E PROOF OF THEOREM 3

Proof of Lemma3.20. Letva ∈V_Ga and letλa(va)=σ=(σ0, . . . ,σ_k). Thenλ_Ga(va) is the marginalρ=(ρ0, . . . ,ρk) ofσonSb×Vb, soλ_Ga is a Borel map fromVaintoN (Sb×Vb).

σis mutually singular, so there are pairwise disjoint Borel setsUi ⊆Θ×Sb×Vbsuch that

σi(Ui)=1 for eachi ≤k. TheG-sectionsWi ={(sb,vb) : (G,sb,vb)∈Ui} are Borel and

pairwise disjoint. Sinceva∈V_Ga⊆C₁a(G),σ({G}×Sb×Vb)=~1. Thereforeρi(Wi)=1 for

eachi≤k, and henceρ∈L_(Sb_×Vb_).

Lemma E.1. LetΘ_{,X,Y be Polish spaces, where Y} ⊆X , and let G∈Θ_{. Then}

(i) For each open U in the topology of{G}×X , there exists an open W in the topology ofΘ×Y such that W =U∩({G}×Y); and

(ii) For each open W in the topology ofΘ×Y , there exists an open U in the topology of

{G}×X such that W =U∩({G}×Y).

Proof of LemmaE.1. Both(i)and(ii)follow immediately from the definition of subspace topology.

Proof of Theorem3.24. Proof of (i).Sinceva∈V_Ga=C₁a(G),λa(va)m({G}×Sb×Vb)=1. Therefore (λa_G(va))(E)=λa(va)(Θ×E)=(λa(va))({G}×E). So ifva assumesE inV_or

{G}×EinV_{G, then} (λ_Ga(va))(E)=( 1, . . . , 1 | {z } 1 or more , 0, . . . , 0 | {z } 0 or more )=(λa(va))({G}×E).

Therefore conditions (a) and (b) hold for assumingE inV_{if and only if they hold for} assuming {G}×E inV_{G. Furthermore, LemmaE.1implies that for every BorelF} ⊆Sb×

V_Gb,F =E∩U 6=_∅for some openU ⊆Sb×V_Gbif and only if {G}×F =W∩({G}×E)6=_∅ for some openW ⊆Θ×Sb×Vb. So condition (c) for assumingE inV_{is equivalent to} condition (c) for assuming {G}×E inV_G.

Proof of (ii). It is quite trivial thatsa maximizes LEU with respect toλa(va) if and only ifsa maximizes LEU with respect toλ_Ga(va). λa_G(va) has full support inSb×V_Gb if and only if va assumesSb×V_Gb inV_{G. By(i), this holds if and only ifλ}a_(va_{) assumes}

Kb(G)={G}×Sb×V_GbinV_{. This proves(ii).}

Proof of (iii). The base case is handled by(ii). Assume the induction hypothesis for

M>1:

∀m≤M, R_ma(G)={G}×R_ma(G,V_G)

By(i)and(ii), for allva∈proj_VaR₁a(G)=proj_Va GR

a

1(G,VG),vaassumes {G}×RMb (G,VG)

inV_{if and only ifv}a_assumesRb

M(G,VG) inVG. Therefore, R_Ma₊₁(G)=Ra_M(G)∩(Θ×Sa×Aa(Rb_M(G))) =£{G}×R_Ma (G,V_G)¤∩hΘ×Sa×Aa({G}×Rb_M(G,V_G))i ={G}× h R_Ma (G,V_G)∩(Sa×Aa(Rb_M(G,V_G)))i ={G}×Ra_M+1(G,V_G). F PROOF OF THEOREM3.25

To prove Theorem3.25, we will first show that it is a consequence of TheoremF.1below. The proof of TheoremF.1is much longer, and is given in the next section.

Fix the finite strategy sets (Sa,Sb). Recall from Section3.4that we identify a game

G with strategy sets (Sa,Sb) with the N-tuple of real numbers that represents the pair (πa,πb) of payoff functions, where N =2· |Sa×Sb|. Therefore, we let the space of all games on (Sa,Sb) beΘ=RN_{. We maintain this definition of}Θ_{throughout this paper.}

TheoremF.1says there is a “Borel family” of type structuresT_{G indexed byG}∈Θ_such

Given Polish spacesX,Y,Z and a Borel functionf :X×Y →Z, we let fy:X →Z be

the Borel function defined byfy(x)=f(x,y).

Theorem F.1. Let Ta,Tbbe uncountable Polish spaces. There exist Borel maps

κa:Ta×Θ→L_(Sb×Tb), κb:Tb×Θ→L_(Sa×Ta)

such that for every G ∈Θ_, T_G =­Sa,Sb,Ta,Tb,κ_Ga,κ_Gb®is a complete one-to-one lexico- graphic type structure such that

proj_SaR_∞a (G,T_G)×proj_SbR_∞b (G,T_G)=Sa_∞(G)×Sb_∞(G).

This result would follow immediately from Theorems3.2and3.4if we only required that the mapκa_G is Borel inta for each fixedG, and similarly forb. The extra difficulty lies in finding mapsκa andκb that are Borel in both variables. Intuitively, {T_G:G∈Θ_}

is a Borel family of type structures indexed byG∈Θ_{. Ann’s beliefs depend on both the}

gameGand a typeta, and Bob’s beliefs depend on both a gameGand a typetb.

To prepare for the proof of Theorem3.25, we first prove an easier intermediate re- sult, in which the requirement thatV_{is complete is omitted. Theorem3.25can then be} proved by carefully embedding thisV_{into a complete type structure so that, for each} gameG∈Θ_{, the set of states in which there is common knowledge ofG remains unal-}

tered.

Theorem F.2. There is a one-to-one lexicographic type structure with nature,

V=­Θ_,Sa_,Sb_,Va_,Vb_,λa_,λb®_{,such that for every game G}∈Θ_, (i) V_{admits common knowledge of G;}

(ii) For every game G∈Θ_,V_{Gis a complete one-to-one lexicographic type structure such}

that

proj_SaR_∞a (G,V_G)×proj_SbR_∞b(G,V_G)=S_∞a (G)×Sb_∞(G); and

(iii) Any pair of types(va,vb)that believes G has common belief of G.

Proof of TheoremF.2from TheoremF.1. LetSa,Sb,Ta,Tb,κa,κb be as they were in The- oremF.1. LetVa=Ta×Θ_andVb=Tb×Θ_{. We will define Borel mapsλ}a_,λb _{so that}

V= ­Θ_,Sa_,Sb_,Va_,Vb_,λa_,λb® _{has the required properties. The plan will be to make} Ta×{G} be the set of types that have common belief ofG.

Define the function

as follows. For each (σ,G)∈L_(Sb×Tb)×Θ_{, letα}b_{(σ,G) be the uniqueµ}∈L₍Θ×Sb×Vb) such that

⊲ _µ({G}×Sb×(Tb×{G}))=~1; and

⊲ _{For each Borel setE}⊆Sb×Tb,µ({G}×E×{G})=σ(E). Note thatE×{G}⊆Sb×Vb, so {G}×E×{G}⊆Θ×Sb×Vb.

Claim.αbis a continuous map.

Proof of Claim:Supposeσn→σinL(Sb×Tb), andGn→GinΘ, where→indicates

weak convergence. We must prove thatαb(σn,Gn)→αb(σ,G). It suffices to prove this

in the case that eachσn andσhave length one, because it would then follow that each

coordinate ofαb(σn,Gn) converges to the corresponding coordinate ofαb(σ,G).

Letβ:Θ→M₍Θ_{) be the mapH}7→δH, whereδH({H})=1. We haveβ(Gn)→β(G),

because for every continuous f :Θ→R_,R_{f dδGn}= f(Gn) converges to

R

f dδG=f(G).

We note that

αb(σn,Gn)=β(Gn)⊗σn⊗β(Gn), αb(σ,G)=β(G)⊗σ⊗β(G).

Therefore, we haveαb(σn,Gn)→αb(σ,G), which proves the claim.

Now, defineλa(ta,G)=αb(κa(ta,G),G). Sinceκa is Borel andαb is continuous,λa is a Borel map, and hence V_{is a type structure with nature. LetG}∈Θ_{. We see from}

the definition ofλa that a type va ∈Va believesG if and only if va =(ta,G) for some

ta ∈Ta. ThusC₁a(G)=Ta×{G}. Moreover,C₂a(G)=C₁a(G), and hence by induction,

C_ma(G)=C₁a(G). Therefore,V_{has the property thatC}a

1(G)=C∞a(G)=VGa, that is, every

vathat believesGhas common belief ofG. It follows thatV_{admits common knowledge} ofG. Finally, the mappingsta 7→(ta,G),tb 7→(tb,G) are topological homeomorphisms from Ta toV_Ga andTb toV_Gb that give an isomorphism from the type structureT_{G of} TheoremF.1toV_{G. Therefore,}V_{G has the same properties as}T_{G. In particular,}V_{G is} a complete one-to-one type structure such that proj_SaR_∞a(G,VG)×projSbRb_∞(G,VG)=

Sa_∞(G)×Sb_∞(G).

Proof of Theorem3.25from TheoremF.1. LetSa,Sb,Ta,Tb,κa,κb be as in TheoremF.1. LetVabe the topological union

Va=[0, 1)⊎([1,∞)×Θ₎⊎(Ta×Θ_),

where the three parts of the union are disjoint and clopen inVa. Note thatVa=[0, 1)⊎ (([1,∞)⊎Ta)×Θ_{). DefineV}b_{analogously. We will define Borel mappingsλ}a_,λb_{so that}

For eachG∈Θ_{, our plan will be to letT}a×{G} be the set of types having common belief ofG; let [m,m+1)×{G} be the set of types havingm-th order, but not (m+1)-th order, belief ofG; and let [0, 1) be the set of types not having belief ofG. We will use the Borel Isomorphism Theorem (PropositionC.5), as we did in the proof of Theorem3.13.

For eachm>0, letT_ma =[m,∞)⊎Ta. ThenT₁a ⊇T₂a ⊇ · · ·, andTa =T_m>0T_ma. Ac- cording to our plan,T_ma×{G} will be the set of types havingm-th order belief ofG. These types will be mapped to the beliefs in setJb_m(G), which we define inductively as follows.

J₁b(G)=

n

σ∈L₍Θ×Sb×Vb) :σ({G}×Sb×Vb)=~1o

J_mb₊₁(G)=nσ∈J_mb(G) :σ({G}×Sb×(T_mb ×{G}))=~1o

Intuitively,J_mb(G) is the set of LPS’s that havem-th order belief ofG. We also write

J₀b=L₍Θ×Sb×Vb) \ [

G∈Θ

J₁b(G).

Now, letαb:L_(Sb_×Tb_)×Θ→L₍Θ×Sb×Vb) be as in the proof of TheoremF.2. We will construct a one-to-one Borel functionλa:Va→L₍Θ×Sb×Vb) such that

(I) λa([0, 1))=J₀b;

(II) For eachm>1 andG∈Θ_,λa_(Tma ×{G})=J_mb(G); and (III) For eachG∈Θ_andta∈Ta,λa(ta,G)=αb(κa(ta,G),G).

Note that(I)and(II)imply that the mapλais onto. Sinceλa will be one-to-one,(I) and(II)will also imply that for eachG∈Θ_,λa_(Ta×{G})=T_m>0J_mb(G).

It is clear that the setJ₀bhas the cardinality of the continuum. We show thatJ₀bis also Borel. To see this, let

γb:L_(Sb×Vb)×Θ→L₍Θ×Sb×Vb)

be the function such thatγb(σ,G) is the uniqueµ∈L₍Θ×Sb×Vb) such thatµ({G}×Sb×

Vb)=~1 and for each Borel setE⊆Sb×Vb,µ({G}×E)=σ(E). Note that

∀G∈Θ_{, γ}b₍L_(Sb×Vb)×{G})=J₁b(G)

Therefore, the range of γb isSG∈ΘJ₁b(G) and the complement of the range ofγb is J₀b.

Arguing as in the proof of the Claim in Theorem F.2, we see that γb is a continuous map. It is also clear thatγb is one-to-one. By Corollary 15.2 inKechris(1995), images of Borels sets under such maps are Borel sets themselves. Therefore, for every Borel

F ⊆L_(Sb×Vb)×Θ_,γb_{(F) is Borel. In particular, the range ofγ}b _{is Borel, and therefore}

By the Borel Isomorphism Theorem, there is a one-to-one and onto Borel mapλa₀ : [0, 1)→J₀b. This will take care of(I)since we will eventually letλa coincide withλa₀ on [0, 1).

For eachG∈Θ _{and eachm} >0, the differenceJ_mb(G) \J_mb₊₁(G) clearly has the car- dinality of the continuum. Moreover, sinceJ_mb(G) is the image underγb of a Borel set,

J_mb(G) is Borel. Hence the difference setsJ_mb(G) \Jb_m+1(G) are Borel as well. By the Borel Isomorphism Theorem, there is a one-to-one Borel function from [m,m+1)×{G} onto this difference. However, since there are uncountably manyG’s, we cannot in general glue these functions together into a single Borel function.

To get around this problem, we introduce mappings that translate the games and keep everything else unchanged. LetG0be the particular game whose payoff functions are everywhere zero. Given two games G,H ∈Θ_{, let G}+H be the game obtained by adding the payoff functions ofG andH pointwise at each strategy profile. Note that

G0+H=H for eachH∈Θ. For eachH ∈Θ, the mapG7→G+H is a homeomorphism fromΘ_{to itself that sendsG0toH.}

ForH∈Θ_{, letψ}a

H:Va→Vabe the map defined by

ψa_H(r)=    (ta,G+H) ifr=(ta,G)∈T₁a×Θ_; r ifr∈[0, 1).

Thenψa_His a homeomorphism fromVatoVaand we have

∀m>0, ψa_H(T_ma×{G0})=Tma ×{H} .

Moreover, (va,H)7→ψa_H(va) is a continuous map fromVa×Θ_ontoVa_.

Letφb_Hbe a function fromL₍Θ×Sb×Vb) to itself such that for eachσ∈L₍Θ×Sb×Vb) and Borel setE⊆Θ×Sb×Vb,

(φb_H(σ))³n(G+H,sb,ψb_H(vb)) : (sb,vb,G)∈Eo´=σ(E).

Thenφb_H is a homeomorphism fromL₍Θ×Sb×Vb) onto itself such that for eachm>1, φb_H(J_mb(G0))=Jmb(H).

By the Borel Isomorphism Theorem, for eachm>0, there is a one-to-one Borel func- tionρm from [m,m+1)×{G0} ontoJmb(G0) \J_mb+1(G0). Let

λa_m: [m,m+1)×Θ→L₍Θ×Sb×Vb)

Borel, and for eachH∈Θ_,

λa_m([m,m+1)×{H})=J_mb(H) \J_mb₊₁(H).

Letλ_∞a :Ta×Θ→L₍Θ×Sb×Vb) be the mapping given by λa_∞(ta,G)=αb(κa(ta,G),G).

It is clear thatλa_∞ is one-to-one. Sinceαa is continuous andκa is Borel, λa_∞is Borel. Also, for eachH∈Θ_,λ∞a _(Ta×{H})=T_mJ_mb(H).

It follows that the unionλa=λa₀∪¡S_mλa_m¢∪λa_∞is a one-to-one Borel mapping from

VaontoL₍Θ×Sb×Vb) that satisfies(I)–(III). ThereforeV_{is a complete one-to-one type} structure with nature.

It follows from(II)that for eachH∈Θ_,Ca

1(H)=T

a

1 ×{H}. We then see by induction that for eachm>0 and H ∈Θ_,Cma_(H)=T_ma ×{H}. Therefore, the set ofva∈Va with common belief ofHis

V_Ha=C_∞a(H)=\

m

T_ma×{H}=Ta×{H} .

SoV_{admits common knowledge of every gameH}∈Θ_{. As in the proof of TheoremF.2,}

for eachG∈Θ_{, the type structure}V_{G is isomorphic to}T_{G. Therefore}V_{G is a complete} one-to-one type structure such that

proj_SaR_∞a (G,VG)×proj_SbR_∞b (G,VG)=Sa∞(G)×S

b

∞(G). By applying Theorem3.24, we get

proj_SaR_∞a (G)×proj_SbR_∞b(G)=S_∞a (G)×Sb_∞(G).