Equilibria in infinite games of incomplete information

The notion of communication equilibrium extends Aumann’s (J Math Econ 1:67–96, 1974, https://doi.org/10.1016/0304-4068(74)90037-8) correlated equilibrium concept for complete information games to the case of incomplete information. This paper shows that this solution concept has the following property: for the class of incomplete information games with compact metric type and action spaces, and with payoff functions jointly measurable and continuous in actions, limits of Bayes-Nash equilibria of finite approximations to an infinite game are communication equilibria (and, in general, not Bayes-Nash equilibria) of the limit game. Stinchcombe’s (J Econ Theory 146:638–655, 2011b, https://doi.org/10.1016/j.jet.2010.12.006) extension of Aumann’s (J Math Econ 1:67–96, 1974, https://doi.org/10.1016/0304-4068(74)90037-8) solution concept to the case of incomplete information fails to satisfy this condition.

1. Alternative notions of an approximation to an infinite game via a sequence of finite games have been considered by Stinchcombe (2005) and Stinchcombe (2005, 2011a), who shows that discretizing action and type spaces, rather than spaces of behavioral strategies, is a rather delicate matter.

2. Stinchcombe (2011b) and Cotter (1991) establish existence of correlated equilibrium within the class \(\mathfrak {G}\). Other authors (see e.g., Milgrom and Weber 1985; Balder 1988; Carbonell-Nicolau and McLean 2018, 2019; He and Yannelis 2016) have proven existence of Bayes-Nash equilibria (and hence communication equilibria) under the additional assumption of diffuse joint information of the players. See Simon (2003) for a proof of the fact that equilibria need not exist if one drops the diffuseness assumption. In related frameworks, such as the state-space framework of Yannelis and Rustichini (1991), Hellman and Levy (2017), and Carbonell-Nicolau and McLean (2020), and the lattice framework of Athey (2001), McAdams (2003) and Reny (2011a), existence results can be proven in which the requirement of diffuse information is replaced by assumptions we do not make here.

3. There are alternative ways of defining the notion of correlated equilibrium (see, e.g., Bergemann and Morris 2016), which are not considered here.

4. See, e.g., Lucchetti and Patrone (1986), Stinchcombe (2005), and Gürkan and Pang (2007).

5. See also the recent extensions in Prokopovych and Yannelis (2019) and He and Sun (2019).

6. Thanks to an anonymous referee for pointing out the limitations of standard approximation results in the context of discontinuous games.

7. While the implication is mathematically correct, the first statement is not, for the sets \(\mathscr {M}/\sim \) and \(\mathfrak M\) differ from one another. Note that, for any \(\mu \in \mathscr {M}\), the equivalence class \([\mu ]\), viewed as a member of \(\mathscr {M}/\sim \), is contained in the corresponding equivalence class from \(\mathfrak M\), but the reverse containment does not hold.

8. Again, the first assertion is an abuse of terminology.

9. See Castaing et al. (2004, ch. 2) for alternative formulations of these topologies.

10. The argument here is based on an example provided by an anonymous referee.

11. Finding conditions under which the topology is compact would be useful to establish the general existence of communication equilibria within the class of games \({\mathfrak {G}}\). As per Exercise 2.48 in Megginson (1998), the topology is compact if and only if (i) \(\vartheta _{{\hat{p}}}(\mathscr {M}/\sim )\) is compact in \(\Delta (T\times X)\) for every \({\hat{p}}\in \varvec{P}\); and (ii) the image of \(\mathscr {M}/\sim \) in \(\prod _{\varvec{P}}\Delta (T\times X)\) under the map \([\mu ]\in \mathscr {M}/\sim \mapsto ({\hat{p}}\otimes \mu )_{{\hat{p}}\in \varvec{P}}\) is closed. While the first condition can be shown to hold, we have not been able to establish the second condition, which requires the following: if \(([\mu ^\alpha ])\) is a net in \(\mathscr {M}/\sim \) such that

    $$\begin{aligned} {\hat{p}}\otimes \mu ^\alpha \xrightarrow [w]{}{\hat{p}}\otimes \mu _{{\hat{p}}}\quad \text {for all }{\hat{p}}\in \varvec{P}, \end{aligned}$$

    where, for each \({\hat{p}}\in \varvec{P}\), \(\mu _{{\hat{p}}}\) is an element of \(\mathscr {M}\), then there exists \(\mu ^{*}\in \mathscr {M}\) such that

    $$\begin{aligned} {\hat{p}}\otimes \mu _{{\hat{p}}}={\hat{p}}\otimes \mu ^{*},\quad \text {for all }{\hat{p}}\in \varvec{P}. \end{aligned}$$

12. It is clear that these functions can be viewed as behavioral strategies that assign a Dirac probability measure to each type.

13. An equivalent topology is the product narrow topology on \(\mathscr {T}\) (see Balder 2001, Definition 1.3), or, more precisely, the product topology induced by the narrow quotient topology on the equivalence classes from each factor \(\mathscr {T}_{i}\) of transition probabilities that only differ on a \(p_{i}\)-null set. This product topology is equivalent to the product weak topology on \(\times _{i}\mathscr {D}_{i}\), where \(\mathscr {D}_{i}:=\{p_{i}\otimes \mu _{i}:\mu _{i}\in \mathscr {T}_{i}\}\) (see Carbonell-Nicolau and McLean 2018, Sect. 5.2).

14. For \(t=(t_1,t_2)\in T\) with \(t_1=t_2\), the sequence of measures

    $$\begin{aligned} \begin{aligned} \delta _{s^1_1(t_1)}\otimes \delta _{s^1_2(t_2)},&\frac{1}{2}(\delta _{s^1_1(t_1)}\otimes \delta _{s^1_2(t_2)})+\frac{1}{2} (\delta _{s^2_1(t_1)}\otimes \delta _{s^2_2(t_2)}),\\&\frac{1}{3}(\delta _{s^1_1(t_1)}\otimes \delta _{s^1_2(t_2)})+\frac{1}{3} (\delta _{s^2_1(t_1)}\otimes \delta _{s^2_2(t_2)}) +\frac{1}{3}(\delta _{s^3_1(t_1)}\otimes \delta _{s^3_2(t_2)}),\ldots \end{aligned} \end{aligned}$$

    (16)

    converges weakly to the measure \(\mu (t)\in \Delta (X)\) defined by \(\mu (\{1,1\}|t)=\mu (\{2,2\}|t)=\frac{1}{2}\). In general, for every \(t=(t_1,t_2)\in T\), the sequence \((\mu ^n(t))\) given in (16) converges weakly to some measure \(\mu (t)\in \Delta (X)\). Applying Theorem 2.6 in Balder (2001), it follows that \({\hat{p}}\otimes \mu ^n\xrightarrow [w]{}{\hat{p}}\otimes \mu \) for all \({\hat{p}}\in \Delta (T)\), and so, in particular, .

15. Recall that \(\delta _{g_{i}(t_{i})}\) denotes the Dirac measure in \(\Delta (T_{i})\) with support \(\{g_{i}(t_{i})\}\).

16. Recall that \(\delta _{g_{i}(t_{i})}\) denotes the Dirac measure in \(\Delta (T_{i})\) with support \(\{g_{i}(t_{i})\}\).

17. The section of a closed (resp. open) subset of a product space is closed (resp. open) (see, e.g., Bourbaki (1989, p. 46, Corollary)).

18. The last inequality in (48) follows from the fact that \(N_{\gamma ''}((B)_{x_{-i}})\subseteq (N_{\gamma ''}(B))_{x_{-i}}\). To see that this containment holds, suppose that \(x_{i}\in N_{\gamma ''}((B)_{x_{-i}})\). Then \(x_{i}\in N_{\gamma ''}(y_{i})\) for some \(y_{i}\in (B)_{x_{-i}}\), implying that \((y_{i},x_{-i})\in B\). Since \(x_{i}\in N_{\gamma ''}(y_{i})\), it follows that \((x_{i},x_{-i})\in N_{\gamma ''}(B)\) and so \(x_{i}\in (N_{\gamma ''}(B))_{x_{-i}}\).

Authors and Affiliations

1. Department of Economics, Rutgers University, New Brunswick, USA

   Oriol Carbonell-Nicolau

Corresponding author

Correspondence to Oriol Carbonell-Nicolau.

Publisher's Note

Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

I would like to thank Philip Reny for very useful discussions, and Maxwell Stinchcombe, Richard McLean, and two anonymous referees for their valuable comments.

Appendix

To begin, we establish an unproven claim made in Sect. 2.3.1.

Let \(\Gamma =\left( T_{i},X_{i},u_{i},p\right) _{i=1}^{N}\) be a Bayesian game, and suppose that the net \((\mu ^\alpha )\) converges uniformly to \(\mu \) in \(\mathscr {M}\). Fix \({\hat{p}}\in \varvec{P}\) and a bounded continuous map \(f:T\times X\rightarrow \mathbb {R}\). We claim that the net of maps

converges uniformly to the map \(t\in T\mapsto \int _Xf(t,x)\mu (dx|t)\).

Prior to proving this claim, we state and prove the following lemma.

Lemma 5

Suppose that Y, Z, and E are metric spaces with Y and Z compact. If \((h^\alpha )\) is a uniformly convergent net of maps \(h^\alpha :Y\rightarrow \ Z\) with limit point \(h:Y\rightarrow Z\), and if \(g:Y\times Z\rightarrow E\) is a continuous map, then the net \((y\in Y\mapsto g(y,h^\alpha (y)))\) converges uniformly to the map \(y\in Y\mapsto g(y,h(y))\).

Proof

Fix \(\epsilon >0\). We must show that there exists \(\alpha ^{*}\) such that, for all \(\alpha \ge \alpha ^{*}\),

Because g is continuous and \(Y\times Z\) is compact, g is uniformly continuous. Consequently, there exists \(\delta >0\) such that, for (y, z) and \((y',z')\) in \(Y\times Z\), \(d_Y(y,y')<\delta \) and \(d_Z(z,z')<\delta \) imply that \(d_E(g(y,z),g(y',z'))<\epsilon \).

From the uniform convergence of the net \((h^\alpha )\) to h, one can choose \(\alpha ^{*}\) such that, for \(\alpha \ge \alpha ^{*}\), \(d_Z(h^\alpha (y),h(y))<\delta \) for all \(y\in Y\). Consequently, for \(\alpha \ge \alpha ^{*}\), one obtains (37). \(\square \)

To see that the net of maps in (36) converges uniformly to the map \(t\in T\mapsto \int _Xf(t,x)\mu (dx|t)\), we apply Lemma 5 with \(Y=T\), \(Z=\Delta (X)\), \(E=\mathbb {R}\), \(h^\alpha =\mu ^\alpha \), \(h=\mu \), and g defined by \(g(t,\mu ):=\int _Xf(t,x)\mu (dx)\). The map g is continuous. Indeed, suppose that \((t^n,\mu ^n)\) is a convergent sequence in \(T\times \Delta (X)\) with limit point \((t,\mu )\in T\times \Delta (X)\). For each n, let \(\delta _{t^n}\) denote the Dirac measure in \(\Delta (T)\) with support \(\{t^n\}\). Because the sequence \((t^n,\mu ^n)\) converges to \((t,\mu )\), the sequence \((\delta _{t^n},\mu ^n)\) converges to \((\delta _t,\mu )\) in \(\Delta (T)\times \Delta (X)\) (see, e.g., Aliprantis and Border 2006, Theorem 15.8). Consequently, by Theorem 2.8(ii) in Billingsley (1999), \(\delta _{t^n}\otimes \mu ^n\xrightarrow [w]{}\delta _t\otimes \mu \), and so the Portmanteau Theorem (see, e.g., Aliprantis and Border 2006, Theorem 15.3) yields

Since \((\mu ^\alpha )\) converges uniformly to \(\mu \), and since g is continuous, Lemma 5 implies that the net of maps in (36) converges uniformly to the map \(t\in T\mapsto \int _Xf(t,x)\mu (dx|t)\), as we sought.

The remainder of this appendix contains the proofs of the lemmas stated in Sect. 6. For the convenience of the reader, each proof is preceded by a restatement of its corresponding lemma.

1.1 Proof of Lemma 4

Lemma 3

Suppose that \(\Gamma =\left( T_{i},X_{i},u_{i},p\right) _{i=1}^{N}\) is a Bayesian game in \({\mathfrak {G}}\). Suppose that \((\mu ,\alpha _{i},\eta _{i})\in \mathscr {M}\times \mathscr {A}_{i}\times \mathscr {D}_{i}\) and

Then there exist \(\alpha ^{*}_{i}\in \mathscr {A}_{i}\) and \(\eta ^{*}_{i}\in \mathscr {D}_{i}\) such that

and the following conditions are satisfied: \(\eta ^{*}_{i}\) is a simple function and there exists a \((\mathscr {B}(T_{i}),\mathscr {B}(T_{i}))\)-measurable map \(g_{i}:T_{i}\rightarrow T_{i}\) such that \(\eta ^{*}_{i}(t_{i})=\delta _{g_{i}(t_{i})}\) for each \(t_{i}\in T_{i}\);^{Footnote 16} the function \(t_{i}\in T_{i}\mapsto \alpha ^{*}_{i}(t_{i},\cdot )\in \Delta (X_{i})^{X_{i}}\) is simple; and, for each \(t_{i}\in T_{i}\), the map \(x_{i}\in X_{i}\mapsto \alpha ^{*}_{i}(t_{i},x_{i})\in \Delta (X_{i})\) is continuous.

Proof

The proof is organized in four steps.

Step 1

There is no loss of generality in assuming that \(\alpha _{i}\) is continuous.

Proof of Step 1

Define \(p'\in \Delta (T)\) by

for all \(A_{i}\times A_{-i}\subseteq T_{i}\times T_{-i}\) in \(\mathscr {B}(T)\), and define \(\rho \in \Delta (T\times X)\) by

for all \(A\times B\subseteq T\times X\) in \(\mathscr {B}(T\times X)\). By Luzin’s Theorem, there is a sequence \((K^n)\) of compact subsets of \(T_{i}\times X_{i}\) such that \(\rho (K^n\times T_{-i}\times X_{-i})\rightarrow 1\) and \(\alpha _{i}|_{K^n}\) is continuous for each n. Applying Theorem 4.1 of Dugundji (1951), each \(\alpha _{i}|_{K^n}\) can be extended to a continuous map \(\widehat{\alpha }^n_{i}:T_{i}\times X_{i}\rightarrow \Delta (X_{i})\). Define \(\vartheta ^n:T\times X\rightarrow \mathbb {R}\) and \(\vartheta :T\times X\rightarrow \mathbb {R}\) by

and

Because \(\vartheta ^n=\vartheta \) on \(K^n\times T_{-i}\times X_{-i}\), for each n, and since \(\rho (K^n\times T_{-i}\times X_{-i})\rightarrow 1\), it follows that

and in light of (18) we conclude that

for some continuous \(\alpha '_{i}\in \mathscr {A}_{i}\). \(\square \)

Step 2

There exists \(\alpha ^{*}_{i}\in \mathscr {A}_{i}\) such that

and the following conditions are satisfied: the function \(t_{i}\in T_{i}\mapsto \alpha ^{*}_{i}(t_{i},\cdot )\in \Delta (X_{i})^{X_{i}}\) is simple, and, for each \(t_{i}\in T_{i}\), the map \(x_{i}\in X_{i}\mapsto \alpha ^{*}_{i}(t_{i},x_{i})\in \Delta (X_{i})\) is continuous.

Proof of Step 2

Let \(\mathscr {C}(X_{i},\Delta (X_{i}))\) represent the set of all the continuous functions from \(X_{i}\) into \(\Delta (X_{i})\), and endow the space \(\mathscr {C}(X_{i},\Delta (X_{i}))\) with the supremum metric. Then \(\mathscr {C}(X_{i},\Delta (X_{i}))\) is a separable metric space (see, e.g., Aliprantis and Border 2006, Lemma 3.99). Define \(\widetilde{\varvec{\alpha }}_{i}:T_{i}\rightarrow \mathscr {C}(X_{i},\Delta (X_{i}))\) by

(recall that \(\alpha _{i}\) can be taken continuous by Step 1). Because \(\alpha _{i}\) is continuous, Theorem 4.55 in Aliprantis and Border (2006) implies that the map \(\widetilde{\varvec{\alpha }}_{i}:T_{i}\rightarrow \mathscr {C}(X_{i},\Delta (X_{i}))\) is \((\mathscr {B}(T_{i}),\mathscr {B}(\mathscr {C}(X_{i},\Delta (X_{i}))))\)-measurable. Consequently, applying Theorem 4.38 in Aliprantis and Border (2006), it follows that \(\widetilde{\varvec{\alpha }}_{i}\) is the pointwise limit of a sequence \((\widetilde{\varvec{\alpha }}^n_{i})\) of \((\mathscr {B}(T_{i}),\mathscr {B}(\mathscr {C}(X_{i},\Delta (X_{i}))))\)-measurable simple functions. Now, for each n, define \(\alpha ^n_{i}:T_{i}\times X_{i}\rightarrow \Delta (X_{i})\) by

Note that it suffices to show that there exists n for which \(\alpha ^{*}_{i}:=\alpha ^n_{i}\) satisfies (38).

Applying Theorem 4.55 in Aliprantis and Border (2006), we see that \((\alpha ^n_{i})\) is a sequence of \((\mathscr {B}(T_{i}\times X_{i}),\mathscr {B}(\Delta (X_{i})))\)-measurable functions. We claim that \((\alpha ^n_{i})\) converges to \(\alpha _{i}\) pointwise. To see this, fix \((t_{i},x_{i})\in T_{i}\times X_{i}\). It must be shown that \(\alpha ^n_{i}(t_{i},x_{i})\xrightarrow [w]{}\alpha _{i}(t_{i},x_{i})\). We know that the sequence \((\widetilde{\varvec{\alpha }}^n_{i})\) converges to \(\widetilde{\varvec{\alpha }}_{i}\) pointwise. Consequently, the sequence \((\widetilde{\varvec{\alpha }}^n_{i}(t_{i}))\) of maps in \(\mathscr {C}(X_{i},\Delta (X_{i}))\) converges uniformly to \(\widetilde{\varvec{\alpha }}_{i}(t_{i})\in \mathscr {C}(X_{i},\Delta (X_{i}))\), i.e., for each \(\epsilon >0\), there exists M such that, for all \(n\ge M\) and \(x_{i}\in X_{i}\), we have

implying that \(\alpha ^n_{i}(t_{i},x_{i})\xrightarrow [w]{}\alpha _{i}(t_{i},x_{i})\).

Next, define \(\theta :T\times X\rightarrow \Delta (X)\) and \(\theta ^n:T\times X\rightarrow \Delta (X)\) by

for all \(B_{i}\times B_{-i}\subseteq X_{i}\times X_{-i}\) in \(\mathscr {B}(X)\), where \(\delta _{x_j}\) denotes the Dirac measure in \(\Delta (X_j)\) with support \(\{x_j\}\). Define \(\eta :T\rightarrow \Delta (T)\) by

for all \(B_{i}\times B_{-i}\subseteq T_{i}\times T_{-i}\) in \(\mathscr {B}(T)\), and \(\mu ^{*}:T\rightarrow \Delta (X)\) by

Because the sequence \((\alpha ^n_{i})\) converges to \(\alpha _{i}\) pointwise, it follows from Theorem 2.8(ii) in Billingsley (1999) that the sequence \((\theta ^n)\) converges to \(\theta \) pointwise. Consequently, applying Theorem 2.6 in Balder (2001), it follows that

where \(p\otimes \mu ^{*}\otimes \theta \in \Delta (T\times X\times X)\) is defined by

for all \(A\times B\times B'\subseteq T\times X\times X\) in \(\mathscr {B}(T\times X\times X)\), and each \(p\otimes \mu ^{*}\otimes \theta ^n\) is defined similarly.

Let \(\nu \in \Delta (T\times X)\) (resp. \(\nu ^n\in \Delta (T\times X)\)) be defined by \(\nu (A\times B):=[p\otimes \mu ^{*}\otimes \theta ](A\times X\times B)\) (resp. \(\nu ^n(A\times B):=[p\otimes \mu ^{*}\otimes \theta ^n](A\times X\times B)\)). By Theorem 2.8(i) in Billingsley (1999), \(\nu ^n\xrightarrow [w]{}\nu \). Therefore, since \(\nu ^n\) and \(\nu \) are members of \(\mathscr {P}^{p'}(T\times X)\), Lemma 2 gives

This, together with (18), gives (38). \(\square \)

Next, observe that

where \(p_{i}\) represents the marginal projection of p into \(\Delta (T_{i})\). Define \(\zeta :T_{i}\times T_{i}\rightarrow \mathbb {R}\) by

Step 3

There exists a \((\mathscr {B}(T_{i}),\mathscr {B}(T_{i}))\)-measurable map \(g:T_{i}\rightarrow T_{i}\) such that

Proof of Step 3

Because the map \(\zeta \) is \((\mathscr {B}(T_{i}\times T_{i}),\mathscr {B}(\mathbb {R}))\)-measurable, Theorem 2 in Brown and Purves (1973) gives, for each n, a \((\mathscr {B}(T_{i}),\mathscr {B}(T_{i}))\)-measurable map \(g^n:T_{i}\rightarrow T_{i}\) such that for every \(t_{i}\in T_{i}\),

This, together with the fact that

gives

Consequently,

Because

(Step 2), it follows that there exists a large enough n for which

\(\square \)

Step 4

There exists a simple map \(\eta ^{*}_{i}\in \mathscr {D}_{i}\) such that

Proof of Step 4

Define \(\lambda \in \Delta (T_{i}\times T_{i})\) by

for all \(A\times B\subseteq T_{i}\times T_{i}\) in \(\mathscr {B}(T_{i}\times T_{i})\), where g is the map from Step 3. Because the map \(\zeta \) defined in (39) is \((\mathscr {B}(T_{i}\times T_{i}),\mathscr {B}(\mathbb {R}))\)-measurable, Luzin’s Theorem gives a sequence \((S^n)\) of compact subsets of \(T_{i}\times T_{i}\) such that \(\lambda (S^n)\rightarrow 1\) and each \(\zeta ^n:=\zeta |_{S^n}\) is continuous. Since \(S^n\) is compact, \(\zeta ^n:S^n\rightarrow \mathbb {R}\) is uniformly continuous, and so there exists \(\delta _n>0\) such that \(d_{i}(t_{i},{\hat{t}}_{i})<\delta _n\) and \(d_{i}(\tau _{i},\hat{\tau }_{i})<\delta _n\) implies that \(|\zeta ^n(t_{i},\tau _{i})-\zeta ^n({\hat{t}}_{i},\hat{\tau }_{i})|<\frac{1}{n}\), where \(d_{i}\) is a compatible metric for \(T_{i}\). In addition, since \(S^n\) is compact, there exists a finite \(\frac{\delta _n}{2}\)-partition \(\{P^{(n,1)},\ldots ,P^{(n,k_n)}\}\) of \(S^n\) (i.e., a partition such that each \(P^{(n,k)}\) has radius less than \(\frac{\delta _n}{2}\)) consisting of sets in \(\mathscr {B}(T_{i}\times T_{i})\).

For each n and \(k\in \{1,\ldots ,k_n\}\), let

(41)

Step 4.1

The partition \(\{P^{(n,1)},\ldots ,P^{(n,k_n)}\}\) of \(S^n\) can be chosen to satisfy the following: if \((t^k_{i},\tau ^k_{i})\in P^{(n,k)}\) and \((t^\kappa _{i},\tau ^\kappa _{i})\in P^{(n,\kappa )}\) for \(\kappa \ne k\), and if \(\tilde{t}^k_{i}\in P^{(n,k)}_1\) and \(\tilde{t}^\kappa _{i}\in P^{(n,\kappa )}_1\), then \((\tilde{t}^k_{i},\tau ^k_{i})\ne (\tilde{t}^\kappa _{i},\tau ^\kappa _{i})\).

Proof of Step 4.1

Note that there exists a finite set \(\{(t^1_{i},\tau ^1_{i}),\ldots ,(t^{k_n}_{i},\tau ^{k_n}_{i})\}\subseteq S^n\) such that

(Here the \(\delta _n/2\)-neighborhoods are neighborhoods in \(T_{i}\).) Now define \(A^{(n,1)},\ldots ,A^{(n,k_n)}\) as follows:

• \(A^{(n,1)}:=N_{\delta _n/2}(t^1_{i})\times N_{\delta _n/2}(\tau ^1_{i})\);

• \(A^{(n,2)}:=\left(N_{\delta _n/2}(t^2_{i})\times N_{\delta _n/2}(\tau ^2_{i})\right) \setminus \left(N_{\delta _n/2}(t^1_{i})\times N_{\delta _n/2}(\tau ^1_{i})\right)\);

• \(A^{(n,3)}:=\left(N_{\delta _n/2}(t^3_{i})\times N_{\delta _n/2}(\tau ^3_{i})\right) \setminus \left[\left(N_{\delta _n/2}(t^1_{i})\times N_{\delta _n/2}(\tau ^1_{i})\right) \cup \left(N_{\delta _n/2}(t^2_{i})\times N_{\delta _n/2}(\tau ^2_{i})\right)\right]\); and so on.

Letting \(P^{(n,k)}:=A^{(n,k)}\cap S^n\), one obtains a \(\delta _n/2\)-partition \(\{P^{(n,1)},\ldots ,P^{(n,k_n)}\}\) of \(S^n\). To see that this partition has the desired property, fix \(({\hat{t}}^k_{i},\hat{\tau }^k_{i})\in P^{(n,k)}\) and \(({\hat{t}}^\kappa _{i},\hat{\tau }^\kappa _{i})\in P^{(n,\kappa )}\) for \(\kappa >k\), and choose \(\tilde{t}^k_{i}\in P^{(n,k)}_1\) and \(\tilde{t}^\kappa _{i}\in P^{(n,\kappa )}_1\). Then the construction of the partition entails that if \(\tilde{t}^k_{i}=\tilde{t}^\kappa _{i}\) then \(\hat{\tau }^k_{i}\ne \hat{\tau }^\kappa _{i}\). \(\square \)

For each n and \(k\in \{1,\ldots ,k_n\}\), let

(42)

Step 4.2

For each n and \(k\in \{1,\ldots ,k_n\}\), the set \({\hat{P}}^{(n,k)}\) belongs to \(\mathscr {B}(T_{i}\times T_{i})\).

Proof of Step 4.2

Because the map g from Step 3 is \((\mathscr {B}(T_{i}),\mathscr {B}(T_{i}))\)-measurable, the graph of g,

belongs to \(\mathscr {B}(T_{i}\times T_{i})\) (see, e.g., Aliprantis and Border 2006, Theorem 12.28). Consequently, since \(P^{(n,k)}\in \mathscr {B}(T_{i}\times T_{i})\) and

it follows that \({\hat{P}}^{(n,k)}\in \mathscr {B}(T_{i}\times T_{i})\). \(\square \)

Step 4.3

For each n and \(k\in \{1,\ldots ,k_n\}\), the set \({\hat{P}}^{(n,k)}_1\) belongs to \(\mathscr {B}(T_{i})\).

Proof of Step 4.3

The assertion follows from Theorem 18.10 in Kechris (1995), together with the facts that \({\hat{P}}^{(n,k)}\in \mathscr {B}(T_{i}\times T_{i})\) (Step 4.2) and that, for each \(t_{i}\in T_{i}\), the set \(\{\tau _{i}:(t_{i},\tau _{i})\in {\hat{P}}^{(n,k)}\}\) is finite (in fact, a singleton). \(\square \)

For each n and \(k\in \{1,\ldots ,k_n\}\), choose \(t^{(n,k)}_{i}\in {\hat{P}}^{(n,k)}_1\) such that

and \(t^{*}_{i}\in T_{i}\), and define \(f^n:T_{i}\rightarrow T_{i}\) by

Step 4.4

The map \(f^n:T_{i}\rightarrow T_{i}\) defined in (43) is \((\mathscr {B}(T_{i}),\mathscr {B}(T_{i}))\)-measurable.

Proof of Step 4.4

The assertion follows from the following facts: \(f^n\) has finite range and (by Step 4.3) the inverse images of the members of the range belong to \(\mathscr {B}(T_{i})\). \(\square \)

Step 4.5

For every \(t_{i}\in {\hat{P}}^{(n,k)}_1\),

Proof of Step 4.5

First, recall that \(d_{i}(t_{i},{\hat{t}}_{i})<\delta _n\) and \(d_{i}(\tau _{i},\hat{\tau }_{i})<\delta _n\) implies that

Now, given \(t_{i}\in {\hat{P}}^{(n,k)}_1\), one has

\(\square \)

Next, define \(\lambda ^n\in \Delta (T_{i}\times T_{i})\) by

for all \(A\times B\subseteq T_{i}\times T_{i}\) in \(\mathscr {B}(T_{i}\times T_{i})\), where \(\delta _{f^n(t_{i})}\in \Delta (T_{i})\) denotes the Dirac measure on \(T_{i}\) with support \(\{f^n(t_{i})\}\).

Now recall the definition of \({\hat{P}}^{(k,n)}\) in (42) and define

where \(\text {Gr}(f^n)\) denotes the graph of \(f^n\) in \(T_{i}\times T_{i}\).

Step 4.6

For each n and \(k\in \{1,\ldots ,k_n\}\), the set \(\tilde{P}^{(n,k)}\) belongs to \(\mathscr {B}(T_{i}\times T_{i})\).

Proof of Step 4.6

Analogous to the proof of Step 4.2. \(\square \)

Step 4.7

For each n and \(k\in \{1,\ldots ,k_n\}\), the set \(\tilde{P}^{(n,k)}_1\) belongs to \(\mathscr {B}(T_{i})\).

Proof of Step 4.7

Analogous to the proof of Step 4.3. \(\square \)

Step 4.8

For each n and \(k\ne \kappa \), \(\tilde{P}^{(n,k)}\cap \tilde{P}^{(n,\kappa )}=\emptyset \).

Proof of Step 4.8

Choose \((t_{i},\tau _{i})\in \tilde{P}^{(n,k)}\) and \(({\hat{t}}_{i},\hat{\tau }_{i})\in \tilde{P}^{(n,\kappa )}\). It suffices to show that \((t_{i},\tau _{i})\ne ({\hat{t}}_{i},\hat{\tau }_{i})\).

Because \((t_{i},\tau _{i})\in \tilde{P}^{(n,k)}\), one has \((t_{i},\tau _{i})=(t_{i},f^n(t_{i}))\) and \(t_{i}\in {\hat{P}}^{(n,k)}_1\), implying (by (43)) that \((t_{i},\tau _{i})=(t_{i},g(t^{(n,k)}_{i}))\). In addition, since \(t_{i}\in {\hat{P}}^{(n,k)}_1\), one has \(t_{i}\in P^{(n,k)}_1\) (recall the definition of \(P^{(n,k)}_1\) in (41)). Summarizing, one has \(t_{i}\in P^{(n,k)}_1\) and \((t_{i},\tau _{i})=(t_{i},g(t^{(n,k)}_{i}))\).

Similarly, one can show that \(({\hat{t}}_{i},\hat{\tau }_{i})=({\hat{t}}_{i},g(t^{(n,\kappa )}_{i}))\) and \({\hat{t}}_{i}\in P^{(n,\kappa )}_1\).

Since \((t^{(n,k)}_{i},g(t^{(n,k)}_{i}))\in P^{(n,k)}\), \((t^{(n,\kappa )}_{i},g(t^{(n,\kappa )}_{i}))\in P^{(n,\kappa )}\), \(t_{i}\in P^{(n,k)}_1\), and \({\hat{t}}_{i}\in P^{(n,\kappa )}_1\), it follows from Step 4.1 that \((t_{i},\tau _{i})=(t_{i},g(t^{(n,k)}_{i}))\ne ({\hat{t}}_{i},g(t^{(n,\kappa )}_{i}))=({\hat{t}}_{i},\hat{\tau }_{i})\). \(\quad \square \)

Step 4.9

For each n and \(k\in \{1,\ldots ,k_n\}\), \({\hat{P}}^{(n,k)}_1=\tilde{P}^{(n,k)}_1\).

Proof of Step 4.9

Suppose that \(t_{i}\in {\hat{P}}^{(n,k)}_1\). Then \((t_{i},f^n(t_{i}))\in \tilde{P}^{(n,k)}\), and so \(t_{i}\in \tilde{P}^{(n,k)}_1\). Hence, \({\hat{P}}^{(n,k)}_1\subseteq \tilde{P}^{(n,k)}_1\). Conversely, suppose that \(t_{i}\in \tilde{P}^{(n,k)}_1\). Then \((t_{i},\tau _{i})\in \tilde{P}^{(n,k)}\) for some \(\tau _{i}\), implying that \(t_{i}\in {\hat{P}}^{(n,k)}_1\), and so \({\hat{P}}^{(n,k)}_1\supseteq \tilde{P}^{(n,k)}_1\). \(\square \)

Step 4.10

For each n and \(k\ne \kappa \), \({\hat{P}}^{(n,k)}_1\cap {\hat{P}}^{(n,\kappa )}_1=\emptyset \).

Proof of Step 4.10

Suppose that \(t_{i}\in {\hat{P}}^{(n,k)}_1\cap {\hat{P}}^{(n,\kappa )}_1\). Then \((t_{i},\tau _{i})\in {\hat{P}}^{(n,k)}\subseteq P^{(n,k)}\) and \((t_{i},\hat{\tau }_{i})\in {\hat{P}}^{(n,\kappa )}\subseteq P^{(n,\kappa )}\) for some \(\tau _{i}\) and \(\hat{\tau }_{i}\), and so \(\tau _{i}=g(t_{i})=\hat{\tau }_{i}\). Hence, \((t_{i},\tau _{i})=(t_{i},\hat{\tau }_{i})\in P^{(n,k)}\cap P^{(n,\kappa )}\), a contradiction. \(\square \)

Step 4.11

For each n,

Proof of Step 4.11

Fix n. Then,

where the first equality follows from Step 4.8; the second equality uses Step 4.9 and Step 4.10; and the third equality uses Step 4.9; \(\square \)

Step 4.12

We have

Proof of Step 4.12

We have

implying (45). Here, the first equality uses the definition of \(\lambda ^n\) in (44); the second equality uses the definition of \(\zeta ^n\) from the first paragraph of the proof of Step 4, Step 4.6, and Step 4.8; the third equality uses Step 4.7, Step 4.9, and Step 4.10; the fourth equality follows from Step 4.9; the inequality follows from Step 4.5; the sixth and seventh equalities follow from Step 4.2 and the definition of \(\lambda \) in (40); and the limit at the end follows from Step 4.11, together with the boundedness of \(\zeta \) and the fact that \(\lambda (S^n)\rightarrow 1\). \(\square \)

Step 4.13

There exists \(n^{*}\) such that

Proof of Step 4.13

The assertion follows immediately from Step 3 and Step 4.12. \(\square \)

Letting \(\eta ^{*}_{i}\in \mathscr {D}_{i}\) be defined by

where \(\delta _{f^{n^{*}}(t_{i})}\) represents the Dirac measure in \(\Delta (T_{i})\) with support \(f^{n^{*}}(t_{i})\), and where \(n^{*}\) is the natural number from Step 4.13, it follows from (46) that

This finishes the proof of Step 4. \(\square \)

Combining Step 4 and the fact that

yields (19). This finishes the proof of Lemma. \(\square \)

1.2 Proof of Lemma 5

Lemma 4

Suppose that \(\Gamma =\left( T_{i},X_{i},u_{i},p\right) _{i=1}^{N}\) is a Bayesian game. Suppose that \((\mu ^n_{i})\) and \((\nu ^n_{i})\) are sequences in \(\mathscr {T}_{i}\). Suppose that

Suppose further that \((\mu ^n_{-i})\) is a sequence in \(\mathscr {T}_{-i}\). Then, for every subsequence \((n_k)\) of (n),

Proof

Suppose that (21) does not hold for some subsequence \((n_k)\) of (n). Then, for some \(t\in T\), and extracting a subsequence if necessary,

for some \(\gamma >0\). Therefore, there exist \(\gamma '>0\) and M such that for each \(m\ge M\),

i.e. (recall (2)),

where \((B)_{x_{-i}}\) (resp. \((N_\epsilon (B))_{x_{-i}}\)) denotes the section of B (resp. \(N_\epsilon (B)\)) in \(X_{i}\) at \(x_{-i}\).^{Footnote 17} Now (20) implies the following:

Therefore,

implying that there exist \(\gamma ''\in (0,\gamma ')\) and K such that for each \(k\ge K\), and for each B closed in \(X_{i}\),

Consequently, for each \(k\ge K\), and for each B closed in X and \(x_{-i}\in X_{-i}\),^{Footnote 18}

implying that for each \(k\ge K\) and B closed in X,

Consequently, there is an \(M'\) such that for each \(m\ge M'\) and each B closed in X,

for some \(\gamma '''\in (0,\gamma ')\). But this implies that for \(m\ge \max \{M,M'\}\), the left-hand side of (47) must be strictly less than \(\gamma '\), a contradiction. \(\square \)

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