemma 4.2.5.

Recall that we use the Borel $\sigma$-fields on $\mathbb{R}^{2}_{\rm c}$,
$\Pi$ and $\mathcal{K}(\Pi)$. These $\sigma$-fields are generated in each case
by the closed (or equivalently, open) subsets. Due to our focus on compactness
it is helpful to work with closed sets whenever possible. Recall that
$d_{\Pi}$, generating the M1 topology on $\Pi$, is defined via Proposition
2.2.1 and the corresponding Hausdorff metric on $\mathcal{K}(\Pi)$ is defined
via (A.1).

###### Lemma A.3.1

The map $(A,z)\mapsto A(z)$ from $\mathcal{K}(\Pi)\times\mathbb{R}^{2}_{\rm
c}\rightarrow\mathcal{K}(\Pi)$ is measurable.

###### Proof.

The proof is rather technical. Recall that $A(z)=\\{f\in A\,;\,z\in H(f)\\}$.
Note that Proposition 2.2.1 gives that $A(z)$ is a closed subset of $\Pi$,
which implies compactness since $A(z)$ is a subset of the compact set $A$. As
both $\mathcal{K}(\Pi)$ and $\mathbb{R}^{2}_{\rm c}$ are separable, in order
to establish measurability of $(A,z)\mapsto A(z)$ it suffices to show that the
marginal maps $A\mapsto A(z)$ and $z\mapsto A(z)$ are both measurable. We
split the proof into these two parts, which will be proved independently. In
each part we will show that the pre-image of a closed subset of
$\mathcal{K}(\Pi)$ is measurable; we will represent this pre-image explicitly
using countably many set operations on measurable subsets. Lemma A.1.2
provides a supply of measurable (in fact, closed) subsets of
$\mathcal{K}(\Pi)$. The fact that a closed graph $H(f)$ is measurable (in
fact, compact) provides a supply of measurable subsets of $\mathbb{R}^{2}_{\rm
c}$.

Measurability of $A\mapsto A(z)$: Fix $z\in\mathbb{R}^{2}_{\rm c}$ and denote
this map by $M_{z}:\mathcal{K}(\Pi)\rightarrow\mathcal{K}(\Pi)$, so
$M_{z}(A)=A(z)$. For $z\in\mathbb{R}^{2}_{\rm c}$ let us write
$\Pi_{z}=\\{f\in\Pi\,;\,z\in H(f)\\}$ and $\mathcal{K}(\Pi_{z})$ for the set
of compact subsets of $\Pi_{z}$. Proposition 2.2.1 implies that $\Pi_{z}$ is a
closed subset of $\Pi$, from which part 1 of Lemma A.1.2 gives that
$\mathcal{K}(\Pi_{z})$ a closed subset of $\mathcal{K}(\Pi)$. Note that
$M_{z}$ maps into $\mathcal{K}(\Pi_{z})$. It therefore suffices to show that
the pre-image of a closed subset of $\mathcal{K}(\Pi_{z})$ is measurable. To
this end, let $\mathscr{C}\subseteq\mathcal{K}(\Pi_{z})$ be closed and let
$\mathscr{C}^{\prime}$ be a dense countable subset of $\mathscr{C}$. We will
show that the following are equivalent:

  1. 1.

$A(z)\in\mathscr{C}$;

  2. 2.

for all $\epsilon>0$ there exists $\delta>0$ and $C\in\mathscr{C}^{\prime}$
such that $\big{(}A\cap\Pi_{z}^{(\delta)}\big{)}\setminus
C^{[\epsilon]}=\emptyset$ and $C\subseteq A^{[\epsilon]}$.

We first give the forwards implication (1)$\Rightarrow$(2). Let us assume
$A(z)\in\mathscr{C}$ and suppose that (2) fails, in preparation for an
argument by contradiction. Then there exists $\epsilon>0$ such that for all
$C\in\mathscr{C}^{\prime}$ and all $\delta>0$ it holds that
$\big{(}A\cap\Pi_{z}^{(\delta)}\big{)}\setminus C^{[\epsilon]}\neq\emptyset$
or $C\nsubseteq A^{[\epsilon]}$. We have $A(z)\in\mathscr{C}$ so we may choose
$C\in\mathscr{C}^{\prime}$ such that

$d_{\mathcal{K}(\Pi)}(C,A(z))\leq\tfrac{\epsilon}{2}.$ (A.5)

Hence $C\subseteq A(z)^{[\epsilon]}$, which implies $C\subseteq
A^{[\epsilon]}$. Taking $\delta=\frac{1}{n}$ we thus have that there exists an
infinite subsequence of $n\in\mathbb{N}$ for which
$\big{(}A\cap\Pi_{z}^{(1/n)}\big{)}\setminus C^{[\epsilon]}\neq\emptyset$. For
such $n$, take $f_{n}\in\big{(}A\cap\Pi_{z}^{(1/n)}\big{)}\setminus
C^{[\epsilon]}$. We thus have $f_{n}\notin C^{[\epsilon]}$, which by (A.5)
implies that $f_{n}\notin A(z)^{[\epsilon/2]}$. We also have that $f_{n}\in
A\cap\Pi_{z}^{(1/n)}$ which, by compactness of $A$ implies that $(f_{n})$ has
a subsequential limit $f\in A(z)$. However, this contradicts the conclusion
reached in the previous sentence. Thus (1)$\Rightarrow$(2).

Let us now establish the reverse implication (2)$\Rightarrow$(1). Take
$\epsilon=\frac{1}{n}$ and take $\delta=\delta_{n}>0$ and
$C=C_{n}\in\mathscr{C}^{\prime}$ as given from (2). That is, we have

$C_{n}\subseteq A^{[1/n]}\qquad\text{ and }\qquad
A\cap\Pi_{z}^{(\delta_{n})}\subseteq C_{n}^{[1/n]}.$ (A.6)

From the first statement in (A.6), since $C_{n}\in\mathcal{K}(\Pi_{z})$ we
have $C_{n}\subseteq A^{[1/n]}\cap\Pi_{z}\subseteq A(z)^{[1/n]}$. It is
automatic that $A(z)\subseteq A\cap\Pi_{z}^{(\delta_{n})}$ so from the second
statement in (A.6) we obtain $A(z)\subseteq C_{n}^{[1/n]}$. Putting these
together, we obtain $d_{K(\Pi)}(A(z),C_{n})\leq\frac{1}{n}$. Thus
$C_{n}\rightarrow A(z)$ as $n\rightarrow\infty$. Since $\mathscr{C}$ is closed
we thus obtain $A(z)\in\mathscr{C}$. Thus (2)$\Rightarrow$(1).

We now have that (1)$\Leftrightarrow$(2). It follows that

$\displaystyle M_{z}^{-1}(\mathscr{C})$
$\displaystyle=\bigcap_{\epsilon>0}\bigcup_{\delta>0}\bigcup_{C\in\mathscr{C}^{\prime}}\left\\{B\in\mathcal{K}(\Pi)\,;\,\left(B\cap\Pi_{z}^{(\delta)}\right)\setminus
C^{[\epsilon]}=\emptyset\right\\}\cap\left\\{B\in\mathcal{K}(\Pi)\,;\,C\subseteq
B^{[\epsilon]}\right\\}$
$\displaystyle=\bigcap_{n\in\mathbb{N}}\bigcup_{m\in\mathbb{N}}\bigcup_{C\in\mathscr{C}^{\prime}}\left\\{B\in\mathcal{K}(\Pi)\,;\,B\cap\left(\Pi_{z}^{(1/m)}\setminus
C^{[1/n]}\right)=\emptyset\right\\}\cap\left\\{B\in\mathcal{K}(\Pi)\,;\,C\subseteq
B^{[1/n]}\right\\}$

Note that we have used the set algebraic identity $X\cap(Y\setminus Z)=(X\cap
Y)\setminus Z$. By parts 1 and 2 of Lemma A.1.2 the last line of the above
consists of countable unions and intersections of closed subsets of
$\mathcal{K}(\Pi)$. Thus $M_{z}$ is measurable.

Measurability of $z\mapsto A(z)$: We will recycle some parts of our notation,
to preserve the symmetry between this argument the above. Fix
$A\in\mathcal{K}(\Pi)$ and let us denote the map in question by
$M_{A}:\mathbb{R}^{2}_{\rm c}\rightarrow\mathcal{K}(\Pi)$, so $M_{A}(z)=A(z)$.
For any $z\in\mathbb{R}^{2}_{\rm c}$ we have that $M_{A}(z)\in\mathcal{K}(A)$,
where $\mathcal{K}(A)$ denotes the set of all compact subsets of $A$. Part 1
of Lemma A.1.2 gives that $\mathcal{K}(A)$ is a closed subset of
$\mathcal{K}(\Pi)$. It is straightforward to check that if
$B\in\mathcal{K}(A)$ then

$\displaystyle M_{A}^{-1}(B)$ $\displaystyle=\left(\bigcap_{f\in
B}H(f)\right)\setminus\left(\bigcup_{f\in A\setminus B}H(f)\right).$ (A.7)

In words, equation (A.7) says that $M_{A}(z)=B$ if and only if $B$ is
precisely the set of $f\in A$ such that $z\in H(f)$. In order to establish
measurability of $M_{A}$ it suffices to fix a closed subset $\mathscr{C}$ of
$\mathcal{K}(A)$ and show that $M_{A}^{-1}(\mathscr{C})$ is a measurable
subset of $\mathbb{R}^{2}_{\rm c}$. Let $\mathscr{C}^{\prime}$ be a dense
countable subset of $\mathscr{C}$. We will show that the following are
equivalent:

  1. 3.

$A(z)\in\mathscr{C}$;

  2. 4.

for all $\epsilon>0$ there exists $C\in\mathscr{C}^{\prime}$ such that
$z\in\left(\bigcap_{f\in
C}H(f)^{[\epsilon]}\right)\setminus\left(\bigcup_{f\in A\setminus
C^{[\epsilon]}}H(f)\right)$.

We first give the forwards implication (3)$\Rightarrow$(4). Suppose that
$A(z)\in\mathscr{C}$, that is $M_{A}(z)\in\mathscr{C}$. Then there exists
$B\in\mathscr{C}$ such that $z\in M_{A}^{-1}(B)$. Choose
$C\in\mathscr{C}^{\prime}$ such that $d_{\mathcal{K}(\Pi)}(B,C)\leq\epsilon$.
From (A.7) we have that $z\in\bigcap_{f\in B}H(f)$. It is straightforward to
check that $\bigcap_{f\in B}H(f)\subseteq\bigcap_{f\in
B^{[\epsilon]}}H(f)^{[\epsilon]}$ and as $C\subseteq B^{[\epsilon]}$ we obtain
that $z\in\bigcap_{f\in C}H(f)^{[\epsilon]}$. Similarly, we have $B\subseteq
C^{[\epsilon]}$, so $A\setminus C^{[\epsilon]}\subseteq A\setminus B$ and from
(A.7) we have $z\notin\bigcup_{f\in A\setminus B}H(f)\supseteq\bigcup_{f\in
A\setminus C^{[\epsilon]}}H(f)$. We have thus obtained that
(3)$\Rightarrow$(4).

Let us now establish the reverse implication (4)$\Rightarrow$(3). Take
$\epsilon=\frac{1}{n}$ and let $C=C_{n}$ be as given from (4). Noting that
$C_{n}\subseteq A$ for all $n\in\mathbb{N}$, Proposition A.2.1 gives that the
sequence $(C_{n})_{n\in\mathbb{N}}$ is relatively compact (as sequence of
elements of $\mathcal{K}(\Pi)$), and thus has a convergence subsequence. With
slight abuse of notation let us pass to this convergent subsequence and set
$B=\lim_{n}C_{n}$. It follows immediately that $B\in\mathscr{C}$.

From (4) we have $z\in\bigcap_{f\in C_{n}}H(f)^{[\epsilon_{n}]}$. For each
$g\in B$ there exists $g_{n}\in C_{n}$ such that $g_{n}\rightarrow g$. As
$g_{n}\in C_{n}$ we have $z\in H(g_{n})^{[\epsilon_{n}]}$. Proposition 2.2.1
gives that $H(g_{n})\rightarrow H(g)$ in $\mathcal{K}(\mathbb{R}^{2}_{\rm
c})$, which implies that $H(g_{n})^{[\epsilon_{n}]}\rightarrow H(g)$. It
follows immediately that $z\in H(g)$, and as $g\in B$ was arbitrary we have
$z\in\bigcap_{g\in B}H(g)$.

Similarly, from (4) we have $z\notin\bigcup_{f\in A\setminus
C_{n}^{[\epsilon_{n}]}}H(f)$ for all $n$. Take $g\in A\setminus B$ and note
that because $B\subseteq\Pi$ is closed we have
$d_{\mathcal{K}(\Pi)}(\\{g\\},B)>0$. As $C_{n}\rightarrow B$ we have also that
$C_{n}^{[\epsilon_{n}]}\rightarrow B$, so for sufficiently large $n$ we have
$g\notin C_{n}^{[\epsilon_{n}]}$. Hence $z\notin H(g)$. As $g\in B$ was
arbitrary we thus have $g\notin\bigcup_{g\in A\setminus B}H(g)$. Putting this
together with the conclusion of the previous paragraph and (A.7) we obtain
that $z\in M_{A}^{-1}(B)$. Thus (4)$\Rightarrow$(3).

We now have that (3)$\Leftrightarrow$(4). It follows that

$\displaystyle M_{A}^{-1}(\mathscr{C})$
$\displaystyle=\bigcap_{\epsilon>0}\bigcup_{C\in\mathscr{C}^{\prime}}\left(\left(\bigcap_{f\in
C}H(f)^{[\epsilon]}\right)\setminus\left(\bigcup_{f\in A\setminus
C^{[\epsilon]}}H(f)\right)\right)$
$\displaystyle=\bigcap_{n\in\mathbb{N}}\bigcup_{C\in\mathscr{C}^{\prime}}\left(\left(\bigcap_{f\in
C}H(f)^{[1/n]}\right)\setminus\left(\bigcup_{f\in A\setminus
C^{[1/n]}}H(f)\right)\right)$ (A.8)

We now examine the two terms in brackets on the right hand side of (A.8). The
set $\bigcap_{f\in C}H(f)^{[1/n]}$ is an intersection of closed sets, and is
therefore closed. Note also that

$\displaystyle\bigcup_{f\in A\setminus
C^{[1/n]}}H(f)=\bigcup_{m\in\mathbb{N}}\left(\bigcup_{f\in A\setminus
C^{(1/n+1/m)}}H(f)\right).$ (A.9)

We claim that $\bigcup_{f\in E}H(f)$ is closed whenever $E\subseteq\Pi$ is
compact; to see this take $x_{n}\in H(f_{n})$ where $f_{n}\in E$ and
$x_{n}\rightarrow x\in\mathbb{R}^{2}_{\rm c}$, pass to a convergence
subsequence $f_{n}\rightarrow f\in E$ and then by Proposition 2.2.1 we have
$H(f_{n})\rightarrow H(f)$ so $x\in H(f)$. In particular, as $C^{(\delta)}$ is
open for all $\delta>0$ we have that $A\setminus C^{(\delta)}$ is compact, so
the right hand side of (A.9) is a countable union of closed sets. We have now
shown that (A.8) represents $M_{A}^{-1}(\mathscr{C})$ using countably many set
operations of measurable subsets of $\mathbb{R}^{2}_{\rm c}$. Thus $M_{A}$ is
measurable. $\blacksquare$

###### Lemma A.3.2

Let $z\in\mathbb{R}^{2}_{\rm c}$. The map $(A,z)\mapsto A|_{z}$ is a
measurable map from $\mathcal{K}(\Pi)\times\mathbb{R}^{2}_{\rm
c}\rightarrow\mathcal{K}(\Pi)$.

###### Proof.

Recall that $A|_{z}=\\{f|_{z}\,;\,f\in A(z)\\}$. As both $\mathcal{K}(\Pi)$
and $\mathbb{R}^{2}_{\rm c}$ are separable, in order to establish
measurability of $(A,z)\mapsto A|_{z}$ it suffices to show that the marginal
maps $A\mapsto A|_{z}$ and $z\mapsto A|_{z}$ are both measurable. Recall
$\Pi_{z}=\\{f\in\Pi\,;\,z\in H(f)\\}$ from the proof of Lemma A.3.1, in which
we showed that $\Pi_{z}\subseteq\Pi$ and
$\mathcal{K}(\Pi_{z})\subseteq\mathcal{K}(\Pi)$ were both closed. By Lemma
A.3.1 the map $(A,z)\mapsto A(z)$ is measurable, and it clear that it maps
into $\mathcal{K}(\Pi_{z})$. Recall that for $f\in\Pi_{z}$, $f|_{z}$ is the
unique $g\subseteq f$ such that $(g(\sigma_{g}-),\sigma_{g})=z$, and
$A|_{z}=\\{f|_{z}\,;\,f\in A\cap\Pi_{z}\\}$. By Lemma A.2.2 the map $f\mapsto
f|_{z}$ defined from $\Pi_{z}$ to itself is continuous. It follows from Lemma
A.1.3 that $A\mapsto A|_{z}$ is continuous on $\mathcal{K}(\Pi_{z})$. Since
$A|_{z}=(A(z))|_{z}$, this completes the proof. $\blacksquare$

###### Lemma A.3.3

The map $A\mapsto A_{\uparrow}$ is a continuous map from $\mathcal{K}(\Pi)$ to
itself.

###### Proof.

Let $A_{n},A\in\mathcal{K}(\Pi)$ with $A_{n}\rightarrow A$. Then $(A_{n})$ is
relatively compact and, noting that $g\subseteq f\Rightarrow
w_{T,\delta}(f)\leq w{T,\delta}(f)$, part 2 of Proposition A.2.1 gives that
$((A_{n})_{\uparrow})$ is relatively compact. Let $B\in\mathcal{K}(\Pi)$ be a
limit point of $((A_{n})_{\uparrow})$. We must show that $B=A_{\uparrow}$.
Without loss of generality let us pass to a subsequence and assume that
$(A_{n})_{\uparrow}\rightarrow B$.

Let $g\in B$. Then there exists $f_{n}\in A_{n}$ with $g_{n}\subseteq f_{n}$
and $g_{n}\rightarrow g$. By Lemma A.1.1 the set $\cup_{n}A_{n}$ is relatively
compact, so we may pass to a further subsequence and assume that
$f_{n}\rightarrow f$. Thus $f\in A$. Lemma A.2.2 gives that $g\subseteq f$, so
$g\in A_{\uparrow}$. Hence $B\subseteq A_{\uparrow}$.

It remains to show the reverse inclusion. Let $g\in A_{\uparrow}$. Then there
exists $f\in A$ with $g\subseteq f$. As $A_{n}\rightarrow A$ there exists
$f_{n}\in A_{n}$ such that $f_{n}\rightarrow f$. From Proposition 2.2.1 (and
the remarks just above it) we have $H(f_{n})\rightarrow H(f)$. In particular,
there exists $z_{n}\in H(f_{n})$ such that
$z_{n}\rightarrow(g(\sigma_{g}-),\sigma_{g})\in H(f)$. Let
$g_{n}=f_{n}|_{z_{n}}$. Lemma A.2.2 gives that $g_{n}\rightarrow g$, so $g\in
B$. Hence $A_{\uparrow}\subseteq B$. Thus $A_{\uparrow}=B$ and the proof is
complete. $\blacksquare$

###  A.4 Proof of Lemma 2.5.1

We must show that the Brownian web $\mathcal{W}_{\rm b}$ satisfies our
definition of a web, in Definition 2.4.1. The argument rests on well known
properties of the Brownian web. We noted in Section 2.2 that
$(\Pi^{\uparrow}_{\rm c},d_{\Pi})$ is the state space that is in common usage
for the Brownian web. As $\mathcal{W}_{\rm b}$ is a
$\mathcal{K}(\Pi^{\uparrow}_{\rm c})$ valued random variable, it is also a
$\mathcal{K}(\Pi^{\uparrow})$ valued random variable. We will now refer to
points (a)-(c) of Theorem 2.3 of Schertzer et al. (2017), which defines
$\mathcal{W}_{\rm b}$.

We first show that $\mathcal{W}_{\rm b}$ is a weave. Let
$D\subseteq\mathbb{R}^{2}_{\rm c}$ be dense and countable. Point (a) of this
definition gives that almost surely, $\mathcal{W}_{\rm b}(z)$ is non-empty at
all $z\in D$, which by Lemma A.2.2 implies that $\mathcal{W}_{\rm b}$ is
almost surely pervasive. Moreover since $\mathcal{W}_{\rm b}(z)$ is almost
surely a singleton, $D$ is almost surely non-ramified. It is well known in the
literature that $\mathcal{W}_{\rm b}$ is almost surely non-crossing; strictly
this follows because point (b) of the definition gives that $\mathcal{W}_{\rm
b}(D)$ is non-crossing, point (c) gives $\mathcal{W}_{\rm
b}\stackrel{{\scriptstyle\rm a.s.}}{{=}}\overline{\mathcal{W}_{\rm b}(D)}$,
and the non-crossing property is preserved by taking limits of continuous
paths (this last implication uses Lemma 3.3.7). We have now shown that
$\mathcal{W}_{\rm b}$ is a weave.

Point (c) of the definition gives that $\mathcal{W}_{\rm
b}\stackrel{{\scriptstyle\rm a.s.}}{{=}}\overline{\mathcal{W}_{\rm b}(D)}$,
for any deterministic dense countable $D\subseteq\mathbb{R}^{2}$. Noting our
remarks immediately above the present lemma regarding $\mathcal{W}_{\rm b}(D)$
and $\mathcal{W}_{\rm b}|D$, we have $\mathcal{W}_{\rm
b}\stackrel{{\scriptstyle\rm a.s.}}{{=}}\overline{(\mathcal{W}_{\rm
b}|_{D})}$. Lemma A.2.2 implies that if $A_{n}\rightarrow A$ in
$\mathcal{K}(\Pi)$ then
$(\mathcal{A}_{n})_{\uparrow}\rightarrow\mathcal{A}_{\uparrow}$. It is
straightforward to combine this fact with the usual system of random walk
approximations to the Brownian web (e.g. Figure 9 of Schertzer et al. (2017))
to show that $\mathcal{W}_{\rm b}\stackrel{{\scriptstyle\rm
a.s.}}{{=}}(\mathcal{W}_{\rm b})_{\uparrow}$. Hence in particular
$\mathcal{W}_{\rm b}|_{D}\stackrel{{\scriptstyle\rm
a.s.}}{{=}}(\mathcal{W}_{\rm b}|_{D})_{\uparrow}$, from which (2.12) gives
$\mathcal{W}_{\rm b}\stackrel{{\scriptstyle\rm
a.s.}}{{=}}\operatorname{web}(\mathcal{W}_{\rm b})$. Theorem 2.4.3 thus gives
that $\mathcal{W}_{\rm b}$ is a web.

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