Unordered forest Topology

+,0×D to(T, d_LGHP).

4.3 Crown of a tree

4.3.1 Unordered forest Topology

The aim of this subsection is to define and study unordered forests of height-labelled trees.

The main result of this chapter is Theorem 4.3.13, giving a filtration(Sh)_h∈R of (T, d_LGHP) adapted to growth process (for example in the case of Galton-Watson and L´evy trees), and a filtration(C−h)_h∈R of(T, d_LGHP) adapted to coalescent processes (for example in the case of Kingman’s orΛ-coalescent).

For every h ∈ R, we set 0_h = ({h}, d_{h}, h,0) ∈T, where d{h} is the only distance over the singleton{h}. We considerT˜C the set of all sequences(Tn, dn, Hn, νn)_n∈N^∗ of S-compact height-labelled trees such thatlim_nd_LGHP(T_n,0_h) = 0for some h∈Rand for alln∈N^∗ but possibly one, min_TnHn = h and if there is a n0 such that min_Tn

0Hn0 6= h then Tn0 =∅ or min_Tn

0H_n0 > h. We define, for(T_n)_n∈N^∗,(T_n⁰)_n∈N^∗∈T˜C, d_LGHP^∞ ((T_n)_n∈N^∗,(T_n⁰)_n∈N^∗) = inf

σ∈S(N^∗)

sup

n∈N^∗

d_LGHP(T_n, T_σ(n)⁰ ), (4.3.1) where S(N^∗) is the set of all permutations of N^∗. The function d_LGHP^∞ is non-negative and satisfies the triangular inequality, so it is a pseudo-distance over T˜C. We define TC the quotient of T˜C by the equivalence relation d_LGHP^∞ ((Tn)_n∈N^∗,(T_n⁰)_n∈N^∗) = 0, and call crowns the equivalence classes. Note that(TC, d_LGHP^∞ ) is separable, but not complete. The spaceTC

is Polish though, since the distance

((T_n)_n∈N^∗,(T_n⁰)_n∈N^∗)7→d_LGHP^∞ ((T_n)_n∈N^∗,(T_n⁰)_n∈N^∗)∨ |h−h⁰|,

where0_hand0_h⁰ are the respective limits of(T_n)n∈N^∗ and(T_n⁰)n∈N^∗induces the same topology and makesTC complete.

4.3. CROWN OF A TREE 95 Remark 4.3.1. We give an intuition of the equivalence relation on T˜C. Take two sequences (T_n)_n∈N^∗ and (T_n⁰)_n∈N^∗ elements ofT˜C. We haved_LGHP^∞ ((T_n)_n∈N^∗,(T_n⁰)_n∈N^∗) = 0if and only if(T_n)_n∈N^∗ and(T_n⁰)_n∈N^∗ have the same limit0_h and the terms different from0_h are the same in both sequences. For example, for(Tn)_n∈N^∗ and 0_h = lim_nTn, the following sequences:

• (0_h, T1, T2, T3, T4, T5, ...),

• (0_h, T₁,0_h, T₂,0_h, T₃, ...),

• (T_φ(n))_n∈N^∗, forφa permutation of N^∗,

are all in the equivalence class of(Tn)_n∈N^∗. But adding or removing any term from(Tn)_n∈N^∗ different from0_h would change the equivalence class.

An element ofTC can always be represented as the class of a sequence(Tn, dn, Hn, νn)_n∈N^∗ such that(sup_T

nH_n)_n∈N^∗ is a non-increasing sequence of elements converging to someh∈R. To ease of notation, we will abusively confuse the classes ofTC with the representents inT˜C. If the terms of (Tn)_n∈N^∗ and (T_n⁰)_n∈N^∗ ∈TC are all compact trees, then we can define

d_GHP^∞ ((Tn)_n∈N^∗,(T_n⁰)_n∈N^∗) = inf

σ∈S(N^∗) sup

n∈N^∗

dGHP(Tn, T_σ(n)⁰ ).

Note that, liked_LGHP^∞ , the function d_GHP^∞ is a pseudo-distance, and that d_GHP^∞ has the same zeros as d_LGHP^∞ thanks to Remark 4.1.16. It follows that d_GHP^∞ defines the same quotient as d_LGHP^∞ , sod_GHP^∞ is a distance on the set of all crowns containing only compact terms.

For h∈R,r >|h|and (T_n)_n∈N^∗ ∈TC withlim_nT_n= 0_h, we define Slice_r((T_n)_n∈N^∗) = (Slice_r(T_n))_n∈N^∗.

Lemma 4.3.2. Forh∈R, r >|h| and(T_n)_n∈N^∗ ∈TC with lim_nT_n= 0_h, we have Slice_r((Tn)_n∈N^∗)∈TC.

Proof. Let (Tn, dn, Hn, νn)_n∈N^∗ be an element of TC. Take n ∈ N^∗ such that min_TnHn ∈ [−r, r]. It follows immediately that Slice_r(T_n) contains the root of T_n, so Slice_r(T_n) is non-empty andmin_Slicer(Tn)H_n= min_TnH_n. Sincemin_TnH_n∈[−r, r], we haveH_n(T_n)∩[−r, r] = Hn(Tn)∩(−∞, r]. It follows that

Slice_r(Tn) ={x∈Tn|H_n(x)≤r}.

For everyx, y∈Slice_r(Tn), we havemax_Jx,yKHn=Hn(x)∨Hn(y)≤r, soJx, yK⊂Slice_r(Tn).

We have proven that if min_TnHn ∈ [−r, r], then Slice_r(T) is a tree and min_Slicer(Tn)Hn = min_TnH_n.

If min_TnHn> r or ifTn=∅thenSlice_r(Tn) =∅.

From these two results, we see that for every n∈ N^∗ such that min_TnHn =h ∈[−r, r], Slice_r(T_n) is a tree andmin_Slicer(Tn)H_n = min_TnH_n=h. If min_TnH_n> h or T_n=∅, which happens for at most one index, then we either have min_Slicer(Tn)Hn = min_TnHn > h or Slice_r(Tn) =∅ and in the former case Slice_r(Tn)is a tree withmin_Slicer(Tn)Hn> h.

We deduce thatSlice_r((T_n)_n∈N^∗ is inTC.

To help prove convergences, we adapt Lemma 3.4.3 to TC and prove the following result.

Lemma 4.3.3. Let (Tn, dn, Hn, νn)_n∈N^∗ and ((T_n^k, d^k_n, H_n^k, ν_n^k)_n∈N^∗)_k∈N^∗ be elements of TC. Take h ∈ R (resp h_k ∈ R) such that lim_nT_n = 0_h (resp. lim_nT_n^k = 0_hk) and (r_k)_k∈N^∗ a sequence of positive real numbers such thatlim_kr_k=∞ and for everyk∈N^∗, r_k>|h| ∨ |h_k|.

If

d_GHP^∞ (Slice_rk((T_n)_n∈N^∗),Slice_rk((T_n^k)_n∈N^∗)) −→

k→∞0, (4.3.2)

then, we have:

d_LGHP^∞ ((T_n)_n∈N^∗,(T_n^k)_n∈N^∗) −→

k→∞0.

Proof. Takeε∈(0,1)andhthe real number such thatlim_nT_n= 0_h. The tree0_h is compact, so according to Remark 4.1.16 there exists ε⁰ ∈ (0,1) such that for every compact height-labelled treeT,

d_GHP(T,0_h)≤ε⁰⇒d_LGHP(T,0_h)≤ ε

2 (4.3.3)

and ε⁰⁰ ∈(0,1)such that for everyS-compact height-labelled tree T, d_LGHP(T,0_h)≤ε⁰⁰⇒(T is compact andd_GHP(T,0_h)≤ ε⁰

2). (4.3.4)

By definition of d_GHP^∞ anddGHP, there exists a sequence of permutations ofN^∗: (σk)_k∈N^∗, such that for everyk,

sup

n∈N^∗

d_GHP(Slice_rk(T_n),Slice_rk(T_σ^k

k(n)))

≤d_GHP^∞ (Slice_rk((T_n)_n∈N^∗),Slice_rk((T_n^k)_n∈N^∗)) + ε⁰

8. (4.3.5) Take n₀ ∈ N^∗ such that for every n ≥ n₀, d_LGHP(T_n,0_h) ≤ ε⁰⁰ ∧ ^ε₂. Thanks to Equation (4.3.4), we have

d_GHP(Tn,0_h)≤ ε⁰

2· (4.3.6)

From the⇐ direction of Lemma 3.4.3, we have that for everyn < n0,

k→∞lim d_LGHP(T_σ^k

k(n), Tn) = 0. (4.3.7)

Combining (4.3.7) line with lim_kr_k = ∞ and (4.3.2), we can take k₀ such that for every k≥k0, we have







∀n < n₀, d_LGHP(T_n, T_σ^k

k(n))≤ε r_k>|h|+ε⁰

d_GHP^∞ (Slice_rk((Tn)_n∈N^∗),Slice_rk((T_n^k)_n∈N^∗))≤ ^ε₄⁰·

Let us prove that fork≥k₀,d_LGHP^∞ ((T_n)_n∈N^∗,(T_n^k)_n∈N^∗)≤ε.By choice ofk₀, we already have dLGHP(Tn, T_σ^k

k(n))≤εforn < n0. For n≥n0, we have by choice of σk and (4.3.5) that sup

n≥n0

dGHP(Slice_rk(Tn),Slice_rk(T_σ^k

k(n)))

≤2d_GHP^∞ (Slice_rk((Tn)_n∈N^∗),Slice_rk((T_n^k)_n∈N^∗))≤ ε⁰

2· (4.3.8)

4.3. CROWN OF A TREE 97 From Equation (4.3.6), we havedGHP(Tn,0_h)≤ ^ε₂⁰, so the label function ofTntakes its values in[h−^ε₂⁰, h+^ε₂⁰]. Sincerk >|h|+ε⁰ and the labels of a tree always span an interval, it follows that Tn = Slice_rk(Tn). Using (4.3.8), we find that the label function of Slice_rk(T_σ^k

k(n)) takes its values in[h−ε⁰, h+ε⁰], soSlice_rk(T_σ^k

k(n)) =T_σ^k

k(n). This gives, using (4.3.5) again, that:

d_GHP(T_σ^k

k(n),0_h)≤d_GHP(T_n, T_σ^k

k(n)) +d_GHP(T_n,0_h)≤ε⁰. With Equation (4.3.3), we have d_LGHP(T_σ^k

k(n),0_h) ≤ ^ε₂· By definition of n₀, we also have dLGHP(Tn,0_h)≤ ^ε₂, sodLGHP(T_σ^k

k(n), Tn)≤ε·

We have for k≥k0:

d_LGHP^∞ ((T_n)_n∈N^∗,(T_n^k)_n∈N^∗)≤ sup

n∈N^∗

d_LGHP(T_σ^k

k(n), T_n)≤ε.

We conclude the proof asεis arbitrary.

4.3.2 Crown of a tree

For (T, d, H, ν) a S-compact height-labelled tree, h ∈ H(T), we define the elements of the skeleton at levelh

I_h(T) ={x∈Skel(T)|H(x) =h} (4.3.9) and the collectionCr_h(T) of sub-trees ofT above levelh as

Cr_h(T) = (C_i^h(T))i∈I_h(T), with fori∈ I_h(T),C_i^h(T)∈Tdefined by:

C_i^h(T) = ({x∈T|xi}, d, H,1_H>h·ν). (4.3.10) Forh⁰> h, recalln^h,h0(T) of Definition 4.2.1, and note with remark 4.2.2 thatn^h,h0(T)is the number of indicesi∈ I_h(T)such thatCi(T)reaches height h⁰. Lemma 4.2.3 tells us that n^h,h0(T)is finite. If C_i(T) doesn’t reachh⁰, thenH(C_i(T))⊂[h, h⁰). By definition, it follows that for such an indexi, its total measure is less thanHν((h, h⁰)), so

d_GHP(Ci(T),0_h)≤(h⁰−h)∨Hν((h, h⁰))^h

0↓h

−→0.

This means that I_h(T) is at most countable, and that if I_h(T) is infinite then, thanks to Lemma 3.4.5,

limn d_GHP(C_in(T),0_h) = 0 and lim

n d_LGHP(C_in(T),0_h) = 0

for every enumeration(i_n)_n∈N^∗ of the elements of I_h(T). This allows us to define an object inT˜C similar to Cr:

Crown_h(T) =











(0_h)_n∈N^∗ if T =∅ or sup_T H≤h;

(T,0_h, ...) if min_T H > h;

(Ci1(T), ..., Cin(T),0_h, ...) if I_h(T) ={i₁, ..., in}with distinct i1, ..., in; (Cin(T))_n∈N^∗ if I_h(T) ={i_n}_n∈N^∗ with distinct (in)_n∈N^∗.

Note that with this definition,Crown_h(T) belongs toT˜C since the sequences converge to0_h. We shall denote by Crown_h(T) its equivalence class in TC whose definition does not depend on the choice of the enumeration ofI_h(T). So the mapCrownis defined onR×Tand takes values in TC.

As T is S-compact, then with Lemma 4.2.3, all the (n^h,h0(T))_h⁰_>h are finite, so we can order (C_i^h(T))_i∈I(T) in some order of non-increasing height, and it will converge to0_h.

Our aim for the rest of this section is to prove, after a series of technical lemmas, that the function (h, T) 7→ Crown_h(T) is measurable from (R×T, d_R×d_LGHP) to (TC, d_LGHP^∞ ), see Proposition 4.3.11. We setD6= the set of all (h, T)∈R×T such thatCrown_h(T)6=(0_h)_n∈N^∗. Looking at the definition of Crown, this is equivalent to “there exists x ∈ T such that H(x)> h”.

Lemma 4.3.4. The set D₆₌ is open in R×T.

Proof. Take (h,(T, d, H, ν)) ∈ D6=. By definition of D6= and Crown_h, there exists x ∈ T such that H(x) > h. Take any element (h⁰,(T⁰, d⁰, H⁰, ν⁰))∈R×T that satisfies |h⁰−h|<

1

2(H(x)−h)and

d_LGHP(T, T⁰)< 1

2 1∧(H(x)−h)e^−|H(x)|. Since ¹₂ 1∧(H(x)−h)<1there exists r >|H(x)|such that

dGHP(Slice_r(T),Slice_r(T⁰))< 1

2(H(x)−h)

by definition of d_LGHP. Note that x is in Slice_r(T). By Proposition 3.4.1, there exists a

1

2(H(x)−h)-correspondence A betweenSlice_r(T) and Slice_r(T⁰). Take x⁰ ∈Slice_r(T⁰) such that(x, x⁰)∈A. We have

H⁰(x⁰)−h⁰ ≥H(x)− |H⁰(x⁰)−H(x)| −h− |h⁰−h|

> H(x)−h−21

2(H(x)−h)

= 0.

We haveH⁰(x⁰)> h⁰. By definition ofCrown,(h⁰_k, T^k)∈D6=. Sinceh⁰ andT⁰ were arbitrary in a small ball, we have proven thatD6= is open.

Lemma 4.3.5. Let (h,(T, d, H, ν)),(h⁰,(T⁰, d⁰, H⁰, ν⁰)) ∈ D6=, r ∈ R+, h⁰⁰ ∈ R and δ > 0 be such that h⁰⁰ > h∨h⁰, 0 < δ < h⁰⁰−h⁰ and r > |h| ∨ |h⁰| ∨ |h⁰⁰|. Set T_r = Slice_r(T), T_r⁰ = Slice_r(T⁰). If A⊂T_r×T_r⁰ is a δ-correspondence between T_r and T_r⁰ such that for every (x, x⁰),(y, y⁰)∈A,

H(x)∧H(y)≥h⁰⁰⇒H(x∧y)≥h⇔H⁰(x⁰∧y⁰)≥h⁰, then

d_GHP^∞ (Slice_r(Crown_h(T)),Slice_r(Crown_h⁰(T⁰)))

≤2(h⁰⁰−(h∧h⁰)) +Hν([h∧h⁰−δ, h⁰⁰+ 2δ]). (4.3.11)

4.3. CROWN OF A TREE 99 Proof. Note that Slice_r(Crown_h(T, d, H, ν)) = Slice_r(Crown_h(T, d, H,1_Tr·ν)) and that

[H(1_Slicer(T)·ν)]([h∧h⁰−δ, h⁰⁰+ 2δ]) =Hν([h∧h⁰−δ, h⁰⁰+ 2δ]).

In the Inequality (4.3.11), if we replace(T, d, H, ν) and(T⁰, d,⁰H⁰, ν⁰) by(T, d, H,1_Tr·ν) and (T⁰, d,⁰H⁰,1_T⁰

r·ν⁰), we do not change the left-hand term, and we replace the right-hand term by a smaller (or equal) upper bound. This means that proving the lemma with the additional assumption [Hν∨H⁰ν⁰] (−∞,−r)∪(r,∞)= 0 is sufficient to prove the lemma. Thus, we assume[Hν∨H⁰ν⁰] (−∞,−r)∪(r,∞)= 0 in the rest of the proof.

Step 1: we handle all the non-empty terms of Slice_r(Crown_h(T)) reaching level h⁰⁰ (if any). Taken∈NandC1, ..., Cn the terms ofSlice_r(Crown_h(T))reaching at least levelh⁰⁰. If n= 0, that is if sup_T H < h⁰⁰, there are none and we can directly go to Step 2. In the rest of Step 1, suppose that n >0, that is sup_TrH ≥h⁰⁰. Take x ∈Tr with H(x) ≥h⁰⁰, and the indexisuch that x∈Ci. Take x⁰ ∈T_r⁰ such that(x, x⁰)∈A. Since A is aδ-correspondence, H⁰(x⁰) ≥ H(x)−δ ≥ h⁰⁰−δ > h⁰ so there exists a sub-tree C_i⁰0 of Crown_h⁰(T⁰) such that x⁰ ∈C_i. Take(y, y⁰)∈AwithH(y)≥h⁰⁰, we find thaty⁰ is in the same sub-tree as x⁰ if and only H⁰(x⁰∧y⁰)≥h⁰, which is equivalent by hypothesis toH(x∧y)≥h, that is if and only ifx and y are in the same sub-treeC_i. Thus, we find thatC_i⁰0 is defined independently from the choice of(x, x⁰)and that for1≤i, j≤n,C_i =C_j ⇔C_i⁰0 =C_j⁰0. Therefore, we shall keep the same indexiand write C_i⁰ instead ofC_i⁰0.

Now, we try to build a correspondence between C_i and C_i⁰. Consider

A_i={(x, x⁰)∈A|x∈C_i, H(x)≥h⁰⁰} ∪ {x∈C_i|H(x)< h⁰⁰} × {x⁰∈C_i⁰|H⁰(x⁰)< h⁰⁰+δ}.

Let us prove that A_i is a correspondence betweenC_i and C_i⁰. For x ∈C_i, it is straight-forward to see that there is x⁰ ∈ C_i⁰ such that (x, x⁰) ∈ Ai. Reciprocally for x⁰ ∈ C_i⁰, if H⁰(x⁰) < h⁰⁰ +δ then x⁰ is in correspondence with elements of Ci thanks to the sec-ond term of A_i. If H⁰(x⁰) ≥ h⁰⁰+δ, then there exists x ∈ T_r such that (x, x⁰) ∈ A, and H(x)≥H⁰(x⁰)−δ≥h⁰⁰, sox∈Ci by definition of C_i⁰, and thus (x, x⁰)is in the first term of Ai. SoAi is a correspondence betweenCi andC_i⁰. We compute the distortion ofAi. We find that

sup

(x,x⁰)∈A_i

|H(x)−H⁰(x⁰)| ≤max(δ, h⁰⁰−h, h⁰⁰+δ−h⁰))≤δ+h⁰⁰−(h∧h⁰)≤2(h⁰⁰−(h∧h⁰)).

Take (x, x⁰),(y, y⁰)∈A_i, we have three cases to check. IfH(x)≥h⁰⁰ and H(y)≥h⁰⁰ then (x, x⁰),(y, y⁰)∈A and we have|d(x, y)−d⁰(x⁰, y⁰)| ≤2δ. If H(x) < h⁰⁰ and H(y)< h⁰⁰, then the distortion is at most

|d(x, y)−d⁰(x⁰, y⁰)| ≤Diam {x∈Ci|H(x)< h⁰⁰}∨Diam {x⁰∈C_i⁰|H⁰(x⁰)< h⁰⁰+δ}

≤2(h⁰⁰+δ−(h∧h⁰)),

since the diameter of a tree is at most twice its height. In the last case, supposeH(y)< h⁰⁰≤ H(x). We have h≤H(x∧y)≤H(y)< h⁰⁰, so

d(x, y) =H(x) +H(y)−2H(x∧y)

∈[H(x)−H(y), H(x) +H(y)−2h]

⊂(H(x)−h⁰⁰, H(x) +h⁰⁰−2h).

Similarly, we haveh⁰ ≤H⁰(x⁰∧y⁰)≤H⁰(y⁰)< h⁰⁰+δ, so d⁰(x⁰, y⁰) =H⁰(x⁰) +H⁰(y⁰)−2H⁰(x⁰∧y⁰)

∈[H⁰(x⁰)−H⁰(y⁰), H⁰(x⁰) +H⁰(y⁰)−2h⁰]

⊂(H⁰(x⁰)−h⁰⁰−δ, H⁰(x⁰) +h⁰⁰+δ−2h⁰).

From those two intervals and the fact that |H(x)−H⁰(x⁰)| ≤δ, we deduce that

|d(x, y)−d⁰(x⁰, y⁰)|

< (H(x) +h⁰⁰−2h)−(H⁰(x⁰)−h⁰⁰−δ)∨ (H⁰(x⁰) +h⁰⁰+δ−2h⁰)−(H(x)−h⁰⁰)

≤ |H(x)−H⁰(x⁰)|+ δ+ 2h⁰⁰−2h∨ δ+ 2h⁰⁰−2h⁰

≤2δ+ 2h⁰⁰−2(h∧h⁰).

In the three cases the distortion is less than 2(δ+h⁰⁰−(h∧h⁰))≤4(h⁰⁰−(h∧h⁰)).

Finally, let us control the measures. Set ν_i the measure ofC_i and ν_i⁰ the measure ofC_i⁰. For any Borel set B₀ ⊂T,A₀ ⊂T ×T⁰, we note (B₀)^A→0 ={x⁰ ∈T⁰|∃x∈ B₀,(x, x⁰) ∈A₀}.

Take B ⊂C_i a measurable set, B≥ =B∩H⁻¹([h⁰⁰,∞)), B_< =B∩H⁻¹((−∞, h⁰⁰)). Using the fact that Ais aδ-correspondence, we have:

νi(B)≤ν_i⁰((B≥)^→A) +δ+νi(B<) =ν_i⁰((B≥)

→

Ai) +δ+Hνi([h, h⁰⁰)).

For any Borel B₀⁰ ⊂T⁰, we note (B⁰₀)^←A ={x ∈T|∃x⁰ ∈ B₀⁰,(x, x⁰) ∈A} and (B₀⁰)^A←i ={x ∈ C_i|∃x⁰ ∈ B₀⁰,(x, x⁰) ∈A_i}. Take B⁰ ⊂C_i⁰ a measurable set, B_≥⁰ =B⁰ ∩(H⁰)⁻¹([h⁰⁰+δ,∞)), B_<⁰ =B⁰∩(H⁰)⁻¹((−∞, h⁰⁰+δ)). Using the fact that Ais aδ-correspondence, we have:

ν_i⁰(B⁰)≤ν_i((B_≥⁰ )^←A) +δ+ν_i⁰(B⁰_<) =ν_i((B_≥⁰ )^A←i) +δ+H⁰ν_i⁰([h⁰, h⁰⁰+δ)).

We find thatAi is aδi-correspondence, with δ_i= max2(h⁰⁰−(h∧h⁰)),1

2 ·4(h⁰⁰−(h∧h⁰)), δ+Hν_i([h, h⁰⁰)), δ+H⁰ν_i⁰([h⁰, h⁰⁰+δ))

= max2(h⁰⁰−(h∧h⁰)), δ+Hν_i([h, h⁰⁰)), δ+H⁰ν_i⁰([h⁰, h⁰⁰+δ)).

Since A_i is a δ_i-correspondence, we get with Proposition 3.4.1 that d_GHP(C_i, C_i⁰) ≤δ_i. We set

δ⁰= max2(h⁰⁰−(h∧h⁰)), δ+Hν([h, h⁰⁰)), δ+H⁰ν⁰([h⁰, h⁰⁰+δ)). We shall use later on that sinceh⁰⁰≥h∨h⁰, we have

|h−h⁰| ≤h∨h⁰−h∧h⁰ ≤h⁰⁰−h∧h⁰ ≤δ⁰/2. (4.3.12) Notice that δ⁰ depends neither oni nor onn. It follows that

1≤i≤nmax dGHP(Ci, C_i⁰)≤ max

1≤i≤nδi

≤ max

1≤i≤n

max2(h⁰⁰−(h∧h⁰)), δ+Hνi([h, h⁰⁰)), δ+H⁰ν_i⁰([h⁰, h⁰⁰+δ))

≤δ⁰.

4.3. CROWN OF A TREE 101 Step 2: we prove the result when there exists an termB0 ofCrown_h(T)or ofCrown_h⁰(T⁰) such that Slice_r(B₀) = ∅. For convenience, assume that B₀ = (T_n, d_n, H_n, ν_n) is a term of Crown_h(T). By Lemma 4.3.2,Slice_r(Crown_h(T))∈TC, soSlice_r(T_n)is the only empty term.

Since there can’t be empty terms inCrown_h(T)by definition, and terms ofCrown_h(T)rooted at heighth have at least a point at heighth∈[−r, r], we havemin_TnH_n> h. A second look at the definition ofCrown_h(T)immediately tells us thatT_n=T.

We have Tr = Slice_r(Tn) = ∅, and dGHP(Tr, T_r⁰) ≤ δ < h⁰⁰ −h⁰ < ∞, so T_r⁰ = ∅. By hypothesis,Crown_h⁰(T⁰)6=(0_h⁰)_n∈N^∗, soT⁰has at least a point at height> h⁰. Sinceh⁰ ∈[−r, r]

andT_r⁰ =∅,T⁰does not have a point at heighth⁰. SinceH⁰(T⁰)is an interval containing a point aboveh⁰⁰> h⁰ but none at heighth⁰, we havemin_T⁰H⁰ > h⁰, soCrown_h⁰(T⁰) = (T⁰,0_h⁰,0_h⁰, ...) and we have

d_GHP^∞ (Slice_r(Crown_h(T)),Slice_r(Crown_h⁰(T⁰))) =d_GHP^∞ ((∅,0_h, . . .),(∅,0_h⁰, . . .))

=|h−h⁰|

≤δ⁰.

Step 3: control of the short sub-trees. Suppose that there are non trivial elements in Slice_r(Crown_h(T)) or Slice_r(Crown_h⁰(T⁰)). Recall C₁, ..., C_n, C₁⁰, ..., C_n⁰ from Step 1. Set (C_i)_i>n and (C_i⁰)_i>n the rest of the sub-trees. None of the sub-trees (C_i)_i>n reache h⁰⁰ while none of the sub-trees (C_i⁰)_i>n reache h⁰⁰+δ. For i > n, we take Ai = Ci ×C_i⁰. Ai is a correspondence betweenC_i and C_i⁰ satisfying

sup

(x,x⁰)∈A_i

|H(x)−H⁰(x⁰)| ≤δ+h⁰⁰−(h∧h⁰)≤δ⁰. Its distortion is less than

Diam(Ci)∨Diam(C_i⁰)≤2(h⁰⁰+δ−(h∧h⁰))≤2δ⁰. We have that for every measurable setsB ⊂C_i,B⁰ ⊂C_i⁰

|ν(B)−ν⁰(B⁰)| ≤max ν(C_i), ν⁰(C_i⁰)

≤maxHν([h, h⁰⁰)), H⁰ν⁰([h⁰, h⁰⁰+δ)

≤δ⁰.

We deduce thatAi is aδ⁰-correspondence. We have proven that fori∈N^∗,d_GHP(Ci, C_i⁰)≤δ⁰ so

d_GHP^∞ (Slice_r(Crown_h(T)),Slice_r(Crown_h⁰(T⁰)))≤ sup

i∈N^∗

dGHP(Ci, C_i⁰)≤δ⁰.

Step 4: conclusion. We only need to proveδ⁰≤2(h⁰⁰−(h∧h⁰))+Hν([h∧(h⁰−δ), h⁰⁰+2δ]).

Recall that we have either[h∧h⁰, h⁰⁰+δ]⊂[−r, r]or[h∧h⁰, h⁰⁰+δ]∩[−r, r] =∅.Recall the assumption that [Hν∨H⁰ν⁰]((−∞,−r)∪(r,∞)) = 0, and

δ⁰= max2(h⁰⁰−(h∧h⁰)), δ+Hν([h, h⁰⁰)), δ+H⁰ν⁰([h⁰, h⁰⁰+δ)).

If[h∧h⁰, h⁰⁰+δ]∩[−r, r] =∅,thenδ+Hν([h, h⁰⁰)) =δ+H⁰ν⁰([h⁰, h⁰⁰+δ)) =δ≤h⁰⁰−h⁰ ≤ h⁰⁰−h∧h⁰, soδ⁰ = 2(h⁰⁰−(h∧h⁰)) and we are done.

Now, suppose[h∧h⁰, h⁰⁰+δ]⊂[−r, r]. SinceA is aδ-correspondence betweenT_r andT_r⁰, we have |H(x)−H⁰(x⁰)| ≤δ for all (x, x⁰)∈A, so

{x⁰ ∈T⁰|H⁰(x⁰)∈[h⁰, h⁰⁰+δ)}

←

A ⊂ {x∈T|H(x)∈[h⁰−δ, h⁰⁰+ 2δ]}.

Since Ais aδ-correspondence between Tr and T_r⁰, we have, with the last inclusion:

H⁰ν⁰([h⁰, h⁰⁰+δ)) =ν⁰ {x⁰∈T⁰|H⁰(x⁰)∈[h⁰, h⁰⁰+δ)}

≤ν {x⁰ ∈T⁰|H⁰(x⁰)∈[h⁰, h⁰⁰+δ)}

← A

+δ

≤ν{x∈T|H(x)∈[h⁰−δ, h⁰⁰+ 2δ]}+δ

=Hν([h⁰−δ, h⁰⁰+ 2δ]) +δ.

Asδ⁰ ≤h⁰⁰−h∧h⁰, this means that

δ⁰ ≤max(2(h⁰⁰−(h∧h⁰)),2δ+Hν([h∧(h⁰−δ), h⁰⁰+ 2δ]))

≤2(h⁰⁰−(h∧h⁰)) +Hν([h∧h⁰−δ, h⁰⁰+ 2δ]).

This ends the proof of the lemma.

The next lemma uses Lemma 4.3.5 to give a sufficient (very technical) criterion for the d_LGHP^∞ -convergence of the crown of a sequence of trees.

Lemma 4.3.6. Take (h,(T, d, H, ν)) ∈ D6= with Hν({h}) = 0, (T^k, d_k, H_k, ν^k)_k∈N^∗ a se-quence of elements of T, (h⁰_k)_k∈N^∗, (h⁰⁰_k)_k∈N^∗ two sequences of real numbers converging to h and satisfyingh⁰⁰_k > h∨h⁰_k,(r_k)_k∈N^∗ a sequence of positive real numbers with limit∞,(δ_k)_k∈N^∗ a sequence of positive real numbers with δ_k< h⁰⁰_k−h⁰_k. If for every k∈N^∗, there exists aδ_k -correspondenceA^k betweenSlice_rk(T)andSlice_rk(T^k) such that for every(x, x⁰),(y, y⁰)∈A^k, H(x)∧H(y)≥h⁰⁰_k⇒H(x∧y)≥h⇔Hk(x⁰∧y⁰)≥h⁰_k, (4.3.13) then

d_LGHP^∞ (Crown_h(T),Crown_h⁰

k(T^k)) −→

k→∞0.

Proof. Recall that by Lemma 4.3.4, D6= is open inR×T. Aslim_kδ_k= 0, we see by Lemma 3.4.3 that

k→∞lim d_LGHP(T^k, T) = 0.

Sincelim_kh⁰_k=hand(h, T)∈D6=, there existsk0 such that for everyk≥k0,(h⁰_k, T^k)∈D6=. By hypothesis, lim_k|h⁰_k|= lim_k|h⁰⁰_k|=|h|, so

sup

k∈N^∗

|h⁰_k| ∨ |h⁰⁰_k|<∞.

Since lim_kr_k=∞, there exists k₀⁰ ≥k₀ such that for every k≥k₀⁰, rk>|h| ∨ sup

k⁰∈N^∗

|h⁰_k0| ∨ |h⁰⁰_k0| ≥ |h| ∨ |h⁰_k| ∨ |h⁰⁰_k|.

4.3. CROWN OF A TREE 103

The next three lemmas apply Lemma 4.3.6 to obtain some form of continuity for the application(h, T)7→Crown_h(T) in three specific settings.

Lemma 4.3.7. For every (h, T)∈D6= such thatHν({h}) = 0, we have: correspondence. According to the proof of Lemma 4.2.7, for all r ≥ 0 the restriction of A¹

k to Slice_r(T)×Slice_r(Trim1

k(T)) provides a ¹_k-correspondence between Slice_r(T) and Slice_r(Trim1

where, at the second line, we used Definition 4.2.5. If x and y are in the same sub-tree of Crown_h(T) thenH(x∧y)≥h, so we have

k(T^k). This means that Condition (4.3.13) holds.

We have h < h⁰_k< h⁰⁰_k→_k→∞h and 0< δk = ¹_k < h⁰⁰_k−h⁰_k. Let(rk)_k∈N^∗ be any sequence of positive real numbers converging to infinity. (h, T) ∈D6=, so we can apply Lemma 4.3.6 with parameters(T_k)_k∈N^∗ = (Trim1

k(T))_k∈N^∗,h⁰_k=h+_k¹,h⁰⁰_k=h+_k³ andδ_k= ¹_k. This gives the result.

Lemma 4.3.8. Take (h,(T, d, H, ν))∈D6= such that Hν({h}) = 0, we have d_LGHP^∞ (Crown_h(T),Crown_h⁰(T))−→

h⁰↑h 0.

Proof. Take(h⁰_k)_k∈N^∗ a non-decreasing sequence of real numbers in(−∞, h) converging toh.

The setA={(x, x)}_x∈T is aδ-correspondence betweenTr= Slice_r(T)andTrfor everyr≥0 and δ >0. Let us find a sequence(h⁰⁰_k)_k∈N^∗_,k≥k0 such that h⁰⁰_k↓_kh and for everyx, y∈T,

H(x)∧H(y)≥h⁰⁰_k ⇒H(x∧y)≥h⇔H(x∧y)≥h⁰_k.

Finding such a sequence will immediately solve our problem, by taking δ_k = ¹₂(h⁰⁰_k −h), r_k =k+ max(|h⁰_k|,|h⁰⁰_k+δ_k|), and by applying Lemma 4.3.6.

Take h⁰⁰ > h. The application h⁰ 7→ n^h0,h00(T) is left-continuous on (−∞, h⁰⁰) and has integer values, so there exists h⁰₀(h⁰⁰) < h such that the map h⁰ 7→ n^h0,h00(T) is constant on [h⁰₀(h⁰⁰), h]. From this and the convergence of(h⁰_k)_k∈N^∗, we can definek_h⁰⁰the smallest positive integer such that for everyk≥kh⁰⁰,n^h0k,h00(T) =n^h,h00(T).

For k ≥ kh+1, set Ek = {n ∈ N^∗|n∨k_h+¹

n

≤ k} and nk = max(Ek). The sequence (n_k)_k≥kh+1 is well-defined since E_k is bounded and non-empty (it contains 1). By definition of k_h+1

n, we see that k7→ {n∈N^∗|n∨k_h+1

n ≤k} is non-decreasing, and that lim_kn_k=∞.

For k ≥ k_h+1 set h⁰⁰_k = h+ _n¹

k· We have h⁰_k ≤ h < h⁰⁰_k and n^h0k,h00k(T) = n^h,h00k(T). So the definition ofk_h⁰⁰

k =k_h+ 1 nk

≤k is consistent with the first part of the proof.

Set δk = ¹₂(h⁰⁰_k−h) and rk =k+ max(|h⁰_k|,|h⁰⁰_k+δk|). The setA ={(x, x)}_x∈Slice

rk(T) is a δ_k-correspondence between Slice_rk(T) and itself. Let us prove that for all x, y ∈ Tr with H(x)∧H(y)≥h⁰⁰_k we have

H(x∧y)≥h⇔H(x∧y)≥h⁰_k. (4.3.14) If there are no points in T_r above level h⁰⁰_k, then (4.3.14) is true. If there are points in Slice_r(T) above level h⁰⁰_k but no point in T at level h⁰⁰_k, then min_TH > h⁰⁰_k ≥ h ≥ h⁰_k, so (4.3.14) is true in this case as well. In the remaining case, T has at least one point at level h⁰⁰_k. Reasoning on the ancestors ofxandyat levelh⁰⁰_k(which might be equal), we can suppose without loss of generality that H(x) = H(y) = h⁰⁰_k. The distance dis ultra-metric on level h⁰⁰_k and the sub-trees for Crown_h(T) (resp Crown_h⁰

k(T)) are the equivalence classes of the relationR_h :d(x, y)≤2(h⁰⁰_k−h)(resp. R_h⁰

k :d(x, y)≤2(h⁰⁰_k−h⁰_k)). Sinceh⁰_k≤hwe naturally have xR_hy⇒ xR_h⁰

ky. It follows that the partition induced by R_h is finer than the partition induced by R_h⁰

k. The partition induced by R_h consists of n^h,h00k(T) equivalence classes, and the partition induced byR_h⁰

k consists of n^h0k,h00k(T) equivalence classes. By choice ofh⁰⁰_k, they have the same number of classes, and one is finer, so the relations are equal. We have

H(x∧y)≥h⇔H(x∧y)≥h⁰. Since (h, T)∈D6=, we get by Lemma 4.3.6 that:

k→∞lim d_LGHP^∞ (Crown_h(T),Crown_h⁰

k(T)) = 0.

4.3. CROWN OF A TREE 105 The sequence (h⁰_k)_k∈N^∗ was arbitrary, so we have the continuous limit by sequential charac-terization:

lim

h⁰↑hd_LGHP^∞ (Crown_h(T),Crown_h⁰(T)) = 0.

Lemma 4.3.9. Let (h,(T, d, H, ν))∈D6= be such that Hν({h}) = 0. If T has no branching point at height h, then the map (h⁰, T⁰)7→ Crown_h⁰(T⁰), taking its values in (TC, d_LGHP^∞ ), is continuous at the point(h,(T, d, H, ν))∈R×T.

Proof. Take (h⁰_k)_k∈N^∗ a sequence of real numbers converging to h, (T^k, d_k, H_k, ν_k)_k∈N^∗ a se-quence of S-compact height-labelled trees converging to T for dLGHP. Thanks to Lemma 3.4.3 and Proposition 3.4.1, there exists (r_k)_k∈N^∗ and (δ_k)_k∈N^∗ two sequences of positive real numbers such that lim_kr_k = ∞, lim_kδ_k = 0 and for every k ∈ N^∗, there exists A_k a δk-correspondence between Slice_rk(T) and Slice_rk(T^k). We shall find an integer k0 and a sequence (h⁰⁰_k)_k∈N^∗_,k≥k0 such that for everyk ∈N^∗, h⁰⁰_k >(h⁰_k∨h) +δ_k,lim_kh⁰⁰_k = h and for every(x, x⁰),(y, y⁰)∈A_k,

H(x)∧H(y)≥h⁰⁰_k⇒(H(x∧y)≥h⇔Hk(x⁰∧y⁰)≥h⁰_k).

Then, we shall use Lemma 4.3.6 to end the proof.

Takeh⁰⁰> h. SetK_h⁰⁰ ={x∈T|H(x) =h⁰⁰}. SinceT isS-compact,K_h⁰⁰ is compact. We setδ(h⁰⁰) = ¹₂inf_x,y∈K

h00|d(x, y)−2(h⁰⁰−h)|ifK_h⁰⁰ is non empty, else takeδ(h⁰⁰) =|h⁰⁰−h|.

Let us prove that we still have0< δ(h⁰⁰)≤h⁰⁰−h. It is true by definition whenKh⁰⁰ is empty so we only prove the case K_h⁰⁰ is non-empty. Take x ∈ K_h⁰⁰, and we see that by definition, δ(h⁰⁰) ≤ ¹₂|d(x, x)−2(h⁰⁰−h)|= |h⁰⁰−h|. A continuous map on a non-empty compact set reaches its minimum, so there exists x0, y0 ∈Kh⁰⁰ such thatδ(h⁰⁰) = ¹₂|d(x₀, y0)−2(h⁰⁰−h)|.

We have

δ(h⁰⁰) = 1

2|d(x₀, y0)−2(h⁰⁰−h)|

= 1

2|H(x₀) +H(y₀)−2H(x₀∧y₀)−2(h⁰⁰−h)|

= 1

2|2h⁰⁰−2H(x0∧y0)−2h⁰⁰+ 2h|

=|h−H(x₀∧y₀)|.

SinceT has no branching points at heighth,H(x0∧y0)6=handδ(h⁰⁰)>0. This means that whether K_h⁰⁰ is empty or not, 0 < δ(h⁰⁰) ≤ h⁰⁰−h. From the convergence of (h⁰_k)_k∈N^∗ and (δ_k)_k∈N^∗, for all h⁰⁰ > h, we can define k_h⁰⁰ the smallest positive integer such that for every k≥k_h⁰⁰,δ(h⁰⁰)>2δ_k+|h⁰_k−h|.

For k ≥ kh+1, set Ek = {n ∈ N^∗|n∨k_h+¹

n ≤ k} and nk = max(Ek). The sequence (n_k)_k≥kh+1 is well-defined since E_k is bounded and non-empty (it contains 1). By definition of k_h+1

n, we see that k7→ {n∈N^∗|n∨k_h+1

n ≤k} is non-decreasing, and that lim_kn_k=∞.

Fork≥k_h+1 seth⁰⁰_k=h+_n¹

k·By constructionk_h⁰⁰

k ≤k. Thus, we haveδ(h⁰⁰_k)>2δ_{`</sub>+|h<sup>0</sup><sub>`}−h|

for all `≥k ask_h⁰⁰

k =k_h+ ¹

nk

≤k. So taking `=kgives:

δ(h⁰⁰_k)>2δ_k+|h⁰_k−h|

and, asδ(h⁰⁰)≤h⁰⁰−h, we have h⁰⁰_k≥h+δ(h⁰⁰_k)> h+|h⁰_k−h|+ 2δk ≥(h∨h⁰_k) +δk. Let us prove that for (x, x_k),(y, y_k)∈A_k, we have

H(x)∧H(y)≥h⁰⁰_k⇒H(x∧y)≥h⇔H_k(x_k∧y_k)≥h⁰_k.

If T doesn’t reach above the level h⁰⁰_k, that is if sup_T H < h⁰⁰_k, there is nothing to do. If it does, let us take(x, x_k),(y, y_k)∈A_k such thatH(x)∧H(y)≥h⁰⁰_k. We have :

|(H_k(x_k∧y_k)−h⁰_k)−(H(x∧y)−h)|

≤ 1

2|2H_k(xk∧yk)−2H(x∧y)|+|h⁰_k−h|

= 1

2|d_k(x_k, y_k)−H_k(x_k)−H(y_k)−(d(x, y)−H(x)−H(y))|+|h⁰_k−h|

≤ 1

2(|d_k(x_k, y_k)−d(x, y)|+|H_k(x_k)−H(x)|+|H(y_k)−H(y))|) +|h⁰_k−h|

≤ 1

2(2δ_k+δ_k+δ_k) +|h⁰_k−h|

= 2δ_k+|h⁰_k−h|

< δ(h⁰⁰_k),

where for the third inequality, we used the fact that (x, x_k),(y, y_k) ∈ A_k and that A_k is a δ_k-correspondence.

Let us prove that δ(h⁰⁰_k)≤ |H(x∧y)−h|. IfH(x∧y)≥h⁰⁰_k then we have

|H(x∧y)−h| ≥h⁰⁰_k−h≥δ(h⁰⁰_k).

If H(x∧y) < h⁰⁰_k, set x⁰ and y⁰ the respective ancestors of x and y at level h⁰⁰_k. We have H(x⁰∧y⁰) =H(x∧y)andx⁰, y⁰ ∈K_h⁰⁰

k, and thus |H(x∧y)−h|=|H(x⁰∧y⁰)−h| ≥δ(h⁰⁰_k) by definition ofδ(h⁰⁰_k). Applying this to the upper bound on|(H_k(x_k∧y_k)−h⁰_k)−(H(x∧y)−h)|, we find

(Hk(xk∧yk)−h⁰_k)−(H(x∧y)−h)< δ(h⁰⁰_k)≤ |H(x∧y)−h|.

For everya, b∈R,|b−a|<|a|implies thataand bhave the same sign, soH_k(x_k∧y_k)−h⁰_k and H(x∧y)−h have the same sign. We have proven that

H(x)∧H(y)≥h⁰⁰_k⇒H(x∧y)≥h⇔Hk(xk∧yk)≥h⁰_k. (h, T)∈D6=,so by Lemma 4.3.6, we get

lim

k→∞d_LGHP^∞ (Crown_h⁰

k(T^k),Crown_h(T)) = 0.

The sequences(h⁰_k)_k∈N^∗ and(T^k)_k∈N^∗ were arbitrary, so we have the continuity ofCrown at(h, T) by sequential characterization.

Definition 4.3.10. For B ⊂R×T, h ∈R, T ∈T and (T_n)_n∈N^∗ a sequence of elements of T, we note

1_(h,T)∈B·(Tn)_n∈N^∗ =

( (0_h)n∈N^∗ if (h, T)∈/B (Tn)_n∈N^∗ if (h, T)∈B.

4.3. CROWN OF A TREE 107 Note that(h, T)7→(0_h)_n∈N^∗is a 1-Lipschitz application fromR×TtoTC, so for any map f fromR×TtoTC, and Borel setB, the application(h, T)7→1_(h,T)∈B·f(h, T)is measurable if and only iff is measurable onB. Recall thatCrown_h(T)is an unordered sequence of elements ofT. Recall the measurable mapgdefined onR×Tby g(h,(T, d, H, ν)) = (T, d, H,1_H>h·ν) from Lemma 3.4.14 which will be useful in the proof of the next Proposition.

Proposition 4.3.11. The map (h, T)7→Crown_h(T) defined on (R×T, d_R×d_LGHP) taking values in (TC, d^∞_LGHP) is measurable.

Proof. Step 1: we prove that f1 : (h, T)7→ 1_(h,T)∈D1 ·Crown_h(T) is measurable, for D1 the measurable (we will prove it in a moment) set of all (h,(T, d, H, ν))∈ D6= such that T has no branching point at heighth. By definition, the terms of Crown_h(T) do not have positive measure at their root, so we haveCrown_h(T) = Crown_h(g(h, T)). Since the measure ofg(h, T) has no mass at height h, we can apply Lemma 4.3.9 to find that for (h, T) ∈D₁, Crown is continuous at (h, g(h, T)). It follows that onD₁,Crown is measurable as the composition of a measurable function by a continuous function. Let us prove that D1 is a Borel set. We have

(R×T)\D₁

=D₌∪ {(h,(T, d, H, ν))∈R×T|∃x₁, x₂∈T, H(x₁)∧H(x₂)> h, H(x₁∧x₂) =h}

=D₌ ^[

ε∈Q^∗+

F_ε,

whereD== (R×T)\D6= and

Fε= (

(h,(T, d, H, ν))∈R×T

∃x₁, x2 ∈T,

H(x1)≥h+ε, H(x2)≥h+ε, H(x1∧x2) =h )

.

Let us prove that forε∈Q^∗+, the setFεis a closed set. Take(h_k,(T^k, d_k, H_k, ν_k))a sequence of elements ofF_ε converging to some(h, T)∈R×T. By hypothesis, we can find a sequence (x_k, y_k)_k∈N^∗ such thatx_k, y_k∈T^k,H_k(x_k)≥h_k+ε,H_k(y_k)≥h_k+εand H_k(x_k∧y_k) =h_k. For every k ∈ N^∗, inf_TkHk ≤ H(xk∧yk) = hk, so we can take x⁰_k and y_k⁰ the respective ancestors of x_k and y_k at height h+ε. Since T is the limit of (T^k)_k∈N^∗, by Lemma 3.4.3 and Proposition 3.4.1, there exists two sequence (r_k)_k∈N^∗, (δ_k)_k∈N^∗ of positive integers such that lim_krk = ∞ and lim_kδk = 0 and for every k ∈ N^∗ some Ak is a δk correspondence betweenSlice_rk(T)andSlice_rk(T^k). Since(h_k)_k∈N^∗ converges,r_k≥ |h_k+ε|forkabove some k₀. For k≥ k₀, we choose x⁰⁰_k, y_k⁰⁰ ∈ T such that (x⁰⁰_k, x⁰_k),(y_k⁰⁰, y_k⁰) ∈A_k. By choice, we have H(x⁰⁰_k), H(y_k⁰⁰)∈[hk+ε−δ_k, hk+ε+δk]. Since(δk)_k∈N^∗ converges to 0,(x⁰⁰_k)_k≥k0 and(y_k⁰⁰)_k≥k0 are bounded inH. AsT isS-compact, up to considering a sub-sequence, we can assume that (x⁰⁰_k, y⁰⁰_k)_k≥k0 converges to some(x, y)∈T ×T. By continuity, we have

H(x) =H(y) = lim

k→∞H(x⁰⁰_k) = lim

k→∞H(y_k⁰⁰) =h+ε

and

d(x, y) = lim

k→∞d(x⁰⁰_k, y⁰⁰_k)

= lim

k→∞dk(x⁰_k, y_k⁰)

= lim

k→∞H_k(x⁰_k) +H_k(y_k⁰)−2H(x⁰_k∧y⁰_k)

= lim

k→∞2(hk+ε)−2hk

= 2ε.

It follows that H(x∧y) = ¹₂(H(x) +H(y)−d(x, y)) = h. Since H(x)∧H(y) ≥h+ε, we deduce that(h, T)∈Fε. We have proven thatFε is closed, andD6= is open by Lemma 4.3.4, soD₁=D6=T

ε∈Q^∗₊F_ε^c is a Borel set. This implies thatf₁ is measurable.

Step 2: we prove that f₂ : (h, T) 7→ 1_(h,T)∈D2 ·Crown_h(T) is measurable, for D₂ the measurable (we will prove it in a moment) set of all (h,(T, d, H, ν))∈ D6= such that T is a discrete tree, see Definition 4.2.9. We prove first thatD2 is a Borel set. Take(T, d, H, ν)∈T.

For every r ∈R+, we set E_r(T) the set of leaves with height in [−r, r] and points at height r. SetT_r = Slice_r(T). Let us prove that for every x∈ T_r, there exists y ∈E_r(T) such that xy. Take x∈Tr, the set{y∈Tr|xy} is a closed set. Since Tr is compact, there exists y₀ ∈ T_r with x y₀ such that H(y₀) = maxH({y ∈ T_r|x y}). This means that y₀ is maximal for in{y∈T_r|xy}, soy₀ is maximal inT_r and y₀ ∈E_r(T).

By definition,T is a discrete tree if and only if for everyr ∈R+,Er(T)is finite. Note that if0≤r⁰< r andEr(T)is finite thenE_r⁰(T) is finite, so T is a discrete tree if and only if for everyr∈N^∗,E_r(T)is finite. Note that for,E_r(T)is the set of all maximal points ofT_r. For everyx1, ..., xn, we can takey1, ..., yn∈Er(T) such that for all i,xi yi. If n≤#(Er(T)), takingy1, ..., yndistinct elements inEr(T)provides a family of non-comparable elements since they are maximal. Ifn >#(E_r(T)), there necessarily existsi6=jsuch thatx_i y_i =y_j x_j. By Remark 4.1.12,xi xj orxj xi. This means thatEr(T) is finite if and only if for every r∈N^∗, there exists n∈N^∗ such that

∀x₁, ..., x_n∈T_r,∃1≤i, j≤n,(i6=j and x_ix_j).

This means that we have

D₂ =D6=∩



R× ^\

r∈N^∗

[

n∈N^∗

(

(T, d, H, ν)∈T

∀x₁, ..., x_n∈T_r,

∃1≤i, j≤n,(i6=j and x_i x_j) )



4.3. CROWN OF A TREE 109 r, but the third equality holds anyway thanks to the intersection overr∈N^∗. The setF_r,nis closed thanks to Lemma 3.4.9 andD6= is an open set by Lemma 4.3.4, soD2 is a Borel set.

Let us prove thatf2is measurable. Recall from the beginning of Step 1 thatCrown_h(T) = Crown_h(g(h, T))and the measure of g(h, T)has no mass at heighth. With Lemma 4.3.8, we a finite number of branching points with heights in[h−1, h]. The treeg(h, T)has the same branching points asT, so for all but a finite number of h⁰ ∈[h−1, h],g(h, T) doesn’t have a

n is measurable. Since f2 and g are measurable and D6= is a Borel set, it is enough to prove that

If (h, T) ∈/ D6=, the result is trivial since for every (Tn)_n∈N^∗ ∈ TC, 1_(h,T)∈D6= ·(Tn)_n∈N^∗ is

The map Crown is the point-wise limit of a sequence of measurable functions, so Crown is measurable.

We recall that B(E) denote the Borelσ-field of a metric space E.

Definition 4.3.12. For every T-valued random variable T, we call (Stump_h(T))h∈R the growth process associated with T, and (Crown_−h(T))_h∈R the coalescent process associated withT.

We also define the filtrationsS = (Sh)_h∈R= (Stump⁻¹_h (B(T)))_h∈RandC = (C−h)_h∈R= (Crown⁻¹_−h(B(TC)))_h∈R.

Theorem 4.3.13. The families S andC are filtrations on(T,B(T)). The growth process is adapted for S, and the coalescent process is adapted for C.

Proof. By Lemma 4.2.11, T 7→ Stump_h(T) is measurable for every h ∈ R, so the family

For T a random variable on T, the process (Stump_h(T))h∈R corresponds to the growth

For T a random variable on T, the process (Stump_h(T))h∈R corresponds to the growth