The structure of the space of weak equivalence classes for general acting groups

TOPOLOGY AND CONVEXITY IN THE SPACE OF ACTIONS MODULO WEAK EQUIVALENCE

2.6 The structure of the space of weak equivalence classes for general acting groups

Moreover, let x ∈ X, t ∈ [0,1] and a,b ∈ A(Γ,X, µ) and consider the action ta+(1−t)b on t X1t (1−t)X2. We have stabta+(1−t)b = staba(x) if x ∈ X1 and stab_b(x)if x ∈ X2. Thus for any H ≤ Γ, {x : stab_ta+(1−t)b(x) = H} = {x ∈ X1 : stab_a(x)= H} t {x ∈ X2: stab_b(x)=H}so for any A ⊆ Sub(Γ)we have

(tµ₁+(1−t)µ₂)({x: stab_ta+(1−t)b(x) ∈ A})

= (tµ1+(1−t)µ2)({x ∈ X1 : staba(x) ∈ A} t {x ∈ X₂: stabb(x) ∈ A})

= tµ({x : staba(x) ∈H})+(1−t)µ({x : stabb(x) ∈ A}).

Therefore type(ta+(1−t)b)=t(type(a))+(1−t)(type(b))and Theorem 2.1.1 fol- lows. Note in particular that ifΓis amenable thenta+(1−t)a∼ a, so for amenable

Letα = sup_y∈Ad(y,B)and let C = {y ∈ A: d(y,B) = α}. ThenC is nonempty, disjoint fromBandC is a face ofA.

Let F be the family of faces ofC, ordered by reverse inclusion. Suppose {Fi}_i∈I is a linearly ordered subset of F and considerÑ

i∈IFi. If x,y ∈ C and 0 < t < 1 are such that t x+ (1− t)y ∈ Ñ

i∈I Fi, then x,y ∈ Fi for each i since each Fi is a face. HenceÑ

i∈I Fi is a face. It is nonempty by compactness. So Zorn’s Lemma guarantees there exist minimal elements ofF. LetF be such a minimal element.

Choose y ∈ F and suppose toward a contradiction that there exists y⁰ ∈ F with y⁰ < cch({y}). Then cch({y}) is a compact convex set, so letting G = z∈ F :d(z,cch({y}))=sup_w∈Fd(w,cch({y})) , G is a nonempty face of F dis- joint from cch({y}), contradiction the minimality of F. So for all y ∈ F we have F ⊆ cch({y}). Fix such a y. Note that cch({y}) = ch({y}). We claim that y is an extreme point of C. Assuming this, sinceC is a face of A we have that y is an extreme point ofAand we have a contradiction to the hypothesis thatC∩B= ∅.

Suppose first that there do not exista,b ∈Cand 0 < t < 1 such thaty= ta+(1−t)b.

Thenyis an extreme point ofCbe definition. So leta,b∈Cand 0< t < 1 be such thaty= ta+(1−t)b. We must show thaty∼ a∼ b. SinceFis a face ofC, we have a,b ∈ F. Thus we can write a = Ín

i=1siy and b = Ík

i=1riy for si,ri ∈ [0,1]. By Proposition 2.4.2 and associativity we havey ∼

Ín

i=1tsiy+Ík

i=1(1−t)riy

. Since 0 < t < 1, iterating this argument we find that for any δ > 0, there is m ∈ Nand (λi)^m_i=1 ⊆ [0,1]such thatλi ≤ δfor alliandy ∼Ím

i=1λiy.

We claim that this impliesy∼ κy+(1−κ)yfor allκ ∈ [0,1]. Note thatκy+(1−κ)y is isomorphic toικ,1−κ×y, whereικ,1−κis the trivial action ofΓon({0,1},mκ)where mκ({0})= κandmκ({1})= 1− κ. Hence yis a factor ofκy+(1− κ)y and it thus suffices to showκy+(1−κ)y ≺ y.

Let X1,X2 be two copies of X, let n,k ∈ N, > 0 and a partition P = (Pi)_i^k₌

1

of X1tX2be given. As before, we get a partition P₁= (P_i¹)_i^k₌

1withP_i¹= Pi∩X1

ofX₁and similarly a partitionP₂ =(P_i²)_i^k₌₁withP_i²= Pi∩X₂ofX₂. Now, choose δ < ₂. Then we can findmand(λp)^m_p=1such thaty ∼Ím

p=1λpyand for somel ≤ m we haveκ−

2 ≤ Íl

p=1λp ≤ κ. Let now X⁰_pbe a copy of X for each p≤ m, and for q ∈ {0,1}letP_i,p^q be the corresponding copy ofP_i^qsitting inX⁰_p. LetQ= (Qi)^k

i=1be

the partition ofÃm

p=1X⁰_pgiven byQi = Ãl

p=1P_i,p¹

t Ãm

p=l+1P_i,p²

. Then fors ≤ n andi,j ≤ kwe have

(κµ+(1−κ)µ)(γ_s^{κy+(1−κ)y}Pi∩Pj) −©

­

«

m

Õ

p=1

λpµª

®

¬

γ^Í_s ^mp=1λpyQi∩Qj

≤

κµ(γ_s^yP_i¹∩P¹_j) −©

­

«

l

Õ

p=1

λpµ(γ_s^yP_i,p¹ ∩P¹_j,p)ª

®

¬

+

(1−κ)µ(γ_s^yP_i²∩P²_j) −©

­

«

m

Õ

p=l+1

λpµ(γ_s^yP_i,p² ∩P²_j,p)ª

®

¬

=

κµ(γ_s^yP_i¹∩P¹_j) −©

­

«

l

Õ

p=1

λpª

®

¬

µ(γ_s^yP_i¹∩P¹_j)

+

(1−κ)µ(γs^yP_i²∩P²_j) −©

­

«

m

Õ

p=l+1

λª

®

¬

µ(γs^yP_i²∩P²_j)

=

©

­

« κ−

l

Õ

p=1

λpª

®

¬

µ(γ_s^yP_i¹∩P¹_j)

+

©

­

«

(1−κ) −

m

Õ

p=l+1

λpª

®

¬

µ(γs^yP_i²∩P²_j)

≤

©

­

« κ−

l

Õ

p=1

λpª

®

¬

+

©

­

«

(1− κ) −

m

Õ

p=l+1

λpª

®

¬

≤ 2 +

2 = . Sincey ∼Ím

p=1λpy,κy+(1−κ)y ≺ yand we are done.

We note that a metrizable topological vector spaceV is locally convex if and only if its topology is induced by a countable family of seminorms | · |^V_n^∞

n=1. Then p(v,w)=Í∞

n=1 1

2ⁿ|v−w|^V_n is a compatible metric onV, which is easily seen to obey Lemma 2.4.1. Thus the technique used to prove Theorem 2.1.2 works to prove the metrizable case of the classical Krein-Milman theorem using only the convex and metric structure ofV, not the vector space structure in the form of linear functionals.

Before proving Theorem 2.1.3, we briefly discuss the ergodic decomposition in the

context of weak equivalence classes. Supposea∈A(Γ,X, µ)anda =∫

Zazdη(z)is the ergodic decomposition of a, that is to say we have a factor mapπ : (X, µ) → (Z, η) such that if µ = ∫

Z µzdη(z)is the disintegration of µover (Z, η) viaπ then µz(π⁻¹(z)) = 1 and Γ y^a (π⁻¹(z), µz) is isomorphic to az. Furthermore, the assignment z 7→ µz from(Z, η) → Ma(X)is Borel, where Ma(X) is the space of a-invariant probablity measures onX(we may assume here thatXis a Polish space).

Recall thatA^∗_∼(Γ)is the space of weak equivalence classes of all measure-preserving actions ofΓ, including those actions on finite space. A^∗_∼(Γ)is topologized using the exact same metric as we use to topologize A∼(Γ,X, µ). We would like to conclude that the assignment z 7→ [az]is measurable from(Z, η)to A^∗_∼(Γ), where[az]is the weak equivalence class ofaz. This is a consequence of the following lemma.

Lemma 2.6.1. Let Γ y^a Y be a Borel action of Γ on a Polish spaceY. Then the map Θ from Ma(Y) to A^∗_∼(Γ) given by ν 7→ [aν] is Borel, where [aν] is the weak equivalence class of the measure preserving actionaν = Γy^a(Y, ν).

Proof. Fix a measureν ∈ Ma(Y)and considerΘ⁻¹(U), where

U = {[a] ∈ A^∗_∼(Γ): dH(Cn,k(aν),Cn,k(a))< for alln,k ≤ N}

for someN ∈Nand > 0, soUis a basic open neighborhood ofΘ(ν)=aν. Since U =

∞

Ø

m=1 N

Ù

n,k=1

[b] ∈ A^∗_∼(Γ):dH(C_n,k(a_ν),C_n,k(b)) ≤ − 1 m

,

it suffices to showΘ⁻¹(V)is Borel for a setV of the form

V = {[b] ∈ A^∗_∼(Γ):dH(Cn,k(aν),Cn,k(b)) ≤r}.

Fixingnandkwe writeC(b)forCn,k(b). Now, letK andLbe compact subsets of a compact Polish spaceW with metricp, letDK be dense inK andDL be dense in L.

We have

dH(K,L) ≤r ⇐⇒ max

x∈K inf

y∈Lp(x,y) ≤r and max

y∈L inf

x∈Kp(y,x) ≤r

⇐⇒ (∀x ∈K)(∀δ >0)(∃y ∈ L)(p(x,y)<r +δ)

∧ (∀y ∈L)(∀δ >0)(∃x ∈ K)(p(y,x)<r +δ)

⇐⇒ (∀x ∈DK)(∀δ >0)(∃y ∈ DL)(p(x,y) <r +δ)

∧ (∀y ∈DL)(∀δ >0)(∃y ∈ DL)(p(y,x)< r+δ).

IfLis a countable algebra generating the Borelσ-algebraB(Y)ofY, thenLis dense in MALG(Y, ρ) for any Borel probability measure ρ onY. Regarding a partition ofY into k pieces as an element of B(Y)^N and consideringL^k, we see that there exists a fixed countable family(A_m)^∞_m=

1 of partitions ofY such that for any Borel probability measure ρ onY, (A_m)^∞_m=1is dense in the set of k-partitions of X with topology inherited from MALG(Y, ρ). We may further assume that each element of eachA_m is clopen. This implies that the set (M^Am(aρ))^∞_m=1 is dense inC(aρ) for any Borel probability measure ρ. Therefore we have

V =

∞

Ù

m=1

∞

Ù

l=1

∞

Ø

i=1

b∈ A^∗_∼(Γ):

M^Am(aν) − M^Ai(b)

<r+ 1 l

!

∩

∞

Ù

m=1

∞

Ù

l=1

∞

Ø

i=1

b∈ A^∗_∼(Γ):

M^Ai(aν) −M^Am(b)

<r + 1 l

! .

Now,|M^Ai(aν) −M^Am(aρ)| < sif and only if|ν(γ^aA_i^j ∩A^t_i) −ρ(γ^aAⁱ_m ∩A^t_m)| < s for all A_i^j,A^t_i ∈ A_i and Aⁱ_m,A^t_m ∈ A_m. Since for any pair J1,J2 ⊆ Y the set {ρ: |ν(J₁) −ν(J₂)| < s}is Borel, we see

Θ⁻¹ b∈ A^∗_∼(Γ):

M^Ai(aν) −M^Am(b)

<r+ 1 l

is Borel and consequentlyΘ⁻¹(V)is Borel.

We now prove Theorem 2.1.3

Proof. (of Theorem 2.1.3)LetΘ: Z → A^∗_∼(Γ)be the map sending each point inz to the weak equivalence class [az], so Θis measurable by Lemma 2.6.1. Suppose towards a contradiction that the theorem fails. Then for every set Z⁰ ⊆ Z with η(Z⁰) =1, there is more than one weak equivalence class in the set{[az]: z ∈ Z⁰}.

Equivalently, the measureΘ∗ηon A^∗_∼(Γ)is not supported on a single point. We can thus split A^∗_∼(Γ) into two disjoint sets Y₁,Y₂ such that 0 < Θ_∗η(Y₁),Θ_∗η(Y₂) < 1.

Letting Ai = Θ⁻¹(Yi), we get disjoint measurable sets A₁,A₂ ⊆ Z such that 0< η(A₁), η(A₂)< 1 and for all z∈ A₁and allw ∈ A₂we have thatz / w.

Recall that for a measure-preserving action bof Γ and n,k ∈ N the setCn,k(a) ⊆ [0,1]^n×k×k was defined in Section 2.3.

Lemma 2.6.2. For any action b of Γ on a probability space (Y, ν), we have cch(Cn,k(b)) ⊆Cn,k(ι×b).

Proof. Write Cn,k(b) = C(b). Suppose x ∈ cch(C(b)). Then we can find points (xi)_i^∞₌

1 such that limi→∞xi = x and each xi has the form xi = Í^j(i)

j=1α_i^jx_i^j for (x_i^j)^j(i)

j=1 ⊆ C(b) and (α_i^j)^j(i)

j=1 ⊆ [0,1] with Í^j(i)

j=1αⁱ_j = 1 for each i. Without loss of generality we may assume that each x_i^j has the form M^Aij(b) for a partition A_i^j =(A_i,l^j )^k_l=

1ofY intokpieces. Fixingiconsider the actionÍ^j(i)

j=1α_i^jbon the space

Ã^j(i)

j=1Yj,Í^j(i)

j=1α_i^jνj

, where each (Yj, νj) is a copy of (Y, ν). Let B = (Bl)^k_l=1 be the partition of Ã^j(i)

j=1Yj given by lettingBl = Ã^j(i)

j=1A_i,l^j , where A_i,l^j sits inside the j copy ofY. For any p ≤ nandl,m ≤ k and x ∈ [0,1]^n×k×k let(x)_p,l,m be thep,l,m coordinate of x. We then have

©

­

« M^B©

­

«

j(i)

Õ

j=1

α_i^jbª

®

¬ ª

®

¬^p,l,m

= ©

­

«

j(i)

Õ

j=1

α_i^jνjª

®

¬

γ^Í

j(i) j=1α_i^jb

p Bl∩Bm

=

j(i)

Õ

j=1

α_i^jνj(γ^b_pA_i,l^j ∩A_i,m^j )

=

j(i)

Õ

j=1

α_i^j

M^Aij(b)

p,l,m.

Therefore

M^B©

­

«

j(i)

Õ

j=1

α_i^jbª

®

¬

=

j(i)

Õ

j=1

α_i^j

M^Aij(b)

= xi. We have shown thatxi ∈C

Íj(i) j=1α_i^jb

. SinceÍj(i)

j=1α_i^jbis a factor ofb×ι, we have xi ∈C(b×ι). Since limi→∞xi= xandC(b×ι)is closed, the lemma follows.

It is clear that for any two measure-preserving actionsb,cwe haveb ≺cif and only ifCn,k(b) ⊆ Cn,k(c)for alln,k. We claim that there are disjoint subsets A₃,A₄ ⊆ Z of positive measure such that for some pairn0,k0, everyz ∈ A3and everyw ∈ A4we haveCn₀,k₀(az) * cch(Cn₀,k₀(aw)). For z ∈ A3let Rz = {w ∈ A2 : az ⊀ aw}. Since azis ergodic,az ≺ aw×ιimpliesaz ≺ aw. ThereforeRz = {w ∈ A2: az ⊀ aw×ι}.

Assume first that there is a setD₃ ⊆ A₁withη(D₃) > 0 such that for eachz ∈ D₃we haveη(Rz) >0. Write ˆKfor cch(K). By Lemma 2.6.2 we can writeRz =Ð∞

n,k=1R_z^n,k whereR_z^n,k = n

w ∈ A₂:Cn,k(az)* Cœn,k(a_w)o

. Thus for eachzthere is a lexicograph- ically least pair(nz,kz)such thatη(Rⁿ_z^z,kz) > 0. Therefore there is a pairn₀,k₀and

a setD₄ ⊆ D₃such thatη(D₄)> 0 and for allz ∈ D₄we haveη(Rⁿ_z^0,k0) > 0. Fixing n0andk0we writeC(b)forCn₀,k₀(b). Let(wj)^∞_j=

1⊆ A2be a sequence of points such that the family

C›(a_wj)∞

j=1is dense in the space n

›C(a_w):w ∈ A₂ o

with respect to the Hausdorff metricdH on the space on compact subsets of[0,1]^n0×k0×k0. Let then Fj,l = n

w ∈ A₂: dH

C›(a_w),C›(a_wj)

< ¹_lo .

Fix z ∈ D4 and choose w ∈ Rⁿz^0,k0. By hypothesis there is > 0 such that C(az) * B

C›(aw)

, whereB(K)denotes the ball of radius aroundK. Then if we choosejso thatdH

C›(aw_j),Cš(w)

< ₂ andlso that¹_l < ₂we havew ∈ Fj,l ⊆ Rⁿz^0,k0. Hence there is a subsetJ ⊆N²such thatRⁿ_z^0,k0 =Ð

(j,l)∈JFj,l. So for eachzwe can choose a lexicographically least pair(jz,lz)such thatη(Fj_z,l_z)> 0 andFj_z,l_z ⊆ Rⁿ_z^0,k0. There is then a pair(j₀,l₀)and a setE₃ ⊆ D₃withη(E₃) >0 such thatη(Fj₀,l₀)> 0 and for all z ∈ E3 and allw ∈ Fj₀,l₀ we haveC(az) * C›(aw). So take A3 = E3and A4 = Fj₀,l₀. Thus we are left with the case η(Rz) = 0 for almost allz ∈ A1. Then for almost allw ∈ A2and almost allz ∈ A1we must haveaw ⊀ az, so a symmetric argument gives the claim.

Given a (real) topological vector space V, we say a hyperplane in V is a set of the form H`,α = {v ∈ V : `(v) = α} for some continuous linear functional` and α ∈ R. Given disjoint compact subsetsW<sub>1</sub>,W<sub>2</sub> ⊆ V we say thatH`,α separatesW₁ fromW₂ifW₁⊆ {v ∈V :`(v)< α} andW<sub>2</sub>⊆ {v ∈V :`(v)> α}.

Lemma 2.6.3. LetS ⊆ Rⁿbe compact. Then there is a countable family(Hi)_i^∞₌

1of hyperplanes such that for anyx ∈Sand any compact convexW ⊆ Sthere isisoHi

separates{x}fromW.

Proof. Let(`j)<sup>∞</sup><sub>j=1</sub>be a countable set of linear functionals which is dense in the sup norm onS. Enumerate Qas(qm)<sup>∞</sup><sub>m=1</sub> and let H<sub>j,m</sub> = {s ∈ S : `j(s) = qm}. Given x and W, by Hahn-Banach find a linear functional ` and α ∈ Rso that H = H<sub>`,α</sub> separatesxfromW. Letr =min

infh∈H ||x−h||,inf_h∈H,

w∈W

||h−w||

sor >0. Then choosemso|qm −α| < ^r₂ and j so sup_s∈S|`(s) −`j(s)| < ^r₂. ThenHj,m separatesx

fromW.

Now takeS =[0,1]^n0×k0×k0and fix a family(Hi)^∞

i=1of hyperplanes as in the lemma.

Since C›(a_w) is compact convex for each w ∈ A₄ and for all z ∈ A₃ we have C(az) * C›(a_w), for each pair(z,w) ∈ A₃× A₄there is an indexi(z,w)and a point

x_z,w ∈C(az)such thatH_i(z,w)separatesx_z,wfromC(a›_w). Fixz ∈ A₃. Taking(wj)^∞_j=1 as before, for (j,l) ∈ N² letGj,l = n

w ∈ A4:dH

C›(aw),C›(aw_j)

< ¹_lo

. Choosing w ∈ A4, let = dH

Cš(w),Hi(z,w)

so >0. Finding jz,w sodH

C›(w_j),Cš(w)

< ₂ and lz,w so ¹_l < ₂ we have w ∈ Gjz,w,lz,w and H_i(z,w) separates xz,w from Cd(u) for all u ∈ Gjz,w,lz,w. Then we have A₄ = Ð

(j_z,w,lz,w):

w∈A₄

Gjz,w,lz,w so we can find w₀ so thatη

Gjz,w0,lz,w0

> 0. Let then Gz = Gjz,w0,lz,w0, xz = xz,w₀ andi(z) = i(z,w0)so that Hi(z) separates xz fromCd(u) for allu ∈ Gz. Since theGz were chosen from a countable family, we can find a set A₅ ⊆ A₃of positive measure such that Gz = G is the same for allz∈ A₅. We can then find an indexiand a set A₆ ⊆ A₅of positive measure such that for all z ∈ K, Hi = H separates xz fromCd(u) for allu ∈ G. H splits[0,1]^n×k×k into two closed convex setsH₊ andH−, whereH₊ contains the xz

andH− contains theC(u).

ForS ⊆ Zwithη(S) > 0 letηS = _η(S)^ηS be normalized measure onS. By Lemma 2.4.2 we haveC∫

GaudηG(u)

⊆ cch(Ð

u∈GC(u)) ⊆ H−. Write A₆ = Ð∞

p=1A^p₆, where A₆^p = n

z ∈ A6:dH(xz,H) ≥ ¹_po

and find p soη(A^p₆) > 0. Letting K = A₆^p, for all z∈ K, xzis an element of the closed convex setH₊^p= {y ∈H₊ :dH(y,H) ≥ ¹_p}and H₊^pis disjoint fromH−. We have∫

KxzdηK(z) ∈C∫

KazdηK(z) and∫

KxzdηK(z) ∈ H₊^p. SinceC∫

GaudηG(u)

⊆ H− we see thatC∫

KazdηK(z)

* C∫

GaudηG(u) and it follows that ∫

KazdηK(z) /

∫

GaudηG(u). Let L₁ = K,L₂ = G then there is i ∈ {1,2}with∫

L_iazdηL_i(z)/ a. Since 0< η(Li)< 1, we can write a=η(Li)

∫

L_i

azdηL_i(z)

+η(Z\Li)

∫

Z\L_i

azdηZ\L_i(z)

,

which contradicts our assumption thatais an extreme point.

We now prove Theorem 2.1.4. Recall that the uniform topology on Aut(X, µ) is given by the metricdu(T,S)= µ({x :T x, S x}). IfP= {P₁, . . . ,Pp}is a partition of a space on which FN acts by an action a, J ⊆ FN is finite and τ : J → p let P_τ^a =Ñ

γ∈Jγ^aPτ(γ).

Proof. (of Theorem 2.1.4)Letabe a free action ofFN. By replacingawitha×ι if necessary, we may assume that for eachn,k the setCn,k(a)is closed and convex.

Fix integers n0 and k0 and > 0. It is enough to find a free ergodic action b of

FN such that for all n ≤ n₀ and k ≤ k₀ we have dH(Cn,k(a),Cn,k(b)) < . Let {γ1, . . . , γn₀} = F0 be the finite subset ofFN under consideration. Let s = s_FN be the Bernoulli shift ofFN acting on 2^FN, ν

whereνis the product measure. For any actioncofFN on(X, µ)andγ ∈FN we have

{(x,y) ∈ X×2^FN :γ^c×s(x,y), γ^a×x(x,y)}= {x ∈ X : γ^cx ,γ^ax} ×Y and hence

(µ×ν)({(x,y) ∈ X×2^FN :γ^c×s(x,y), γ^a×x(x,y)})= µ({x ∈ X :γ^cx , γ^ax}). Assume du(γ^a, γ^c) < ₁₆ for all γ ∈ F0. Then for any measurable partition A = A₁, . . . ,Ak of X×2^FN, allγ ∈ F₀and alli, j ≤ k we have

|(µ×ν)(γ^a×sAi∩Aj) − (µ×ν)(γ^c×sAi∩Aj)| <

16 for allγ ∈ F₀. In the notation of Section 2.3,ρ

M_n,k^A(a×s),M_n,k^A(c×s)

< ₁₆ where ρis the supremum metric on[0,1]^n×k×k. Choose a finite collectionLof measurable subsets of X×2^FN such that for every measurable partitionA of X×2^FN there is a partitionB ⊆ L such that ρ

M_n,k^A(a×s),M_n,k^B (a×s)

< ₁₆ . Then for every such Athere existsB ⊆ L such thatρ

M_n,k^A(c×s),M_n,k^B (c×s)

< ³₁₆.

Forγ ∈FN letπγ : 2^FN → 2 be projection onto theγ coordinate. Fori ∈ {0,1}let Si =π⁻_e¹({i})and putS ={S₁,S₂}. Choose now a finite partitionR = {R₁, . . . ,Rr} of X and a finite subset F ⊆ FN containingF₀such that for every A ∈ L there are setsRj with 1 ≤ j ≤ r and a family of functions(τj)^t

j=1withτj : F →2 such that µ©

­

«

©

­

«

t

Ä

j=1

Rj×S_τ^s_jª

®

¬ 4Aª

®

¬

<

16.

WriteP= R × S. We can identify a functionθ : F →r×2 with a pair(σ, τ)where σ : F →r andτ: F →2 so

P_θ^c×s = Ù

γ∈F

γ^bP_θ(γ)^c×s = Ù

γ∈F

γ^cR_σ(γ)

!

× Ù

γ∈F

γ^sS_τ(γ)

!

= R^c_σ×S_τ^s.

Note that for any j ≤ r, Rj×S_τ^sis a finite disjoint union of sets of the formR_σ^c ×S_τ^s, hence any A ∈ L is within ₁₆ of finite disjoint union of sets of the form P_θ^c×s for θ : F →r×2.

Letδ= _4(2r)2|F|. Fix an ergodic actioncofFNsuch thatdu(γ^a, γ^c)< ^δ2

32|F|²(2r)^|F|2 for allγ ∈F. (For example use the fact that the ergodic automorphisms are uniformly dense in Aut(X, µ)to move one of the generatorsγ ofFN so it acts ergodically but is still sufficiently close toγ^a). Then clearly dH(Cn,k(a),Cn,k(c)) < ₂ for alln≤ n0

and k ≤ k₀. Let b = c×s. Sincec is ergodic and s is free and mixing, b is free and ergodic. Thus it is sufficient to show dH(Cn,k(c),Cn,k(b)) < ₂ for alln ≤ n₀, k ≤ k₀. Sincec ≺ b, it is sufficient to show that for every partitionA of X×2^FN there is a partition C of X such that ρ

M_n,k^A(b),M_n,k^C (c)

< ₂. By our previous reasoning, for each partition A = (A₁, . . . ,Ak) of X × 2^FN there is a partition B whose pieces are disjoint unions of sets of the formP_θ^bforθ : F →r ×2 such that ρ

M_n,k^A(b),M_n,k^B (b)

< ₄.

Claim 2.6.1. There is a partition Q of X indexed by r × 2 such that for every θ : J →r ×2with J ⊆ F₀F we have|(µ×ν)(P_θ^b) − µ(Q^c_θ)| < δ.

Suppose the claim holds. Regard FN as acting on Ð

J⊆FN{θ : J → 2× r} by shift, γ· θ(γ⁰) = θ(γ⁻¹γ⁰). Thus the domain dom(γ ·θ)= γdom(θ). Then for any θ, κ :F → 2×r andγ ∈F0we have

γ^bP_θ^b∩P_κ^b=











P_γ·θ∪κ^b ifγ·θandκ are compatible,

∅ if not.

and similarly

γ^cQ^c_θ∩Q^c_κ =











Q^c_γ·θ∪κ ifγ·θ andκare compatible,

∅ if not.

Therefore the claim gives |(µ× ν)(γ^bP_θ^b∩P_κ^b) − µ(γ^cQ^c_θ ∩Q^c_κ)| < δ for all θ, κ : F → r ×2. So if B = {B₁, . . . ,Bk} is a partition such that Bi = Ãt

s=1P_θ^b

i(s) for functionsθi(s): F →r×2 and we letCi = Ãt

s=1Q^c_θ

i(s), then we have

|(µ×ν)(γ^bBi∩Bj) − µ(γ^cCi∩Cj)| =

(µ×ν)

t

Ä

s,s⁰=1

γ^bP_θ^b

i(s) ∩P_θ^b

j(s⁰)

!

− µ

t

Ä

s,s⁰=1

γ^cQ^c_θ

i(s)∩Q^c_θ

j(s⁰)

!

≤ t²δ ≤ (2r)^2|F|δ <

4,

sincet ≤ (2r)^|F|. TakingC =(Ci)^k

i=1we getρ

M_n,k^B (b),M_n,k^C (c)

< ₄, which implies the theorem.

It remains to show Claim 2.6.1. This part of the argument follows the proof of Theorem 1 in [2] and the extensions of these ideas developed in [74]. LetG= F₀F. Assume without loss of generality thatGis closed under taking inverses. Note that it suffices to prove the claim forθdefined on all ofG. In order to findQwe will find a partition T = {T₁,T₂} and setQi,j = Ri∩Tj for 1 ≤ i ≤ r, 1 ≤ j ≤ 2. Thus we are looking forT = {T₁,T₂} such that for all(τ, σ)withσ :G→ r andτ :G→ 2 we have

|(µ×ν)(R_σ^c ×S_τ^s) −µ(R_σ^c ∩T_τ^c)| < δ.

Note thatν(S_τ^s)=2^−|G|for any suchτso we are looking forTsuch that

2^−|G|µ(R^c_σ) − µ(R_σ^c ∩T_τ^c) < δ.

The idea is that a randomTshould have this property.

Without loss of generality we may assume X is a compact metric space with a compatible metric p. Forη >0 let

D_η= {x ∈ X : for allγ, γ⁰∈G, γ₁ ,γ₂implies p(γ₁^cx, γ₂^cx) > η}

and

Eη = {(x,x⁰) ∈ D_η²: for allγ1, γ2 ∈G,p(γ₁^cx, γ₂^cx⁰) > η}. Lemma 2.6.4. There isη > 0such that µ(Dη) > 1− ^δ2

16(2r)^|F|2 and µ²(X²\Eη) <

δ² 16(2r)^2|F|.

Proof. Clearly ifη₁ < η₂ then Dη₂ ⊆ Dη₁. We haveX \Ð

η>0Dη = {x ∈ X : for someγ₁ , γ₂ ∈ G, γ₁^cx = γ₂^cx}. Now sincea is free, ifγ₁^cx = γ^c₂x then we must haveγ_i^cx , γ_i^ax for somei ∈ {1,2}. Eachγ ∈ Gis a product f1f2for f1 ∈ F0and

f2∈ F, thus for anyγ ∈Gwe have

du(γ^c, γ^a)< du(f₁^a, f₁^c)+du(f₂^a, f₂^c) < δ² 16|F|²(2r)^|F|2 since fi ∈ F. Therefore

µ({x : for someγ ∈G, γ^cx , γ^ax})< |G| δ²

16|F|²(2r)^|F|2 < δ² 16(2r)^2|F|

and hence µ X\Ð

η>0Dη

< ^δ2

16(2r)^|F|2. So we can find η = η0 such that Dη₀

satisfies the lemma. Now for anyη > 0, D²_η0 \Ø

η>0

Eη ={(x,x⁰) ∈ D²_η0 : for allη > 0

there existγ₁, γ₂∈Gsuch that p(γ₁x, γ₂x⁰)< η}

={(x,x⁰) ∈ D²_η0 : there existγ₁, γ₂ ∈Gsuch thatγ₁x =γ₂x⁰}. For a fixedx, {(x,x⁰) ∈ D²_η0 : there existγ₁, γ₂ ∈Gsuch thatγ₁x = γ₂x⁰}is finite so µ

D_η²₀\Ð

η>0Eη

has measure 0 by Fubini and hence we have the lemma for

E_η.

LetY = {Y₁, . . . ,Ym}be a partition of X into pieces with diameter < ^η₄. For x ∈ X letY(x)be the unique l ≤ m such that x ∈ Yi. Let κ be the uniform (= product) probability measure on 2^m and for eachω ∈2^m define a partitionZ(ω)={Z₁^ω,Z₂^ω} by letting x ∈ Z_i^ω if and only if ω(Y(x)) = i. Thus we have a random variable Z : (2^m, κ) → MALG(X, µ)² given by ω 7→ Z(ω). Fix now τ : G → 2 and an arbitrary subset A ⊆ X. We compute the expected value of µ(Z(ω)_τ ∩A). Let χB

be the characteristic function ofB.

E[µ(Z_τ∩A)]=∫

2^m

µ(Z(ω)_τ∩A)dκ(ω)

=∫

2^m

∫

X

χ_Z(ω)τ∩A(x)dµ(x)dκ^m(ω)

=

∫

A

∫

2^m

χZ(ω)τ(x)dκ(ω)dµ(x)

=∫

Dη∩A

∫

2^m

χZ(ω)τ(x)dκ(ω)dµ(x)+∫

A\D_η

∫

2^m

χZ(ω)τ(x)dκ(ω)dµ(x). (2.6) Now if x ∈ D_η then for allγ₁ , γ₂ ∈Gwe have p(γ₁^cx, γ₂^cx) ≥ ηso thatY(γ^c₁x) , Y(γ₂^cx) and hence the events ω(Y(γ₁^cx)) = i and ω(Y(γ₂^cx)) = j are independent.

We have x ∈ γ^cZ(ω)_τ(γ) if and only if ω(Y((γ⁻¹)^cx)) = τ(γ), so if x ∈ Dη and γ₁ , γ₂ ∈ G the events x ∈ γ^cZ(ω)_τ(γ1) and x ∈ γ^cZ(ω)_τ(γ2) are independent. So for x ∈Dη,

∫

2^m

χ_Z(ω)τ(x)dκ(ω)= κ({ω : x ∈γ^cZ(ω)_τ(γ) for allγ ∈G})

=Ö

γ∈G

κ

ω :ω(Y((γ⁻¹)^cx))= τ(γ)

=2^−|G|. (2.7)

Since µ(X \Dη)< ^δ2

16(2r)^|F|2, we have 2^−|G|

µ(A) − ^δ2

16(2r)^|F|2

≤ (6) ≤ 2^−|G|µ(A)+

δ²

16(2r)^|F|2 and thus

E[µ(Zτ∩A)] −µ(A)2^−|G|

< ^δ2

16(2r)^|F|2. We now compute the second moment of µ(Zτ∩A), in order to estimate its variance.

Eµ(Zτ ∩A)²

=∫

2^m

µ(Zτ(ω) ∩A)²dκ(ω)

=∫

2^m

∫

A

χ_Zτ(ω)(x)dµ(x) 2

dκ(ω)

=

∫

2^m

∫

A²

χZ_τ(ω)(x₁)χZ_τ(ω)(x₂)dµ²(x₁,x₂)dκ(ω)

=∫

A²

∫

2^m

χZ_τ(ω)(x₁)χZ_τ(ω)(x₂)dκ(ω)dµ²(x₁,x₂)

=∫

A²∩E_η

∫

2^m

χ_Zτ(ω)(x₁)χ_Zτ(ω)(x₂)dκ(ω)dµ²(x₁,x₂) +∫

A²\E_η

∫

2^m

χ_Zτ(ω)(x1)χ_Zτ(ω)(x2)dκ(ω)dµ²(x1,x2).

(2.8) Now if(x₁,x₂) ∈ Eη then for any pairγ₁, γ₂ ∈Gwe havep(γ₁^cx₁, γ₂^cx₂)> ηso that Y(γ₁^cx₁),Y(γ₂^cx₂)and thus for a fixed pair(x₁,x₂)the eventsω(Y(γ⁻¹)^cx₁)= τ(γ) for allγ ∈ Gandω(Y(γ⁻¹)^cx2) = τ(γ)for allγ ∈ Gare independent. Hence for a fixed(x1,x2) ∈ Eηwe have

∫

2^m

χ_Zτ(ω)(x₁)χ_Zτ(ω)(x₂)dκ(ω)

= κ({ω: x₁ ∈γ^cZ(ω)_τ(γ) andx₂ ∈γ^cZ(ω)_τ(γ)for allγ ∈G})

= κ({ω:ω(Y((γ⁻¹)^cx₁)= τ(γ)andω(Y((γ⁻¹)^c)x₂)= τ(γ)for allγ ∈G})

= κ

ω:ω(Y((γ⁻¹)^cx₁))= τ(γ)for allγ ∈G

·κ

ω :ω(Y((γ⁻¹)^cx2))=τ(γ)for allγ ∈G

=2^−2|G|

by(7)and the fact thatEη ⊆ D_η². Since µ²(A\Eη)< ^δ2

16(2r)^|F|2 we see

µ(A)²− δ² 16(2r)^|F|2

2^−2|G| ≤ (8) ≤ 2^−2|G|µ(A)²+ δ² 16(2r)^|F|2 and hence

E[µ(Zτ∩A)²] − µ(A)²2^−2|G|

< ^δ2

16(2r)^|F|2. Therefore Var(µ(Z_τ∩A))=E[µ(Z_τ∩A)²] −E[µ(Z_τ ∩A)]²

≤

E[µ(Zτ ∩A)²] −µ(A)²2^−2|G|

+µ(A)²2^−2|G|

−

−

E[µ(Zτ∩A)] −µ(A)2^−|G|

+ µ(A)2^−|G|

2

≤ δ²

16(2r)^|F|2 +µ(A)²2^−2|G|−

− δ²

16(2r)^|F|2 +µ(A)2^−|G|

2

= δ²

16(2r)^|F|2 − δ⁴

(16(2r)^|F|2)² +2µ(A)2^−|G| δ²

16(2r)^|F|2 ≤ δ² 8(2r)^|F|2. Therefore Chebyshev’s inequality forµ(Z_τ∩A)gives

κ ω :|µ(Z_τ(ω) ∩A) −E[µ(Z_τ∩A)]| ≥ δ

2 ≤ Var(µ(Zτ∩A))

δ 2

2

≤ 1

2(2r)^|F|2. Now since

E[µ(Zτ∩A)] −µ(A)2^−|G|

< ^δ₂ we have κ n

ω :

µ(Zτ(ω) ∩A) − µ(A)2^−|G| ≥ δo

≤ 1

2(2r)^|F|2. Since this is true for eachτ∈2^G we have

κ n ω:

µ(Zτ(ω) ∩A) − µ(A)2^−|G|

| ≥ δ for someτ:G→ 2o

≤ 1

2r^|F|2.

Finally, letting Arange over the setsRσforσ ∈r^G we get

κ n ω :

µ(Zτ(ω) ∩R_σ^c) − µ(R^c_σ)2^−|G|

≥ δfor someτ :G →2 andσ :G→ r o

≤ 1 2.

Then any member of the nonempty complement of nω:

µ(Zτ(ω) ∩R^c_σ) −µ(R_σ^c)2^−|G|

≥ δ for someτ:G→ 2 andσ :G→r o

works asT. This completes the proof of Theorem 2.1.4.

We note that the proof of Theorem 2.1.4 goes through for any groupΓsuch that an arbitrary free action can be approximated in the uniform topology by ergodic actions - for example any group of the formZ∗H. Such an approximation is impossible if Γ has property(T), and in this case the extreme points of FR∼_s(Γ,X, µ)are closed.

Therefore the following question is natural.

Question 2.6.1. LetΓbe a group without property(T). Can every free action of Γ be approximated in the uniform topology of A(Γ,X, µ)by ergodic actions?