Measurability of the ancestral process

h⁰↓h0.

This concludes Step 2, as we have proven that h7→S_h(T) is right-continuous onR.

Using Lemma 3.4.13, the measurability in T and right-continuity inh imply the measur-ability of(T, h)7→Sh(T).

4.2.4 Measurability of the ancestral process

In this section, we give a parametrization of some trees which will be used in the next chapter.

We define the vertical deformation of a tree, and give its action on the parametrization.

Definition 4.2.12. For (T, d, H, ν) a height-labelled tree such that ν(T) is finite and f a non-decreasing continuous function from R to R, we call vertical deformation of T by f the 4-uple (T⁰, d⁰, H⁰, ν⁰), where

• H⁰⁰=f ◦H,

• d⁰(x, y) =H⁰⁰(x) +H⁰⁰(y)−2H⁰⁰(x∧y),

• T⁰⁰ is the quotient of T by the relationd⁰(·,·) = 0,

• T⁰ is the completion of T⁰⁰ for d⁰,

• ρ(x) is the natural projection of T into T⁰,

• ν⁰ =ρ◦ν,

• H⁰ is the 1-Lipschitz extension ofH⁰⁰ over T⁰.

Remark 4.2.13. The hypothesis ν(T) < ∞ is only used when f is bounded, to prevent the accumulation of an infinite measure at finite height for ν⁰. We could alternatively suppose that f is surjective on R. Similarly, the completion step in the definition of T⁰ comes into play whenf is bounded, and allowsT⁰ to beS-compact (this is proven in Lemma 4.2.14).

Lemma 4.2.14. The vertical deformation (T⁰, d⁰, H⁰, ρ◦ν) of aS-compact tree (T, d, H, ν), withν(T)finite, by a non-decreasing continuous functionf is still aS-compact height-labelled tree.

Proof. Step 1: we prove that (T⁰⁰, d, H⁰,0)is a height-labelled tree. Set ⁰ the relation such x^{0 0} y⁰ ⇔d⁰(x⁰, y⁰) =H⁰(y⁰)−H⁰(x⁰). With Proposition 4.1.14, we just have to prove that T⁰⁰= (ρ(T), H⁰,⁰) is a coded tree. Let us prove the four conditions in Definition 4.1.11 and that⁰ is an order.

Condition 1: f is continuous, so H⁰(T⁰⁰) =f◦H(T) is an interval.

Condition 2: if x^{0 0} y⁰ and x⁰ 6= y⁰, then H⁰(y) −H⁰(x) = d⁰(x⁰, y⁰) > 0, so H⁰ is increasing.

We now prove that the relation ⁰ is an order on T⁰. The reflexivity is obvious since d⁰(x⁰, x⁰) =H⁰(x)−H⁰(x⁰) = 0; Condition 2 implies the anti-symmetry; and forx^{0 0} y⁰ and y⁰⁰ z⁰, we use the fact thatH⁰ is 1-Lipschitz for d⁰ to prove that

H⁰(z⁰)−H⁰(x⁰)≤d⁰(z⁰, x⁰)

≤d⁰(z⁰, y⁰) +d⁰(y⁰, x⁰)

=H⁰(z⁰)−H⁰(y⁰) +H⁰(y⁰)−H⁰(x⁰)

=H⁰(z⁰)−H⁰(x⁰),

and deduce that z^{0 0} x⁰, so⁰ is transitive. We have proven that⁰ is an order onT⁰⁰. Condition 3: take x⁰ ∈T⁰⁰ and h⁰ ∈H⁰(T⁰⁰) such that h⁰ ≤H⁰(x⁰) and let us prove that there exists y^{0 0} x⁰ such that H⁰(y⁰) = h⁰. Choose x ∈ T, h ∈ H(T) such that ρ(x) = x⁰ and f(h) = h⁰. If h⁰ =H⁰(x⁰) then the result is obvious. If h⁰ < H⁰(x⁰), then h < H(x), so there existsy x such that H(y) =h. Sety⁰ =ρ(y), we haveH⁰(y⁰) =h⁰. Since y =y∧x, we have d⁰(x⁰, y⁰) = H⁰(x⁰) +H⁰(y⁰)−2H⁰(ρ(y ∧x)) = H⁰(x⁰) −H⁰(y⁰), so y^{0 0} x⁰. We have proven the existence. Now, suppose that there existsy₁⁰, y⁰₂⁰ x⁰ withH⁰(y₁⁰) =H⁰(y₂⁰).

Take x, y₁, y₂ ∈T some respective antecedents of x⁰, y₁⁰ and y⁰₂ by ρ. x∧y₁ and x∧y₂ are ancestors ofx, so they are comparable by Remark 4.1.12. Suppose without loss of generality that x∧y1 x∧y2. This means that x∧y1 is a common ancestor ofy1 and y2 and thus H(x∧y₁)≤H(y₁∧y₂). We have

d⁰(y⁰₁, y₂⁰) =d⁰(y₁, y₂)

=H⁰(y1) +H⁰(y2)−2H⁰(y1∧y2)

≤H⁰(y1) +H⁰(y2)−2H⁰(y1∧x)

=H⁰(y₂) +d⁰(x, y₁)−H⁰(x)

=H⁰(y₂⁰)−H⁰(x⁰) +d⁰(x⁰, y⁰₁)

=H⁰(y₂⁰)−H⁰(x⁰) +H⁰(x⁰)−H⁰(y⁰₁)

= 0,

where for the first and fourth equality, we used the fact thatρ preservesd⁰ and H⁰; the fifth equality comes from the fact that y⁰₁⁰ x⁰. We have proven Condition 3.

Condition 4: let us prove that for every x⁰, y⁰ ∈T⁰⁰, the set of all pointsz⁰ ∈T⁰⁰ such that z^{0 0} x⁰ and z^{0 0} y⁰ has a maximal element. With Condition 2, if we find z⁰ maximizing H⁰(z⁰),z⁰ is automatically maximal. Takex, yantecedents ofx⁰, y⁰ forρ. For anyz⁰ such that z^{0 0} x⁰ and z^{0 0} y⁰, we have H⁰(x⁰)−2H⁰(z⁰) +H⁰(y⁰) =d⁰(x⁰, z⁰) +d⁰(z⁰, y⁰)≥d⁰(x⁰, y⁰) = H⁰(x⁰)−2H⁰(x∧y) +H⁰(y⁰), so H⁰(z⁰) ≤ H⁰(x∧y). It follows that taking z⁰ = ρ(x∧y) provides a maximal element.

We have proven that(T⁰⁰, H⁰,⁰)is a coded tree, so with Proposition 4.1.14,(T⁰⁰, d⁰, H⁰,0) is a height-labelled tree.

Step 2: We prove that (T⁰, d⁰, H⁰, ν⁰) is an S-compact height-labelled tree. The space (T⁰, d⁰, H⁰,0)is the completion of(T⁰⁰, d⁰, H⁰,0)which is a labelled tree, so it is a height-labelled tree by Lemma 4.1.4. We also haveν⁰(T⁰) =ν⁰(T⁰⁰) =ν(T)<∞, so(T⁰, d⁰, H⁰, ν⁰)is

4.2. SOME MEASURABLE MAPS OVERT 87 a height-labelled tree. To prove that (T⁰, d⁰, H⁰, ν⁰) is S-compact, we will use Lemma 4.2.3.

Takeh⁰₁ < h⁰₂∈Rand0< ε < ¹₂(h⁰₂−h⁰₁). Suppose thatn^h01,h02(T⁰)≥kfor some integerk >1.

Since H⁰(T⁰) is the closure of the intervalH⁰(T) andh⁰₁< h⁰₁+ε < h⁰₂−ε < h⁰₂, there exists h1(ε)< h2(ε)∈H(T)such thatf(h1(ε)) =h⁰₁+εandf(h2(ε)) =h⁰₂−ε. By definition, there existsx⁰₁, ..., x⁰_k∈T⁰ such that for everyi6=j,H⁰(x⁰_i) =h⁰₂ and d⁰(x⁰_i, x⁰_j)>2(h⁰₂−h⁰₁). Take ε >0. Since T⁰⁰ =ρ(T) is dense inT⁰, there exists x₁, ..., x_k ∈T such that for every i6=j,

|H⁰(xi)−h⁰₂|< εandd⁰(xi, xj)>2(h⁰₂−h⁰₁). Sinced⁰(xi, xj) =H⁰(xi) +H⁰(xj)−2H⁰(xi∧x_j), we have

H⁰(xi∧xj) = 1

2 H⁰(xi) +H⁰(xj)−d⁰(xi, xj)< h⁰₂+ε−(h⁰₂−h⁰₁) =h⁰₁+ε.

It follows that for everyi6=j,H⁰(x_i)> h⁰₂−εand H⁰(x_i∧x_j)< h⁰₁+ε. Recalling the choice of h1(ε),h2(ε) and since f is non-decreasing, we deduce that for every i6=j, H(xi)> h2(ε) and H(x_i∧x_j)< h₁(ε). This means thatn^{h1(ε),h2(ε)}(T)≥k.

We have proven that for for every k > 1, n^h01,h02(T⁰) ≥ k ⇒ n^{h1(ε),h2(ε)}(T) ≥ k, so we haven^h01,h02(T)⁰≤1∨n^{h1(ε),h2(ε)}(T). SinceT isS-compact, so with Lemma 4.2.3, we get that n^{h1(ε),h2(ε)}(T)is finite. We have proven thatn^h01,h02(T⁰) is finite with arbitraryh₁< h₂. Since T⁰ is complete, we know from Lemma 4.2.3 that T⁰ is S-compact.

Now, we give a construction similar to the ancestral processes defined in [8], that is a tree with all the leaves at the same height, a measure concentrated on the leaves and a characterization of the tree by the coalescence times.

SetR^N

∗

+,0the set of non-increasing sequences of non-negative real numbers converging to 0, andR^ω_+,0 ⊂R^N

∗

+,0 the set of non-increasing sequences of non-negative real numbers containing only a finite number of positive terms. We set

D={(u_n)_n∈N^∗ ∈(0,1)^N∗|∀i < j, u_i 6=u_j}.

The spacesR^ω+,0andR^N

∗

+,0are equipped with the norm||·||_∞of uniform convergence, for which R^ω+,0 is dense inR^N

∗

+,0. The spaceDequipped with the topology of the pointwise convergence, for which it is a Borel set of the Polish space[0,1]^N∗.

Definition 4.2.15. For h ∈R, (ζn)_n∈N^∗ ∈R^N

∗

+,0 and (un)_n∈N^∗ ∈D, we define E = (0,1)× (−∞, h], and for(x, y),(x⁰, y⁰)∈E, H(x, y) =y,

d((x, y),(x⁰, y⁰)) =y+y⁰−2(y∧y⁰∧ inf

x≤un<x⁰(h−ζ_n)),

with the convention thatinf∅= +∞. Letν be the 1-dimensional Lebesgue measure on(0,1)×

{h}and T the quotient of E by the relation d(·,·) = 0. We call τ(h,(ζn)_n∈N^∗,(un)_n∈N^∗)) the space (T⁰, d⁰, H⁰, ν⁰), where (T⁰, d⁰) is the completion of (T, d), H⁰ the 1-Lipschitz extension of H to T, and ν⁰ the projection of ν onto the quotient.

Lemma 4.2.16. Let h ∈ R, (ζ_n)n∈N^∗ ∈R^N

∗

+,0, and (u_n)n∈N^∗ ∈D. The space τ(h,(ζ_n)n∈N^∗, (un)_n∈N^∗)) is well-defined, and is a S-compact tree.

Proof. In the proof, we use the notations from Definition 4.2.15. Let us define the relation , such that for every(x, y)(x⁰, y⁰)∈E = (0,1)×(−∞, h]we have

(x, y)(x⁰, y)(x⁰, y⁰)⇔y≤y⁰∧ inf

x∧x⁰≤un<x∨x⁰(h−ζn).

H

to−∞ to−∞

h ν

(u1, h−ζ1) (u₂, h−ζ₂)

(u3, h−ζ3)

· · · ...

u2 (u₃−u₁) (1−u₃)

Figure 4.5: Example of the construction.

Step 1. We will prove thatd(·,·) = 0is an equivalence relation, and thatis an order on the quotient ofE byd(·,·) = 0. The relation is reflexive. Let us prove that it is transitive.

Suppose(x⁰⁰, y⁰⁰)(x⁰, y⁰)(x, y). We need to consider the order of the abscissas, there are three cases to consider, depending on which abscissa is in the middle. Since the demonstration is the same in each case, we only do the casex⁰⁰≤x⁰ ≤x. In this case, we havex⁰⁰≤x and

y∧( inf

x⁰⁰≤u_n<x(h−ζ_n)) =y∧( inf

x⁰≤u_n<x(h−ζ_n))∧( inf

x⁰⁰≤u_n<x⁰(h−ζ_n))

≥y⁰∧( inf

x⁰⁰≤un<x⁰(h−ζn))

≥y⁰⁰,

where we used the definition of(x⁰, y⁰)(x, y) for the first inequality and (x⁰⁰, y⁰⁰)(x⁰, y⁰) for the second. We have proven that(x⁰⁰, y⁰⁰)(x, y), sois transitive. Finally, let us prove that(x⁰, y⁰)(x, y)and (x, y)(x⁰, y⁰) if and only ifd((x, y),(x⁰, y⁰)) = 0. Suppose without loss of generality thatx≤x⁰. We see easily that

( (x⁰, y⁰)(x, y) (x, y)(x⁰, y⁰) ⇔

( y⁰≤y∧(inf_x≤un<x⁰(h−ζn)) y≤y⁰∧(inf_x≤un<x⁰(h−ζ_n))

⇔ y=y⁰ ≤ inf

x≤un<x⁰(h−ζ_n)

⇔ y+y⁰−2 y∧y⁰∧( inf

x≤u_n<x⁰(h−ζ_n))= 0

⇔ d (x, y),(x⁰, y⁰)= 0.

Since is transitive and reflexive, we have proven thatd(·,·) = 0 is an equivalence relation, so the quotientT ofE by this relation is well-defined. This means thatis defined without ambiguity and is anti-symmetric onT, sois an order on T.

In particular, note that if d (x, y),(x⁰, y⁰)= 0, then y =y⁰, so H is also defined on T.

This means that (T, d, H) is well-defined.

4.2. SOME MEASURABLE MAPS OVERT 89 Step 2. Let us prove that(T, H,) is a coded tree, by checking the four conditions from Definition 4.1.11.

Condition 1: the direct imageH(T)is equal to(−∞, h]by choice ofE, so it is an interval.

Condition 2: by definition of , the map H: (x, y)7→y is strictly increasing.

Condition 3: we work on E. Take (x₀, y₀) ∈E,h₀ ∈(−∞, y₀], so that the point (x₀, h₀) satisfies (x0, h0) (x0, y0) and H(x0, h0) = h0. Let us prove that for every (x, y) such that (x, y) (x₀, y₀) andH(x, y) =h₀, we have d((x, y),(x₀, h₀)) = 0. Suppose thatx₀ ≤x (the demonstration is the same for x < x₀), we have

y≤y0∧( inf

x0≤u_n<x(h−ζn)) by definition of. It follows that, by definition ofd,

d((x₀, h₀),(x, y)) =h₀+y−2 h₀∧y∧( inf

x0≤un<x(h−ζ_n))

=h0+y−2(h0∧y).

Since we have y = H(x, y) = h0, the last line gives that d((x0, h0),(x, y)) = 0. Since ρ (the projection from E to T) is surjective ontoT, we have proven that ρ(x₀, h₀) is the only ancestor ofρ(x₀, y₀) at heighth₀, so(T, d, H) satisfies Condition 3.

Condition 4: once again we work on E. Take (x, y), (x⁰, y⁰) ∈E. From Step 1 and the proof of Condition 3, we see that (x⁰⁰, y⁰⁰) is an ancestor of (x, y) at heighty⁰⁰ if and only if it is equivalent to(x, y⁰⁰) and y⁰⁰ ≤y. It follows that (x⁰⁰, y⁰⁰) is a common ancestor of (x, y) and(x⁰, y⁰)if and only ifd((x, y⁰⁰),(x⁰⁰, y⁰⁰)) = 0,d((x⁰, y⁰⁰),(x⁰⁰, y⁰⁰)) = 0andy⁰⁰ ≤y∧y⁰. This means that there exists a common ancestor of (x, y) and (x⁰, y⁰) at height y⁰⁰ if and only if d((x, y⁰⁰),(x⁰, y⁰⁰)) = 0 and y⁰⁰≤y∧y⁰. Supposing without loss of generality that x≤x⁰, we have

d((x, y⁰⁰),(x⁰, y⁰⁰)) = 0⇔ y⁰⁰+y⁰⁰−2 y⁰⁰∧y⁰⁰∧( inf

x≤un<x⁰(h−ζ_n))= 0

⇔ y⁰⁰≤( inf

x≤un<x⁰(h−ζn)). (4.2.4)

It follows that x, y∧y⁰∧inf_x≤un<x⁰(h−ζn) is the MRCA of (x, y) and (x⁰, y⁰). Noticed that:

d((x, y),(x⁰, y⁰) =y+y⁰−2 y∧y⁰∧ inf

x≤un<x⁰(h−ζ_n)

=H((x, y)) +H(x,⁰, y⁰))−2H((x, y)∧(x⁰, y⁰)).

Thus (4.1.4) holds. We have proven that(T, d, H)is a coded tree, so with Proposition 4.1.14, (T, d, H,0)is a height-labelled tree.

Step 3 : Using Lemma 4.2.3, we prove that (T⁰, d⁰, H⁰, ν⁰) is S-compact height-labelled tree. Thanks to Lemma 4.1.4,(T⁰, d⁰, H⁰,0)is a complete height-labelled tree, andν⁰ is finite, so, with Lemma 4.2.3 we only have to prove thatn^h1,h2(T⁰)<∞ for everyh1 < h2 ∈R. For everyh₁< h₂ ≤h we have n^h1,h(T⁰)≤n^h1,h2(T⁰), so it is enough to prove it for h₂ 6=h. We have H⁰(T⁰) =H(T) = (−∞, h], so, if h₂ > h, we have n^h1,h2(T⁰) = 0<∞. We can suppose h2 < hwithout loss of generality.

Setn(h2) the number of integersn∈N^∗ such that h−ζn < h2. Since lim_nζn= 0,n(h2) is finite, and Equivalence (4.2.4) implies thatT has exactlyn(h₂) + 1 points(z₁, ..., z_n(h2)+1) at height h₂. In E, every point (x, y) with y ≤ h₂ satisfies (x, y) (x, h₂). This means that {z ∈ T|H(z) ≤ h2} is covered by the sets {z ∈ T|z zi} for 1 ≤ i≤ n(h2) + 1. For every i, H is an bijective isometry from {z ∈ T|z z_i} to(−∞, h₂], which is complete, so {z∈T|∃i, zz_i} is a closed set ofT⁰. It follows that{x∈T|H(x) ≤h₂} is a closed subset of T⁰, so all the points ofT⁰\T are at height at leasth2. In particular, the points at height h₁ inT⁰ are all inT, so they are all ancestors of{z₁, ..., z_n(h2)}. By Condition 3, there are at mostn(h₂) + 1points at height h₁. Since n^h1,h2(T⁰) is the number of ancestors at heighth₁ of the points at level h₂. We get n^h1,h2(T⁰) ≤n(h₂) + 1 <∞. Since the restriction h₂ 6= h was done without loss of generality, we have proven that for allh1 < h2 ≤h,n^h1,h2(T⁰)<∞, so(T⁰, d⁰, H⁰, ν⁰) is aS-compact height-labelled tree.

Lemma 4.2.17. Let (h,(ζ_n)_n∈N^∗,(u_n)_n∈N^∗) ∈ R× R^N

∗

+,0 × D and f a continuous non-decreasing map from R to R. Set (T, d, H, ν) = τ(h,(ζ_n)_n∈N^∗,(u_n)_n∈N^∗). Then f(h)− f(h−ζn))_n∈N^∗ belongs to R^N

∗

+,0 and the vertical deformation of (T, d, H, ν) by f is equal to the only non-0_h⁰ term of

Crown_h⁰τ(f(h),(f(h)−f(h−ζ_n))_n∈N^∗,(u_n)_n∈N^∗), where h⁰ = lim_r→−∞f(r).

Proof. It suffices to carefully consider the height of the leaves (they are all at height f(h)) and of the branching points (they were at height (h−ξ_n)_n∈N^∗ in T, so they are at height (f(h−ξ_n))n∈N^∗ inT⁰).

Lemma 4.2.18. The function (h,(ζn)_n∈N^∗,(un)_n∈N^∗) 7→ τ(h,(ζn)_n∈N^∗,(un)_n∈N^∗) from R× R^N

∗

+,0×D to Tis measurable.

Proof. Step 1: we prove that for every (ζn)_n∈N^∗ ∈ R^N

∗

0 and (un)_n∈N^∗ ∈ D, the map h 7→

τ(h,(ζ_n)_n∈N^∗,(u_n)_n∈N^∗) is continuous. Take h ∈ R, (T, d, H, ν) = τ(h,(ζ_n)_n∈N^∗,(u_n)_n∈N^∗) and (δ_k)_k∈N^∗ a sequence of real numbers converging to 0 such that sup_k∈

N^∗|δ_k| ≤ 1. Note that replacingh byh+δk only introduce a shift inH, such that we have the simple relation τ(h +δ_k,(ζ_n)_n∈N^∗,(u_n)_n∈N^∗) = (T, d, H +δ_k, ν), where H +δ_k represents the map x 7→

H(x) +δ_k. We shall now prove that (T, d, H+δ_k, ν) converges to (T, d, H, ν) when δ_k goes to 0.

For all k∈N^∗, let us setE^k= Slice_|h|+k+1(T, d, H, ν)and G^k= Slice_|h|−δ

k+k+1(T, d, H+ δ_k, ν). We have−(|h|+k+ 1)< h= max_TH <|h|+k+ 1, soE^k is non-empty and we have E^k={x∈T|H(x)≥ −(|h|+k+ 1)}. Sincek≥1, we have

−(|h| −δk+k+ 1)≤ −(|h|+k)≤h+δk= max

T (H+δk)≤ |h|+k≤ |h| −δk+k+ 1.

This implies thatG^k is non-empty and that

G^k={x∈T|H(x) +δ_k≥ −(|h| −δ_k+k+ 1)}={x∈T|H(x)≥ −(|h|+k+ 1)}.

This means that G^k is just the shift in height of E^k, so dGHP(E^k, G^k)≤ |δ_k| −→

k→∞0.

4.2. SOME MEASURABLE MAPS OVERT 91 The setsE^kandG^kare both compact, and we havemin(|h|+k+ 1,|h| −δk+k+ 1)≥ |h|+k, soSlice_|h|+k(T, d, H, ν)⊂E^k⊂(T, d, H, ν) and Slice_|h|+k(T, d, H+δ_k, ν)⊂G^k⊂(T, d, H+ δ_k, ν). The sequence(|h|+k)k∈N^∗ goes to ∞, so, by Lemma 3.4.2,

d_LGHP((T, d, H+δ_k, ν),(T, d, H, ν)) −→

k→∞0.

We have proven that the sequence of trees(τ(h+δn,(ζn)n∈N^∗,(un)n∈N^∗))_k∈N^∗ converges to the treeτ(h,(ζn)_n∈N^∗,(un)_n∈N^∗)ford_LGHP. The choiceh and(δ_k)_k∈N^∗ ∈[−1,1]^N∗ was arbitrary, soh7→τ(h,(ζ_n)_n∈N^∗,(u_n)_n∈N^∗) is continuous by sequential characterization.

Step 2: we prove that (ζn)_n∈N^∗ 7→τ(h,(ζn)_n∈N^∗,(un)_n∈N^∗) is 1-Lipschitz, that is:

d_LGHP τ(h,(ζ_n)n∈N^∗,(u_n)n∈N^∗), τ(h,(ζ_n⁰)n∈N^∗,(u_n)n∈N^∗)≤ sup

n∈N^∗

|ζ_n−ζ_n⁰|. (4.2.5) Set δ = sup_n∈

N^∗|ζ_n−ζ_n⁰|. Take E = [0,1]×(−∞, h], equipped with ν_E the 1-dimensional Lebesgue measure on [0,1]× {h} ⊂ E. We note (T, d, H, ν) ⊂ τ(h,(ζ_n)_n∈N^∗,(u_n)_n∈N^∗) the quotient ofE by the pseudo-distance

d((x, y),(x⁰, y⁰)) =y+y⁰−2(y∧y⁰∧ inf

x≤un<x⁰h−ζ_n),

and (T⁰, d⁰, H⁰, ν⁰)⊂τ(h,(ζ_n⁰)_n∈N^∗,(u_n)_n∈N^∗)the quotient of E by the pseudo-distance d⁰((x, y),(x⁰, y⁰)) =y+y⁰−2(y∧y⁰∧ inf

x≤un<x⁰h−ζ_n⁰).

Let us callρ the projection ofE to the quotientT, andρ⁰ the projection ofE to the quotient T⁰. Note thatτ(h,(ζ_n)_n∈N^∗,(u_n)_n∈N^∗)is the completion of(T, d, H, ν)and thatρis surjective from E toT. We setA={(ρ(x, y), ρ⁰(x, y))}_(x,y)∈E.

Take r ∈R+, and let us prove thatA induces a δ-correspondence between Slice_r(T) and Slice_r(T⁰). For every(x, y)∈E, we haveH(ρ(x, y)) =H⁰(ρ⁰(x, y)), so ρ(x, y)∈Slice_r(T) ⇔ ρ⁰(x, y) ∈ Slice_r(T⁰). Since ρ and ρ⁰ are surjective, A induces a correspondence between Slice_r(T) and Slice_r(T⁰). Recall Conditions (3.4.1)-(3.4.4) to be a δ-correspondence, and let us prove that they are satisfied by A. Take (x, y),(x⁰, y⁰) ∈ E, and let us compute the distortion ofA. We have

|d(ρ(x, y), ρ(x⁰, y⁰))−d⁰(ρ⁰(x, y), ρ⁰(x⁰, y⁰))|

=

y+y⁰−2(y∧y⁰∧ inf

x≤un<x⁰h−ζn)−y+y⁰−2(y∧y⁰∧ inf

x≤un<x⁰h−ζ_n⁰)

= 2(y∧y⁰∧ inf

x≤un<x⁰h−ζ_n)−(y∧y⁰∧ inf

x≤un<x⁰h−ζ_n⁰)

≤2( sup

x≤u_n<x⁰

ζn)−( sup

x≤u_n<x⁰

ζ_n⁰)

≤2 sup

x≤un<x⁰

|ζ_n−ζ_n⁰| ≤2δ.

This implies thatA satisfies Condition (3.4.1) with εreplaced byδ. For(x, y)∈E, we have H(ρ(x, y)) =H⁰(ρ⁰(x, y)), so Asatisfies Condition (3.4.2). Finally, forB ⊂T a Borel set, we defineB⁰ ={ρ⁰(x, y), ρ(x, y)∈B}=ρ⁰◦ρ⁻¹(B). We have

ν(B) =νE(ρ⁻¹(B))≤νE(ρ⁰⁻¹(B⁰)) =ν⁰(B⁰).

This proves Condition (3.4.3). SinceT and T⁰ play symmetric roles here, we have Condition (3.4.4) as well. We have proven that A induces a δ-correspondence between Slice_r(T) and Slice_r(T⁰) for everyr∈R.

The spaces T⁰⁰ = τ(h,(ζn)_n∈N^∗,(un)_n∈N^∗) and T⁰⁰⁰ = τ(h,(ζ_n⁰)_n∈N^∗,(un)_n∈N^∗) are the respective completions of T and T⁰. Set A⁰ the closure of A in T⁰⁰×T⁰⁰⁰, and let us prove thatA⁰ is aδ-correspondence. For everyx∈T⁰⁰, there exists a sequence(x_n)n∈N^∗ of elements of T converging to x. Since A is a correspondence, there exists a sequence (x⁰_n)_n∈N^∗ of elements of T⁰ such that (x_n, x⁰_n) ∈ A. By definition of A, we have H⁰(x⁰_n) = H(x_n), so lim_nH(x⁰_n) =H(x). SinceT⁰⁰⁰ isS-compact, it follows that(x⁰_n)_n∈N^∗ has an adherence value x⁰ ∈T⁰⁰⁰. By choice ofx⁰, we have immediatelyH⁰(x⁰) =H(x)and(x, x⁰)∈A⁰. SinceT⁰⁰ and T⁰⁰⁰ have symmetric roles, we have proven that A⁰ is an height-preserving correspondence.

Since T⁰⁰\T is ν-negligible and T⁰⁰⁰ \T⁰ is ν⁰-negligible, A⁰ still satisfies Conditions (3.4.1)-(3.4.4). For every r ∈ R+, since A⁰ is an height preserving δ-correspondence, it induces a δ-correspondence between Slice_r(T⁰⁰) and Slice_r(T⁰⁰⁰). With Proposition 3.4.1, we have

d_GHP(Slice_r(T⁰⁰),Slice_r(T⁰⁰⁰))≤δ.

By definition of d_LGHP, this yields

d_LGHP(T⁰⁰, T⁰⁰⁰)≤δ.

SinceT⁰⁰andT⁰⁰⁰are the completions ofT andT⁰, we haveT⁰⁰=τ(h,(ζ_n)_n∈N^∗,(u_n)_n∈N^∗)and T⁰⁰⁰ =τ(h,(ζ_n⁰)_n∈N^∗,(u_n)_n∈N^∗). Since δ =||(ζ_n)_n∈N^∗ −(ζ_n⁰)_n∈N^∗||_∞,we have proven that the application(ζn)_n∈N^∗7→τ(h,(ζn)_n∈N^∗,(un)_n∈N^∗) is 1-Lipschitz.

Step 3: we prove that if(ζ_n)_n∈N^∗ ∈R^ω+,0,(u_n)_n∈N^∗ 7→τ(h,(ζ_n)_n∈N^∗,(u_n)_n∈N^∗) is continu-ous fromDwith the pointwise convergence topology toT. Takeε >0andn0 ∈N^∗ such that for everyn > n0,ζn= 0. Set δ= _2(n^ε

By surjectivity ofρandρ⁰,Ais a correspondence as well. Now, we find a simpler expression of

4.2. SOME MEASURABLE MAPS OVERT 93 dandd⁰. For(z1, z₁⁰),(z2, z₂⁰)∈A, there exists((x1, y1),(x⁰₁, y₁⁰)),((x2, y2),(x⁰₂, y⁰₂))∈AE such that (z₁, z₁⁰) = (ρ(x₁, y₁), ρ⁰(x⁰₁, y⁰₁)) ∈ A and (z₂, z₂⁰) = (ρ(x₂, y₂), ρ⁰(x⁰₂, y₂⁰)). By definition of A_E, there exists i₁, i₂ between 0 and n₀ such that for j ∈ {1,2}, u_σ(ij) < x_j ≤u_σ(ij+1). Supposex1 ≤x2, we have

d(z₁, z₂) =y₁+y₂−2(y₁∧y₂∧( inf

n∈N^∗ x1≤un<x2

h−ζ_n))

=y₁+y₂−2(y₁∧y₂∧( min

1≤i≤n₀ x1≤u_σ(i)<x2

h−ζ_σ(i))∧( inf

n>n0

x1≤u_n<x2

h−ζ_n))

=y₁+y₂−2(y₁∧y₂∧( min

i1<i≤i₂h−ζ_σ(i))).

For the last equality, we used the fact that for n > n₀, h−ζ_n =h ≥ y₁, and the fact that since forj∈ {1,2},u_σ(ij)< xj ≤u_σ(ij+1), we have u_σ(i)< xj if and only ifi≤ij. Note that the distance depends only on(i1, y1) and (i2, y2), so we have

d(z₁, z₂) =y₁+y₂−2(y₁∧y₂∧( min

i1<i≤i₂h−ζ_σ(i))) as soon asi₁ ≤i₂. Similarly, we have, supposing thati₁ ≤i₂, that:

d⁰(z₁⁰, z₂⁰) =y₁⁰ +y₂⁰ −2(y₁⁰ ∧y₂⁰ ∧( min

i1<i≤i2

h−ζ_σ(i))).

By definition ofA_E, we havey₁=y⁰₁andy₂=y₂⁰, so we have proven thatd(z₁, z₂) =d⁰(z⁰₁, z₂⁰) andH(z1) =H⁰(z₁⁰). This means thatAinduces a height-preserving isometry betweenT and T⁰, and satisfies Conditions (3.4.1) and (3.4.2).

We set, for all 0 ≤ i ≤ n₀, B_i = {h} ×(u_σ(i), u_σ(i+1)] and B_i⁰ = {h} ×(u⁰_σ(i), u⁰_σ(i+1)].

For every 0≤i≤n0,ν(Bi) =u_σ(i+1)−u_σ(i) and ν(B_i⁰) =u⁰_σ(i+1)−u⁰_σ(i). According to our previous calculation,ρ is constant on eachB_i andρ⁰ on eachB_i⁰, and theB_i form a partition ofSupp(ν). TakeB ⊂T. Since all the mass is at heighthandApreserves the height, we can neglect the part ofB situated strictly under heighth. Sinceρis constant on theB_i,ρ⁻¹(B)is of the form ^S_i∈IB_i forI some subset of{0, ..., n₀}. SetB⁰ ={z⁰ ∈T⁰ | ∃z∈B,(z, z⁰)∈A}.

By definition of A, it means that B⁰ =ρ⁰(^S_i∈IB_i⁰). It follows that ν(B) =^X

i∈I

ν(B_i)

=^X

i∈I

(u_σ(i+1)−u_σ(i))

≤^X

i∈I

(u⁰_σ(i+1)−u⁰_σ(i)+ 2δ)

=^X

i∈I

ν(B_i⁰) + 2(n0+ 1)δ

≤ν(B⁰) +ε.

We have proven Condition (3.4.3) for A. By symmetry of the roles for T and T⁰, we have Condition (3.4.4) as well, soAis aε-correspondence. SinceApreserves the labels,Ainduces a ε-correspondence between Slice_r(T) and Slice_r(T⁰) for every r ∈ R+. Using Proposition

3.4.1, we have dGHP(Slice_r(T),Slice_r(T⁰)) ≤ ε for every r ∈ R+, so by Definition 3.1.12 we have d_LGHP(T, T⁰) ≤ ε. Since ε was arbitrary in R^∗+ and (u⁰_n)_n∈N^∗ was arbitrary in a ball of center (u_n)_n∈N^∗ and radius δ > 0, we have proven that if (ζ_n)_n∈N^∗ ∈ R^ω_+,0, the map (un)_n∈N^∗ 7→τ(h,(ζn)_n∈N^∗,(un)_n∈N^∗) is continuous.

Conclusion: Since τ(h,(ζ_n)_n∈N^∗,(u_n)_n∈N^∗) is 1-Lipschitz in(ζ_n)_n∈N^∗ and lim_n∈N^∗ζ_n = 0, the map(un)_n∈N^∗ 7→τ(h,(ζn)_n∈N^∗,(un)_n∈N^∗) is the uniform limit of

(un)_n∈N^∗ 7→τ(h,(1_n≤n0ζn)_n∈N^∗,(un)_n∈N^∗)

when n0 goes to ∞. With the result of Step 3, (un)_n∈N^∗ 7→ τ(h,(ζn)_n∈N^∗,(un)_n∈N^∗) is the uniform limit of continuous functions, hence is continuous.

The map τ(h,(ζ_n)n∈N^∗,(u_n)n∈N^∗) is continuous in (u_n)n∈N^∗ and 1-Lipschitz in (ζ_n)n∈N^∗

over its domain, so it is continuous in((ζn)_n∈N^∗,(un)_n∈N^∗). The map h, (ζn)_n∈N^∗,(un)_n∈N^∗7→τ(h,(ζn)_n∈N^∗,(un)_n∈N^∗)

is continuous in its two variables h and ((ζ_n)_n∈N^∗,(u_n)_n∈N^∗), and R is separable, so, using Lemma 3.4.13 the mapτ is measurable from R×R^N

∗

+,0×D to(T, d_LGHP).

Dans le document Une topologie pour les arbres labell´ es, application aux arbres al´ eatoires (Page 85-94)