Essential equilibria of large generalized games

Abstract

We characterize the essential stability of games with a continuum of players, where strategy profiles may affect objective functions and admissible strategies. Taking into account the perturbations defined by a continuous mapping from a complete metric space of parameters to the space of continuous games, we prove that essential stability is a generic property and every game has a stable subset of equilibria. These results are extended to discontinuous large generalized games assuming that only payoff functions are subject to perturbations. We apply our results in an electoral game with a continuum of Cournot-Nash equilibria, where the unique essential equilibrium is that only politically engaged players participate in the electoral process. In addition, employing our results for discontinuous games, we determine the stability properties of competitive prices in large economies.

Appendix

Lemma 1

The set of messages $\widehat{M}$ is non-empty and compact.

Proof

Since $t \in T_1 \twoheadrightarrow K_t$ is measurable, from Aliprantis and Border (2006, Lemma 18.2, and Theorem 18.6), we know that this correspondence has an ${\mathcal {A}}\times {\mathcal {B}}(\widehat{K})$-measurable graph. It follows from Aumann’s Selection Theorem [see Aliprantis and Border (2006, Theorem 18.26, page 608)] that there exists an ${\mathcal {A}}$-measurable function $g:T_{1}\rightarrow \widehat{K}$ such that, $g(t)\in K_t,\,\forall t \in T_1$. Hence, the compactness of $T_1$ and $\widehat{K}$ and the continuity of $H$ guarantee that the map $t \rightarrow H(t, g(t))$ is bounded and measurable. Therefore, $\widehat{M}$ is non-empty.

Since $H$ is continuous, $\mu$ is a finite measure, and the sets $T_1$ and $\widehat{K}$ are compact, the correspondence $t\in T_1 \twoheadrightarrow H(t,\widehat{K})$ is integrable bounded and has closed values. Thus, as $\widehat{M}={\int _{T_1} } H(t, \widehat{K}) d\mu$, it follows from Aumann (1965, Theorem 4) that $\widehat{M}$ is compact. $\square$

Let us define some notations. Given a metric space $(S,d)$, consider the sets

$$\begin{aligned} A(S)&= \{K\subseteq S: K\,\,\text{ is } \text{ non-empty } \text{ and } \text{ compact }\},\\ A_c(S)&= \{C\subseteq A(S): C\,\,\text{ is } \text{ convex }\}. \end{aligned}$$

Denote by $d_H$ the Hausdorff metric induced by the metric of $S$. If $S$ is compact, then $(A(S), d_H)$ is a complete metric space. Also, when $S$ is compact and convex, $(A_c(S),d_H)$ is complete.²⁶

Given a set $X$, let ${\mathcal {U}}(X)$ be the collection of bounded functions $u:X\rightarrow {\mathbb {R}}$ endowed with the sup norm topology, i.e., the topology determined by the metric $d(u_1,u_2)=\sup _{x\in X} \vert u_1(x) - u_2(x)\vert .$

Proof of Proposition 1

Let $\{{\mathcal {G}}_n\}_{n\in {\mathbb {N}}}$, with ${\mathcal {G}}_n= {\mathcal {G}}_n((K_{n,t},\varGamma _{n,t},u_{n,t})_{t\in T_1 \cup T_2})$, be a Cauchy sequence on $({\mathbb {G}},\rho )$. It follows from definition of ${\mathbb {G}}$ and $\rho$ that for any non-atomic player $t\in T_1, \{K_{n,t}\}_{n\in {\mathbb {N}}}$ is a Cauchy sequence on $(A(\widehat{K}),d_H)$. Also, for any atomic player $s \in T_2, \{K_{n,s}\}_{n\in {\mathbb {N}}}$ is a Cauchy sequence on $(A_c(\widehat{K}_s),d_{H,s})$. Hence, there are sets $\{\overline{K}_t\}_{t\in T_1\cup T_2}$ such that (i) $(\overline{K}_t,\overline{K}_s)\in A(\widehat{K}) \times A_c(\widehat{K}_s),\,\forall (t,s)\in T_1\times T_2;$ and (ii) for any $(t,s)\in T_1\times T_2,$ we have that $\lim _{n\rightarrow +\infty } d_H(K_{n,t},\overline{K}_t)= \lim _{n\rightarrow +\infty } d_{H,s}(K_{n,s},\overline{K}_s)=0.$

The definition of the metric $\rho$ ensures that, for any $t\in T_1$ and $(m,a)\in \widehat{M}\times \widehat{\mathcal {F}}^2$, the sequence $\{\varGamma _{n,t}(m,a)\}_{n\in {\mathbb {N}}}\subseteq A(\widehat{K})$ is Cauchy and, therefore, there exists a set $K_t(m,a)\in A(\widehat{K})$ such that $d_H(\varGamma _{n,t}(m,a),K_t(m,a))$ converges to zero as $n$ goes to infinity. Let $\overline{\varGamma }_t: \widehat{M}\times \widehat{\mathcal {F}}^2 \twoheadrightarrow \widehat{K}$ be the set-valued mapping defined by $\overline{\varGamma }_t(m,a)=K_t(m,a).$ It follows that correspondences $(\overline{\varGamma }_t)_{t\in T}$ are continuous.²⁷ By analogous arguments, we can obtain that for any $s\in T_2$, there is a continuous correspondence $\overline{\varGamma }_s: \widehat{M}\times \widehat{\mathcal {F}}^2_{-s} \twoheadrightarrow \widehat{K}_s$ such that, for each $(m,a_{-s})\in \widehat{M}\times \widehat{\mathcal {F}}^2_{-s}$ both $\overline{\varGamma }_{s}(m,a_{-s})\in A_c(\widehat{K}_s)$ and $d_{H,s}(\varGamma _{n,s}(m,a_{-s}),\overline{\varGamma }_s(m,a_{-s}))$ converges to zero as $n$ increases.

Since $\{{\mathcal {G}}_n\}_{n\in {\mathbb {N}}}$ is Cauchy on $({\mathbb {G}},\rho )$, there is a bounded function $U:T_1\times \widehat{K}\times \widehat{M}\times \widehat{\mathcal {F}}^2 \rightarrow {\mathbb {R}}$ such that, for each $t\in T_1$, the sequence $\{u_{n,t}\}_{n\in {\mathbb {N}}}\subseteq {\mathcal {U}}(\widehat{K}\times \widehat{M}\times \widehat{\mathcal {F}}^2)$ converges uniformly to $\overline{u}_t:=U(t,\cdot )$ and, therefore, $\overline{u}_t$ is continuous. Analogously, for any $t\in T_2$, the sequence $\{u_{n,t}\}_{n\in {\mathbb {N}}}\subseteq {\mathcal {U}}( \widehat{M}\times \widehat{\mathcal {F}}^2)$ converges to some continuous function $\overline{u}_t\in {\mathcal {U}}(\widehat{M}\times \widehat{\mathcal {F}}^2)$ that is quasi-concave on $a_t$.

Let $\overline{\mathcal {G}}= \overline{\mathcal {G}}((\overline{K}_{t},\overline{\varGamma } _{t},\overline{u}_{t})_{t\in T_1 \cup T_2})$. It follows from arguments above that $\lim _{n\rightarrow +\infty } \rho ({\mathcal {G}}_n, \overline{\mathcal {G}})=0$. Thus, to conclude that $({\mathbb {G}},\rho )$ is complete, it is sufficient to guarantee that

• (i) for each $(m,a)\in \widehat{M} \times \widehat{\mathcal {F}}^2, (t,x) \in T_1 \times \widehat{K} \rightarrow \overline{u}_t(x,m,a)$ is measurable;

• (ii) for each $(m,a)\in \widehat{M}\times \widehat{\mathcal {F}}^2, t\in T_1 \twoheadrightarrow \overline{\varGamma }_t(m,a)$ is measurable;

• (iii) the correspondence $t\in T_1 \twoheadrightarrow \overline{K}_t$ is measurable.

Fix $(m,a) \in \widehat{M} \times \widehat{\mathcal {F}}^2$, the definition of $\rho$ ensures that measurable functions $(t,x) \in T_1\times \widehat{K} \rightarrow u_{n,t}(x,m,a)$ converge to the mapping $(t,x) \in T_1 \times \widehat{K} \rightarrow \overline{u}_t(x,m,a)$. Since $(T_1\times \widehat{K}, {\mathcal {A}}\times {\mathcal {B}}(\widehat{K}))$ is a measurable space, item (i) holds [see Aliprantis and Border (2006, Lemma 4.29, page 142)]. Furthermore, since for every $n\in {\mathbb {N}}$ the correspondence $t\in T_1 \twoheadrightarrow \varGamma _{n,t}(m,a)$ is measurable, it follows from Aliprantis and Border (2006, Theorem 18.10, page 598) that the function $\varTheta _{n,(m,a)}:T_1\rightarrow A(\widehat{K})$ defined by $\varTheta _{n,(m,a)}(t)=\varGamma _{n,t}(m,a)$ is Borel measurable. Also, the sequence $\{\varTheta _{n,(m,a)}\}_{n\in {\mathbb {N}}}$ converges to $\overline{\varTheta }_{(m,a)}:T_1\rightarrow A(\widehat{K})$, where $\overline{\varTheta }_{(m,a)}(t)=\overline{\varGamma }_t(m,a)$. By Aliprantis and Border (2006, Lemma 4.29), $\overline{\varTheta }_{(m,a)}$ is a Borel measurable function. Thus, $t\in T_1 \twoheadrightarrow \overline{\varGamma }_t(m,a)$ is measurable [see Aliprantis and Border (2006, Theorem 18.10)]. By analogous arguments, we obtain item (iii). $\square$

Proof of Theorem 1

The proof is a direct consequence of the following steps.

Step 1. $\varLambda :{\mathbb {G}} \twoheadrightarrow \widehat{M} \times \widehat{\mathcal {F}}^2$ is upper hemicontinuous with compact values.

Since $\widehat{M}\times \widehat{\mathcal {F}}^2$ is compact and non-empty, we only need to prove that $\text{ Graph }(\varLambda )$ is closed, where $\text{ Graph }(\varLambda ):=\left\{ (\mathcal {G},(m,a))\in \mathbb {G}\times \widehat{M}\times \widehat{\mathcal {F}}^2: (m,a)\in \varLambda ({\mathcal {G}})\right\}$. Let $\{(\mathcal {G}_n,(m_n,a_n))\}_{n \in {\mathbb {N}}}\subset \text{ Graph }(\varLambda )$ be a sequence converging to $(\overline{\mathcal {G}},(\overline{m},\overline{a}))\in \mathbb {G}\times \widehat{M}\times \widehat{\mathcal {F}}^2$, where $\overline{\mathcal {G}}= \overline{\mathcal {G}}((\overline{K}_{t},\overline{\varGamma }_{t},\overline{u}_{t})_{t\in T_1 \cup T_2})$ and, for every $n \in {\mathbb {N}}, {\mathcal {G}}_n= {\mathcal {G}}_n((K^n_{t},\varGamma ^n_{t},u^n_{t})_{t\in T_1 \cup T_2})$.

We aim to ensure that $(\overline{m},\overline{a}) \in \varLambda ({\overline{\mathcal {G}}})$. Since for any $n \in {\mathbb {N}}, (m_n,a_n)\in \varLambda ({\mathcal {G}_n})$, there exists $f_n \in \widehat{\mathcal {F}}^1$ such that (i) the function $g_n:T_1\rightarrow {\mathbb {R}}^m$ given by $g_n(t)=H(t,f_n(t))$ is measurable and $m_n=m(f_n)$; and (ii) for almost all $t \in T_1$ both $f_n(t)\in \varGamma ^n_t(m_n,a_n)$ and

$$\begin{aligned} u_t^n(f_n(t), m_n,a_{n})=\max \limits _{x\in \varGamma _t^n(m_n, a_n)}{u_t^n(x, m_n,a_{n})}. \end{aligned}$$

Claim A

There exists $\overline{f}\in \widehat{\mathcal {F}}^1$ such that $\overline{m}=\int _{T_1}{H(t,\overline{f}(t))d\mu }$.

Proof

Since $H$ is continuous, $T_1$ is compact and $\{f_n\}_{n \in {\mathbb {N}}} \subset \widehat{\mathcal {F}}^1$, it follows that the sequence $\{g_n\}_{n\in \mathbb {N}}$ is uniformly integrable [see Hildenbrand (1974, page 52)]. In addition, $\{\int _{T_1}{g_n(t)d\mu }\}_{n \in {\mathbb {N}}} \subset {\mathbb {R}}^m$ converges to $\overline{m}$ as $n$ goes to infinity , and therefore, the multidimensional version of Fatou’s Lemma [see Hildenbrand (1974, page 69)] guarantees that there is $g:T_1 \rightarrow {\mathbb {R}}^m$ integrable such that, ²⁸

• (1) $\overline{m}=\lim \limits _{n\rightarrow \infty }{\int _{T_1}{g_n(t)d\mu }}=\int _{T_1}{g(t)}d\mu$;

• (2) there exists a full measure set $\tilde{T}_1\subseteq T_1$ such that, for any $t \in \tilde{T}_1, g(t)\in L_S(g_n(t)),$ where $L_S(g_n(t))$ is the set of cluster points of $\{g_n(t)\}_{n\in \mathbb {N}}$.²⁹

Fix $t\in \tilde{T}_1$. Then, there is a subsequence $\left( g_{n_k}(t)\right) _k$ converging to $g(t)$. Since $\{f_{n_k}(t)\}_{k\in \mathbb {N}}\subseteq \widehat{K}$, by taking a subsequence if it is necessary, we can ensure that there exists $f(t)\in \widehat{K}$ such that both $f_{n_k}(t) \rightarrow f(t)$ and $g(t)=\lim \limits _{k\rightarrow \infty }{H(t,f_{n_k}(t))}=H(t,f(t))$ hold.

Let $\overline{f}:T_1 \rightarrow {\widehat{K}}$ be a function such that

$$\begin{aligned} \overline{f}(t)\in \left\{ \begin{array}{ll} \{ f(t)\},&{}\quad \text{ if }\;\;t\in \;\tilde{T}_1, \\ \overline{\varGamma }_t(\overline{m},\overline{a}),&{}\quad \text{ if }\;\; t\in \; T {\setminus } \tilde{T}_1. \end{array} \right. \end{aligned}$$

Then, it follows that

$$\begin{aligned} \overline{m}=\lim _{n \rightarrow \infty } m_n= \lim \limits _{n\rightarrow \infty }{\int \limits _{T_1} g_n(t) d\mu }=\int \limits _{T_1} g(t) d\mu =\int \limits _{T_1}H(t,\overline{f}(t))d\mu , \end{aligned}$$

where the last equality follows from the fact that $T_1 {\setminus } \tilde{T}_1$ has zero measure. $\square$

Claim B

For almost all $t\in T_1, \overline{f}(t)\in \overline{\varGamma }_t(\overline{m},\overline{a})$. In addition, for any $t \in T_2, \overline{a}_t \in \overline{\varGamma }_t(\overline{m},\overline{a}_{-t})$.

Proof

Following the notation of the proof of the previous claim, fix $t \in \tilde{T}_1$ and let $\{f_{n_k}(t)\}_{k \in {\mathbb {N}}}$ be the sequence converging to $\overline{f}(t)$ and that was obtained in the previous claim. We know that, for any $k \in \mathbb {N}, f_{n_k}(t)\in \varGamma _t^{n_k}(m_{n_k},a_{n_k})$. Therefore,

$$\begin{aligned} d(\overline{f}(t),\overline{\varGamma }_t(\overline{m},\overline{a}))&\le \widehat{d}(\overline{f}(t), f_{n_k}(t)) +d_H(\varGamma _t^{n_k}(m_{n_k},a_{n_k}),\overline{\varGamma }_t(m_{n_k},a_{n_k}))\\&+d_H(\overline{\varGamma }_t(m_{n_k},a_{n_k}),\overline{\varGamma }_t(\overline{m},\overline{a}))\\&\le \widehat{d}(\overline{f}(t) , f_{n_k}(t))+\rho ({\mathcal {G}}_{n_k}, \overline{\mathcal {G}}) + d_H(\overline{\varGamma }_t(m_{n_k},a_{n_k}),\overline{\varGamma }_t(\overline{m},\overline{a})), \end{aligned}$$

where $\widehat{d}$ denotes the metric of the compact metric space $\widehat{K}$. Since $\overline{\varGamma }_t$ is continuous, by taking the limit as $k$ goes to infinity, we obtain the first property.

On the other hand, for any $(t,n) \in T_2 \times {\mathbb {N}}, a_{n,t}\in \varGamma _t^n(m_n,a_{n,-t})$, which implies that

$$\begin{aligned} d(\overline{a}_t,\overline{\varGamma }_t(\overline{m},\overline{a}_{-t}))&\le \widehat{d}_t(\overline{a}_t ,a_{n,t})+d_{H,t}(\varGamma _t^n(m_n,a_{n,-t}),\overline{\varGamma }_t(m_n,a_{n,-t}))\\&+d_{H,t}(\overline{\varGamma }_t(m_n,a_{n,-t}),\overline{\varGamma }_t(\overline{m},\overline{a}_{-t}))\\&\le \widehat{d}_t(\overline{a}_t , a_{n,t})+\rho ({\mathcal {G}}_n, \overline{\mathcal {G}}) + d_{H,t}(\overline{\varGamma }_t(m_n,a_{n,-t}),\overline{\varGamma }_t(\overline{m},\overline{a}_{-t})), \end{aligned}$$

where $\widehat{d}_t$ denotes the metric of $\widehat{K}_t$. Taking the limit as $n$ goes to infinity, we obtain the result. $\square$

Claim C

The following properties hold:

• (i) For almost all $t \in T_1, \overline{f}(t) \in \mathop {\hbox {argmax}}\nolimits _{x \in \overline{\varGamma }_t(\overline{m},\overline{a})}\,\,\overline{u}_t(x,\overline{m},\overline{a}).$

• (ii) For any $t\in T_2, \overline{a}_t\in \mathop {\hbox {argmax}}\nolimits _{x\in \overline{\varGamma }_t(\overline{m},\overline{a}_{-t})}\overline{u}_t(\overline{m},x,\overline{a}_{-t}).$

Proof

(i) Given $t \in \tilde{T}_1$, we have that

$$\begin{aligned} d_H(\varGamma _t^{n_k}(m_{n_k},a_{n_k}),\overline{\varGamma }_t(\overline{m},\overline{a}))&\le \rho (\mathcal {G}_{n_k},\overline{\mathcal {G}})+d_H(\overline{\varGamma }_t(m_{n_k},a_{n_k}),\overline{\varGamma }_t(\overline{m},\overline{a})). \end{aligned}$$

Then, $\varGamma _t^{n_k}(m_{n_k},a_{n_k})\longrightarrow _{k}\,\, \overline{\varGamma }_t(\overline{m},\overline{a})$. Since $u^{n_k}_t$ converges uniformly to $\overline{u}_t$, it follows from Yu (1999, Lemma 2.5) and Aubin (1982, Theorem 3, page 70) that

$$\begin{aligned} u_t^{n_k}(f_{n_k}(t),m_{n_k},a_{n_k})=\max _{x\in \varGamma _t^{n_k}(m_{n_k},a_{n_k})}{u_t^{n_k}(x,m_{n_k},a_{n_k})}\longrightarrow _k\,\, \max _{x\in \overline{\varGamma }_t(\overline{m},\overline{a})}{\overline{u}_t(x,\overline{m},\overline{a})} \end{aligned}$$

On the other hand,

$$\begin{aligned} |u_t^{n_k}(f_{n_k}(t),m_{n_k},a_{n_k})-\overline{u}_t(\overline{f}(t),\overline{m},\overline{a})| \\ \le \rho (\mathcal {G}_{n_k},\overline{\mathcal {G}})+ |\overline{u}_t(f_{n_k}(t),m_{n_k},a_{n_k})-\overline{u}_t(\overline{f}(t),\overline{m},\overline{a})|. \end{aligned}$$

Taking the limit as $k$ goes to infinity, we obtain that $u_t^{n_k}(f_{n_k}(t),m_{n_k},a_{n_k})\rightarrow \overline{u}_t(\overline{f}(t),\overline{m},\overline{a})$. Hence, it follows from Claim B that $\overline{f}(t) \in \mathop {\hbox {argmax}}\limits _{x\in \overline{\varGamma }_t(\overline{m},\overline{a})}{\overline{u}_t(x,\overline{m},\overline{a})}$.

(ii) Given $t \in T_2$, analogous arguments to those made in the previous item ensure that

$$\begin{aligned} d_{H,t}(\varGamma _t^n(m_n,a_{n,-t}),\overline{\varGamma }_t(\overline{m},\overline{a}_{-t}))&\le \rho (\mathcal {G}_n,\mathcal {G})+d_{H,t}(\overline{\varGamma }_t(m_n,a_{n,-t}),\overline{\varGamma }_t(\overline{m},\overline{a}_{-t})), \end{aligned}$$

which implies that $\varGamma _t^n(m_n,a_{n,-t})$ converges to $\overline{\varGamma }_t(\overline{m},\overline{a}_{-t})$ as $n$ goes to infinity. Hence, Yu (1999, Lemma 2.5) ensures that

$$\begin{aligned} u_t^n(m_n,a_{n})=\max _{x\in \varGamma _t^n(m_n,a_{n,-t})}{u_t^n(m_n,x,a_{n,-t})}\longrightarrow \max _{x\in \overline{\varGamma }_t(\overline{m},\overline{a}_{-t})}{\overline{u}_t(\overline{m},x,\overline{a}_{-t})}. \end{aligned}$$

Since $\lim \limits _{n \rightarrow +\infty }\,\,u_t^n(m_n,a_{n})=\overline{u}_t(\overline{m},\overline{a})$,³⁰ $\overline{a}_t\in \mathop {\hbox {argmax}}\limits _{x\in \overline{\varGamma }_t(\overline{m},\overline{a}_{-t})}\overline{u}_t(\overline{m},x,\overline{a}_{-t}).$ $\square$

It follows from Claims A and C that $(\overline{m},\overline{a})\in \varLambda ({\overline{\mathcal {G}}})$. Thus, we ensure that $\varLambda$ is an upper hemicontinuous correspondence with compact values.

Step 2. There is a dense residual set $Q\subseteq {\mathbb {G}}'$ where $\varLambda$ is lower hemicontinuous.

Since ${\mathbb {G}}'$ is a closed subset of ${\mathbb {G}}$, it follows that $({\mathbb {G}}',\rho )$ is a complete metric space and, therefore, it is a Baire space. Since the correspondence $\varLambda$ is non-empty, compact-valued, and upper hemicontinuous, it follows from Lemmas 5 and 6 in Carbonell-Nicolau (2010) [see also Fort (1949) and Jiang (1962)] that there exists a dense residual subset $Q$ of $\mathbb {G}'$ in which $\varLambda$ is lower hemicontinuous.

Step 3. If $\mathcal {G}\in {\mathbb {G}}'$ is a point of lower hemicontinuity of $\varLambda$, then $\mathcal {G}$ is essential with respect to ${\mathbb {G}}'$.

Fix $(f^*,a^*) \in \text{ CN }({\mathcal {G}})$. Then, for any open neighborhood $O\subseteq \widehat{M}\times \widehat{\mathcal {F}}^2$ of $(m(f^*),a^*)$, we have $\varLambda (\mathcal {G})\cap O\ne \emptyset$, and therefore, by the lower hemicontinuity, we have that $\left\{ \mathcal {G}'\in \mathbb {G}'\,:\;\;\varLambda (\mathcal {G}')\cap O\ne \emptyset \right\}$ contains a neighborhood of ${\mathcal {G}}$, that is, for some $\epsilon >0$ and for any ${\mathcal {G}}'\in {\mathbb {G}}'$ such that $\rho ({\mathcal {G}}', {\mathcal {G}})<\epsilon$, we have $\varLambda (\mathcal {G}')\cap O\ne \emptyset$. Hence, all Cournot-Nash equilibria of ${\mathcal {G}}$ are essential with respect to ${\mathbb {G}}'$.

It follows from Steps 2 and 3 that any game in $Q$ is essential.

Finally, suppose that for some ${\mathcal {G}}\in {\mathbb {G}}'$ the set $\varLambda (\mathcal {G})$ is a singleton. Then, the upper hemi-continuity of $\varLambda$ guarantees that it is continuous at $\mathcal {G}$. Finally, Step 3 implies that ${\mathcal {G}}$ is an essential generalized game with respect to ${\mathbb {G}}'$. $\square$

Proof of Theorem 2

(i) Existence of a minimal essential set.

Fix $\mathcal {G}\in \mathbb {G}'$. Let $\mathcal {S}$ be the family of essential subsets of $\varLambda (\mathcal {G})$ with respect to ${\mathbb {G}}'$ ordered by set inclusion. Since $\varLambda$ is upper hemicontinuous, $\varLambda (\mathcal {G})\in \mathcal {S}$ and, hence, $\mathcal {S}\ne \emptyset$. As any essential set is non-empty and compact, each totally ordered subset of $\mathcal {S}$ has a lower bound. By Zorn’s Lemma, $\mathcal {S}$ has a minimal element and, by definition, it is an essential set of $\varLambda (\mathcal {G})$ with respect to ${\mathbb {G}}'$.

(ii) If there are connected essential sets, then there are essential components.

Suppose that there is a connected essential set of $\varLambda (\mathcal {G})$ with respect to ${\mathbb {G}}'$, denoted by $c({\mathcal {G}})$. Since $c({\mathcal {G}})$ is non-empty, fix $(\widehat{m},\widehat{a})\in c(\mathcal {G})$ and consider the set $\varLambda _{(\widehat{m},\widehat{a})}(\mathcal {G})$ defined as the union of all connected subsets of $\varLambda (\mathcal {G})$ that contains $(\widehat{m},\widehat{a})$. By definition, $\varLambda _{(\widehat{m},\widehat{a})}(\mathcal {G})$ is a component of $\varLambda (\mathcal {G})$. As the closure of a connected set is connected and $\varLambda ({\mathcal {G}})$ is compact, it follows that $\varLambda _{(\widehat{m},\widehat{a})}(\mathcal {G})$ is compact. Hence, the essentiality of $c(\mathcal {G})\subseteq \varLambda _{(\widehat{m},\widehat{a})}(\mathcal {G})$ with respect to ${\mathbb {G}}'$ ensures that the component $\varLambda _{(\widehat{m},\widehat{a})}(\mathcal {G})$ is also an essential subset of $\varLambda (\mathcal {G})$ with respect to ${\mathbb {G}}'$.

(iii) Connectedness of minimal essential sets in normed spaces.

Suppose that $\widehat{K}$ is a convex subset of a normed space and it is equipped with a metric induced by a norm. Fix a minimal essential set of $\varLambda ({\mathcal {G}})$ with respect to ${\mathbb {G}}'$, denoted by $m(\mathcal {G})$. Suppose, by contradiction, that $m(\mathcal {G})$ is disconnected.

Claim A

There are open sets $U_1, U_2 \subseteq \widehat{M} \times \widehat{\mathcal {F}}^2$ such that $m({\mathcal {G}})\subset U_1 \cup U_2$ and $\overline{U}_1\cap \overline{U}_2=\emptyset$.

Proof

Since $m(\mathcal {G})$ is disconnected, there are closed and non-empty subsets $A_1, A_2 \subseteq \varLambda (\mathcal {G})$ such that $A_1 \cap A_2 = \emptyset$ and $m(\mathcal {G})=A_1\cup A_2$. Since $m(\mathcal {G})$ is minimal, neither $A_1$ nor $A_2$ are essential with respect to ${\mathbb {G}}'$. Hence, for each $i \in \{1,2\}$, there exists an open set $U_i$ such that $A_i\subset U_i$ and for all $\epsilon >0$, there exists $\mathcal {G}^i_\epsilon \in \mathbb {G}'$ such that both $\rho (\mathcal {G},\mathcal {G}^i_\epsilon )<\epsilon$ and $\varLambda (\mathcal {G}^i_\epsilon )\cap U_i=\emptyset$. Since $A_i$ is compact, we can assume that $\overline{U}_1\cap \overline{U}_2=\emptyset$. $\square$

Claim B

There are large generalized games ${\mathcal {G}}_1, {\mathcal {G}}_2 \in {\mathbb {G}}'$ such that

$$\begin{aligned} \varLambda ({\mathcal {G}}_i) \cap U_i =\emptyset \quad \wedge \quad \varLambda ({\mathcal {G}}_i) \cap U_j \ne \emptyset ,\,\,\,\,\quad \forall i,j \in \{1,2\}: i \ne j. \end{aligned}$$

In addition, there is a continuous map $G:\widehat{M}\times \widehat{\mathcal {F}}^2\rightarrow {\mathbb {G}}$ such that, for every $(m,a)\in \widehat{M}\times \widehat{\mathcal {F}}^2$,

$$\begin{aligned} \varLambda (G(m,a))\cap (U_1 \cup U_2)\ne \emptyset ,\quad \quad (G(m,a)={\mathcal {G}}_i \Longleftrightarrow (m,a)\in \overline{U}_i),\,\forall i \in \{1,2\}. \end{aligned}$$

Proof

As $m(\mathcal {G})$ is essential with respect to ${\mathbb {G}}'$, there exists $\nu >0$ such that for every $\mathcal {G}'\in \mathbb {G}'$ with $\rho (\mathcal {G},\mathcal {G}')<\nu$, we have $\varLambda (\mathcal {G}')\cap (U_1 \cup U_2)\ne \emptyset$. Following the notation of the previous claim, for each $i \in \{1,2\}$, set ${\mathcal {G}}_i={\mathcal {G}}^i_{\nu /3}$. Hence, for $i \ne j$, we obtain $\varLambda (\mathcal {G}_i)\cap U_j\ne \emptyset$.

Let $G:\widehat{M}\times \widehat{\mathcal {F}}^2\rightarrow {\mathbb {G}}$ be the function³¹

$$\begin{aligned} G(m,a)=\lambda (m,a) \mathcal {G}_1+(1-\lambda (m,a))\mathcal {G}_2,\,\,\,\,\, \forall (m,a)\in \widehat{M}\times \widehat{\mathcal {F}}^2, \end{aligned}$$

where $\lambda :\widehat{M}\times \widehat{\mathcal {F}}^2\rightarrow [0,1]$ is the continuous function given by,

$$\begin{aligned} \lambda (m,a)=\frac{d((m,a),\overline{U}_2)}{d((m,a),\overline{U}_1)+d((m,a),\overline{U}_2)}. \end{aligned}$$

By construction, for each $i \in \{1,2\}, G(m,a)=\mathcal {G}_i$ if and only if $(m,a)\in \overline{U}_i$.

Since metric spaces $\widehat{K}$ and $\{\widehat{K}_t\}_{t \in T_2}$ are contained in normed vectorial spaces and their metrics are induced by norms, for any $(m,a) \in \widehat{M}\times \widehat{\mathcal {F}}^2$, we can ensure that $G(m,a)$ is well defined and

$$\begin{aligned} \rho (G(m,a), {\mathcal {G}}_1)&= \rho \left( \lambda (m,a) {\mathcal {G}}_1 + (1-\lambda (m,a)) {\mathcal {G}}_2, \lambda (m,a) {\mathcal {G}}_1 + (1-\lambda (m,a)) {\mathcal {G}}_1\right) \\&= \lambda (m,a) \rho ({\mathcal {G}}_1, {\mathcal {G}}_1) + (1-\lambda (m,a)) \rho ({\mathcal {G}}_2, {\mathcal {G}}_1) \\&\le \rho ( \,{\mathcal {G}}_2, {\mathcal {G}}_1) \le \rho ( \,{\mathcal {G}}_2, {\mathcal {G}}) + \rho ( \,{\mathcal {G}}, {\mathcal {G}}_1) < \frac{2\nu }{3}, \end{aligned}$$

which implies that $\rho ({\mathcal {G}}, G(m,a))\le \rho ( \,{\mathcal {G}}, {\mathcal {G}}_1) + \rho ( \,{\mathcal {G}}_1, G(m,a)) < \nu .$ Hence, for each $(m,a) \in \widehat{M}\times \widehat{\mathcal {F}}^2, \varLambda (G(m,a))\cap (U_1 \cup U_2)\ne \emptyset$. $\square$

Given a large generalized game ${\mathcal {G}}\in {\mathbb {G}}$, let $\varPhi _{\mathcal {G}}:\widehat{M}\times \widehat{\mathcal {F}}^2 \twoheadrightarrow \widehat{M}\times \widehat{\mathcal {F}}^2$ be the correspondence defined by $\varPhi _{{\mathcal {G}}}(m,a)=\left( \varOmega ^{{\mathcal {G}}}(m,a),(B^{_{{\mathcal {G}}}}_t(m,a_{-t}))_{t\in T_2}\right)$, where

$$\begin{aligned} \varOmega ^{{\mathcal {G}}}(m,a)&:= \left\{ \int \limits _{T_1} H(t,f(t)) d\mu : H(\cdot , f(\cdot )) \,\text{ is } \text{ integrable } \wedge f(t)\in B^{\mathcal {G}}_t(m,a),\,\forall t \in T_1\right\} \!; \\ B^{{\mathcal {G}}}_t(m,a)&:= \mathop {\hbox {argmax}}\limits _{x_t\in \varGamma _t(m,a)}{u_t(x_t,m,a)},\,\,\,\,\,\,\forall t \in T_1;\\ B^{{\mathcal {G}}}_t(m,a_{-t})&:= \mathop {\hbox {argmax}}\limits _{x_t\in \varGamma _t(m,a_{-t})}{u_t(x_t, m,a_{-t})},\,\,\,\,\,\,\forall t \in T_2. \end{aligned}$$

We affirm that $\varPhi _{{\mathcal {G}}}$ is upper hemicontinuous and has non-empty, compact, and convex values. Note that, Berge’s Maximum Theorem [see Aliprantis and Border (2006, Theorem 17.31, page 570)] guarantees that for any $t \in T_1 \cup T_2$, the correspondence $B^{\mathcal {G}}_t$ is upper hemicontinuous and has non-empty and compact values. Also, as atomic players have convex strategy sets and quasi-concave payoff functions, correspondences $(B^{\mathcal {G}}_t)_{t\in T_2}$ have convex values. Hence, we want to prove that $\varOmega ^{\mathcal {G}}(\cdot )=\int _{T_1} H(t,B^{\mathcal {G}}_t(\cdot ))d\mu$ is upper hemicontinuous and has non-empty, compact, and convex values. Aumann (1965, Theorem 1) guarantees that $\varOmega ^{\mathcal {G}}$ has convex values. Fix $(m,a) \in \widehat{M}\times \widehat{\mathcal {F}}^2$. Since $t\in T_1 \twoheadrightarrow \varGamma _t(m,a)$ is measurable, it follows from Aliprantis and Border (2006, Lemma 18.2, page 593) that $t\in T_1 \twoheadrightarrow \varGamma _t(m,a)$ is weakly measurable. The Measurable Maximum Theorem (Aliprantis and Border (2006, Theorem 18.19, page 605)) implies that $t\in T_1 \twoheadrightarrow H(t,B^{\mathcal {G}}_t(m,a))$ has a measurable selector. Since $H$ is continuous, the compactness of $T_1$ and $\widehat{K}$ guarantees that $t\in T_1 \rightarrow H(t,B^{\mathcal {G}}_t(m,a))$ is bounded and, therefore, its measurable selectors are integrable. We conclude that $\varOmega ^{\mathcal {G}}(m,a)$ is non-empty. Since $T_1$ has finite measure and $t\in T_1 \twoheadrightarrow H(t,B^{\mathcal {G}}_t(m,a))$ is bounded and has closed values, it follows from Aumann (1965, Theorem 4) that $\varOmega ^{\mathcal {G}}(m,a)$ is compact. Finally, as $H$ is continuous, $\widehat{K}$ is compact, and for every $t \in T_1$ the correspondence $(m,a) \twoheadrightarrow B^{\mathcal {G}}_t(m,a)$ has closed graph; it follows that $(m,a) \twoheadrightarrow H(t,B^{\mathcal {G}}_t(m,a))$ is upper hemicontinuous for each $t \in T_1$. Thus, it follows from Aumann (1965, Corollary 5.2) that $\varOmega ^{\mathcal {G}}$ is upper hemicontinuous.

Therefore, Kakutani’s Fixed Point Theorem implies that the set of fixed points of $\varPhi _{{\mathcal {G}}}$ is non-empty and compact. Note that $(f^*,a^*)$ is a Cournot-Nash equilibrium of ${\mathcal {G}}$ if and only if $(m^*,a^*)\in \widehat{M}\times \widehat{\mathcal {F}}^2$ is a fixed point of $\varPhi _{{\mathcal {G}}}$, where $m^*=\int _{T_{1}}H(t,f^{*}(t))d\mu$.³²

Claim C

There exists $(\overline{m},\overline{a}) \in U_1$ such that $(\overline{m},\overline{a}) \in \varLambda (G(\overline{m},\overline{a}))$.

Proof

Given a compact, convex, and non-empty set $\tilde{A}_1\subset U_1$, let $\varTheta : \tilde{A}_1 \times \tilde{A}_1 \twoheadrightarrow \tilde{A}_1 \times \tilde{A}_1$ be the correspondence defined by $\varTheta ((m_1,a_1), (m_2,a_2))=\left( \varPhi _{G(m_1,a_1)}(m_2,a_2) \cap \tilde{A}_1\right) \times \{(m_1,a_1)\}.$ If the set-valued map $\varTheta _1: \tilde{A}_1 \times \tilde{A}_1 \twoheadrightarrow \tilde{A}_1$ given by $\varTheta _1((m_1,a_1), (m_2,a_2))=\varPhi _{G(m_1,a_1)}(m_2,a_2) \cap \tilde{A}_1$ has closed graph, then $\varTheta$ is upper hemicontinuous and has non-empty, compact, and convex values. Thus, applying Kakutani’s Fixed Point Theorem, we could find $(\overline{m},\overline{a}) \in \tilde{A}_1 \subset U_1$ such that $(\overline{m},\overline{a}) \in \varLambda (G(\overline{m},\overline{a}))$.

Therefore, to prove the claim, it is sufficient to ensure that $\varTheta _1$ has closed graph. Fix a sequence $\{(z^n_1,z^n_2, (m^n,a^n))\}_{n \in {\mathbb {N}}}\subset \text{ Graph }(\varTheta _1)$ that converges to $(\tilde{z}_1, \tilde{z}_2, (\tilde{m},\tilde{a}))$. We aim to guarantee that $(\tilde{m},\tilde{a}) \in \varTheta _1(\tilde{z}_1, \tilde{z}_2)$.

For convenience of notations, assume that ${\mathcal {G}}_i={\mathcal {G}}_i((K^i_{t},\varGamma ^i _{t},u^i_{t})_{t\in T_1 \cup T_2}),\,\forall i \in \{1,2\}$. Given $t \in T_2$, let $\gamma _t: (\widehat{M} \times {\widehat{\mathcal {F}}}^2_{-t})\times \tilde{A}_1 \twoheadrightarrow \widehat{K}_t$ be the correspondence defined by

$$\begin{aligned} \gamma _t((m, a_{-t}), z )=\mathop {\hbox {argmax}}\limits _{x \in \varPsi ((m,a_{-t}),z) } v_t(x,(m,a_{-t}),z), \end{aligned}$$

where³³

$$\begin{aligned}&\varPsi ((m,a_{-t}),z)=\lambda (z)\varGamma _t^1(m, a_{-t})+ (1-\lambda (z)) \varGamma _t^2(m,a_{-t}), \\&v_t(x,(m,a_{-t}),z)= \lambda (z) u_t^1(m,x,a_{-t})+ (1-\lambda (z)) u_t^2(m,x,a_{-t}). \end{aligned}$$

It follows that $\gamma _t$ is upper hemicountinuous with non-empty and compact values.$^{33}$ Therefore, the correspondence $\gamma : (\widehat{M} \times {\widehat{\mathcal {F}}}^2)\times \tilde{A}_1 \twoheadrightarrow \Pi _{t\in T_2 \,}\widehat{K}_t$ given by $\gamma ((m,a),z_2)=\prod _{t \in T_2} \gamma _t((m,a_{-t}), z)$ is upper hemicontinuous with compact and non-empty values. In particular, $\gamma$ has closed graph.

Since $\{(z^n_1,z^n_2,a^n)\}_{n \in {\mathbb {N}}}\subset \text{ Graph }(\gamma )$, it follows that $\tilde{a} \in \gamma (\tilde{z}_1,\tilde{z}_2)$.

On the other hand, for each $n \in {\mathbb {N}}$, there exists $f_n:T_1 \rightarrow \widehat{K}$ such that, $m_n=\int _{T_1} H(t,f_n(t))d\mu$ and, for almost all $t \in T_1, f_n(t) \in \xi _t(z^n_1,z^n_2):=\mathop {\hbox {argmax}}\nolimits _{x \in \varPsi (z^n_1,z^n_2) } v_t(x,z^n_1,z^n_2),$ where we use notations analogous to those described above. Note that, for all $t \in T_1, \xi _t$, it has a closed graph.

Since $m^n \rightarrow \tilde{m}$, analogous arguments to those made in Theorem 1 (Claim A) ensure that, applying the multidimensional Fatou’s Lemma [see Hildenbrand (1974, page 69)], there exists a full measure set $\dot{T}_1\subseteq T_1$ and a function $\overline{f}:T_1 \rightarrow \widehat{K}$ such that,

• (i) For any $t \in \dot{T}_1$, there is a subsequence of $\{f_n(t)\}_{n \in {\mathbb {N}}}$ that converges to $\overline{f}(t)$;

• (ii) For any $t \in T_1 {\setminus } \dot{T}_1, \overline{f}(t)\in \xi _t(\tilde{z}_1,\tilde{z}_2)$;

• (iii) $\tilde{m}=\int _{T_1} H(t, \overline{f}(t)) d\mu .$

Since for any $t\in \dot{T}_1$ the correspondence $\xi _t$ is closed, it follows from item (i) that $\overline{f}(t) \in \xi _t(\tilde{z}_1,\tilde{z}_2)$. Items (ii) and (iii) jointly with the fact that $\tilde{a} \in \gamma (\tilde{z}_1,\tilde{z}_2)$ imply that $(\tilde{m }, \tilde{a}) \in \varTheta _1(\tilde{z}_1, \tilde{z}_2)$. $\square$

Since $(\overline{m},\overline{a}) \in U_1$, it follows that $G(\overline{m},\overline{a})={\mathcal {G}}_1$. Hence, Claim B implies that $\varLambda (G(\overline{m},\overline{a}))\cap U_1 =\emptyset$. This is a contradiction, since both $(\overline{m},\overline{a}) \in U_1$ and $(\overline{m},\overline{a}) \in \varLambda (G(\overline{m},\overline{a}))$. Therefore, the minimal essential set $m({\mathcal {G})}$ is connected.

(iv) If $\varLambda ({\mathcal {G}})$ is finite, then ${\mathcal {G}}$ has at least one essential equilibrium.

Suppose that $\widehat{K}$ is a convex subset of a normed space with a metric induced by a norm. It follows from (iii) that for every ${\mathcal {G}}\in {\mathbb {G}}'$ there is a minimal essential set of $\varLambda ({\mathcal {G}})$ that is connected. On the other hand, as $\varLambda ({\mathcal {G}})$ is finite, minimal essential sets are singletons. $\square$

Proof of Theorem 4

Given ${\mathcal {X}}\in {\mathbb {X}}$, the ${\mathcal {T}}$-essential subsets of $\varLambda (\kappa ({\mathcal {X}}))$ are stable.

It follows from Definition 4 that it suffices to guarantee that minimal essential sets are stable in the sense of Definition 8. Let $\varLambda _m({\mathcal {T}}, \mathcal {X})$ be the collection of minimal ${\mathcal {T}}$-essential subsets of $\varLambda (\kappa ({\mathcal {X}}))$.

By contradiction, assume that there is $A\in \varLambda _m({\mathcal {T}}, \mathcal {X})$ and $\epsilon _0>0$ such that, for any $\delta >0$, there is $\mathcal {X}_\delta \in \mathbb {X}$ with $\tau (\mathcal {X},\mathcal {X}_\delta )<\delta$ and $A'\cap B[\epsilon _0, A]^c \ne \emptyset ,\,\,\forall A'\in \varLambda _m({\mathcal {T}}, \mathcal {X}_\delta )$, where $B[\epsilon _0, A]^c:= (\widehat{M} \times \widehat{\mathcal {F}}^2) {\setminus } B[\epsilon _0, A]$. Since $A$ is ${\mathcal {T}}$-essential, there is $\delta _0>0$ such that, for any $\mathcal {X}'\in \mathbb {X}$ with $\tau (\mathcal {X},\mathcal {X}')<\delta _0$, we have that $\varLambda (\kappa ({\mathcal {X}}')) \cap C(\epsilon _0, A)\ne \emptyset$, where $C(\epsilon _0,A)=\{(m,a)\in \widehat{M}\times \widehat{\mathcal {F}}^2:\,\inf \limits _{(m',a')\in A} \widehat{\sigma }((m,a),(m',a'))<\epsilon \}.$ It follows that $\varLambda (\kappa ({\mathcal {X}}_{\delta _0})) \cap B[\epsilon _0, A]$ is a non-empty and closed set contained in $B[\epsilon _0, A]$. Therefore, $\varLambda (\kappa ({\mathcal {X}}_{\delta _0})) \cap B[\epsilon _0, A]$ is not an essential subset of $\varLambda (\kappa ({\mathcal {X}}_{\delta _0}))$.

Hence, there exists $\epsilon _1>0$ such that, for any $n\in {\mathbb {N}}$, there is ${\mathcal {X}}_n \in {\mathbb {X}}$ with $\tau (\mathcal {X}_{\delta _0},\mathcal {X}_n)<\frac{\delta _1}{n}$ and $C(\epsilon _1, \varLambda (\kappa ({\mathcal {X}}_{\delta _0})) \cap B[\epsilon _0, A]) \cap \varLambda (\kappa ({\mathcal {X}}_n))=\emptyset$, where $\delta _1>0$ is chosen in such form that, for any ${\mathcal {X}}''\in {\mathbb {X}}$, if $\tau ({\mathcal {X}}_{\delta _0}, {\mathcal {X}}'')<\delta _1$, then $\tau ({\mathcal {X}}, {\mathcal {X}}'')<\delta _0$. The last property ensures that $\tau ({\mathcal {X}}, {\mathcal {X}}_n)<\delta _0$ for any $n \in {\mathbb {N}}$, which implies that $\varLambda (\kappa ({\mathcal {X}}_n)) \cap C(\epsilon _0, A)$ is non-empty. Take a sequence $\{(m_n,a_n)\}_{n\in {\mathbb {N}}}$ such that $(m_n,a_n) \in \varLambda (\kappa ({\mathcal {X}}_n)) \cap C(\epsilon _0, A),\,\forall n \in {\mathbb {N}}$. Without the loss of generality, there is $(m_0,a_0)\in B[\epsilon _0,A]$ such that $(m_n,a_n)\rightarrow _{n} (m_0,a_0)$. The upper hemicontinuity of $(\varLambda \circ \kappa )$ ensures that $(m_0,a_0) \in \varLambda (\kappa ({\mathcal {X}}_{\delta _0}))$, that is, $(m_0,a_0) \in \varLambda (\kappa ({\mathcal {X}}_{\delta _0}))\cap B[\epsilon _0,A].$ However, as for any $n \in {\mathbb {N}}$, we have that $(m_n,a_n) \in \varLambda (\kappa ({\mathcal {X}}_n))$ and $C(\epsilon _1, \varLambda (\kappa ({\mathcal {X}}_{\delta _0})) \cap B[\epsilon _0, A]) \cap \varLambda (\kappa ({\mathcal {X}}_n))=\emptyset$, it follows that $(m_n,a_n)\notin C(\epsilon _1, \varLambda (\kappa ({\mathcal {X}}_{\delta _0})) \cap B[\epsilon _0, A]),\,\forall n\in {\mathbb {N}}$. Thus, $(m_0,a_0) \notin \varLambda (\kappa ({\mathcal {X}}_{\delta _0}))\cap B[\epsilon _0,A],$ which is a contradiction.

If $\mathbb {X}$ is a convex subset of a normed space and $\tau$ is induced by a norm, then for each ${\mathcal {X}}\in {\mathbb {X}}$, the ${\mathcal {T}}$-essential components of $\varLambda (\kappa ({\mathcal {X}}))$ are strongly stable.

Since $\widehat{M}\subset {\mathbb {R}}^m$ is compact and $\widehat{K}_t$, with $t\in T_2$, are compact subsets of normed spaces with metrics induced by norms, it follows that $\varLambda (\kappa ({\mathcal {X}}))\subseteq \widehat{M} \times \widehat{\mathcal {F}}^2$ is a locally connected and compact space. Therefore, $\varLambda (\kappa ({\mathcal {X}})$ as a finite number of connected components.³⁴ For this reason, given a ${\mathcal {T}}$-essential component $A\subseteq \varLambda (\kappa (\mathcal {X}))$, there exists $\pi >0$ such that $B[\pi ,A ]\cap B[\pi ,\varLambda (\kappa (\mathcal {X})){\setminus } A]=\emptyset$.

Furthermore, it follows from the proof of Theorem 2(i) that there is $A_m\in \varLambda _m({\mathcal {T}}, \mathcal {X})$ such that $A_m \subseteq A$. By the previous item, for each $\epsilon >0$, there is $\delta _1>0$ such that, given $\mathcal {X}'\in \mathbb {X}$ with $\tau (\mathcal {X},\mathcal {X}')<\delta _1$, there exists $A'_m\in \varLambda _m({\mathcal {T}}, \mathcal {X}')$ for which $A'_m\subseteq B[\epsilon , A_m]\subseteq B[\epsilon , A]$. Since $\mathbb {X}$ is a convex subset of a normed space and $\tau$ is induced by a norm, it follows from Theorem 3(iv) that minimal essential sets are connected, and therefore, following analogous arguments to those made in the proof of Theorem 2(ii), we can ensure that for any $\mathcal {X}'\in \mathbb {X}$ with $\tau (\mathcal {X},\mathcal {X}')<\delta _1$, there is an essential component $A' \in \varLambda _c({\mathcal {T}},{\mathcal {X}}')$ which contains $A'_m$, where $\varLambda _c({\mathcal {T}},{\mathcal {X}}')$ is the set of ${\mathcal {T}}$-essential components of $\varLambda (\kappa ({\mathcal {X}}'))$. We want to prove that, for ${\mathcal {X}}'$ closely enough to ${\mathcal {X}}, A' \subseteq B[\epsilon , A].$

Since the correspondence $\varLambda \circ \kappa$ is upper hemicontinuous, there is $\delta _2>0$ such that for any $\mathcal {X}'\in \mathbb {X}$ with $\tau (\mathcal {X},\mathcal {X}')<\delta _2$ we have that $\varLambda (\kappa ({\mathcal {X}}'))\subset C(\epsilon , \varLambda (\kappa ({\mathcal {X}}))) \subset B[\epsilon , A] \cup B[\epsilon , \varLambda (\kappa ({\mathcal {X}})){\setminus } A]$.

Note that $\varLambda (\kappa ({\mathcal {X}})){\setminus } A$ is a compact set.³⁵ Let $\delta =\min \{\delta _0, \delta _1\}$ and fix $\mathcal {X}'\in \mathbb {X}$ with $\tau (\mathcal {X},\mathcal {X}')<\delta$. If $A'\cap B[\epsilon , A]^c\ne \emptyset$, then $A' \cap B[\epsilon , \varLambda (\kappa ({\mathcal {X}})){\setminus } A]\ne \emptyset$ and $A' \cap B[\epsilon , A]\ne \emptyset$. In addition, when $\epsilon <\pi$, it follows that $B[\epsilon , A] \bigcap B[\epsilon , \varLambda (\kappa ({\mathcal {X}})){\setminus } A]=\emptyset$. Since $A$ and $\varLambda (\kappa ({\mathcal {X}})){\setminus } A$ are compact sets, both $B[\epsilon , A]$ and $B[\epsilon , \varLambda (\kappa ({\mathcal {X}})){\setminus } A]$ are closed. Thus, we obtain a partition of the connected set $A'$ into two non-empty and disjoint closed sets, $A' \cap B[\epsilon , \varLambda (\kappa ({\mathcal {X}})){\setminus } A]$ and $A' \cap B[\epsilon , A]$, which is a contradiction. Therefore, for any $\mathcal {X}'\in \mathbb {X}$ with $\tau (\mathcal {X},\mathcal {X}')<\delta$, we have that $A'\subset B[\epsilon , A]$. $\square$

Lemma 2

Let ${\mathcal {G}}={\mathcal {G}}( (K_{t},\varGamma _{t},u_{t})_{t\in T_1 \cup T_2})$ be a generalized payoff secure and upper semicontinuous game. Then, ${\mathcal {G}}$ satisfies continuous security.

Proof

Given $(m,a)\notin \varLambda ({\mathcal {G}})$, generalized payoff security guarantees that, for any $\epsilon >0$, there exists $(U^\epsilon , (\varphi ^\epsilon _t)_{t \in T_1\cup T_2}, \alpha ^\epsilon )$ satisfying item (i) of Definition 10. Thus, to guarantee that ${\mathcal {G}}$ is continuous secure, it is sufficient to prove that $(U^\epsilon , \alpha ^\epsilon )$ satisfies Definition 10 (ii) for some $\epsilon >0$. Suppose, by contradiction, that for any $n \in {\mathbb {N}}$, there is $(f_n,a_n)\in \widehat{\mathcal {F}}^1\times \widehat{\mathcal {F}}^2$ satisfying,

• (a) $(m(f_n),a_n)\in U^{\frac{1}{n}}$,

• (b) $f_n(t) \in \varGamma _t(m(f_n),a_n)$ for almost all $t \in T_1$,

• (c) $a_{n,t} \in \varGamma _t(m(f_n), a_{n,-t})$ for all $t \in T_2$,

• (d) for almost all $t \in T_1, u_t(f_n(t), m(f_n),a_n)\ge \alpha ^{\frac{1}{n}}(t)$,

• (e) for any $t \in T_2, u_t(m(f_n), a_{n,t}, a_{n, -t})\ge \alpha ^{\frac{1}{n}}(t)$.

Since we can assume that $\bigcap _n U^{\frac{1}{n}}=\{(m,a)\},$ it follows from (a) that $(m(f_n),a_n) \rightarrow _n (m,a)$. Conditions (b)-(c) guarantee, by using analogous arguments to those made in the proof of Theorem 1 (Claim A), that we can find a strategy profile $f\in {\mathcal {F}}^1((K_t)_{t\in T_1})$ such that $m=m(f)$ and there is a full measure set $T_1'\subseteq T_1$ such that $f(t)\in L_S(f_n(t)),\,\forall t \in T'_1$.

In addition, as correspondences of admissible strategies have closed graph, it follows that (i) for almost all $t \in T_1, f(t)\in \varGamma _t(m(f),a)$; (ii) for all $k\in T_2, a_k \in \varGamma _k(m(f),a_{-k}).$

Hence, as $(m,a)\notin \varLambda ({\mathcal {G}})$, there is a non-negligible set of agents that are sub-optimizing, i.e., there exists $\delta >0$ such that either for a positive measure set $T''_1\subseteq T_1$,

$$\begin{aligned} u_t(f(t), m, a) \, + \, \delta \, < \sup _{x \in \varGamma _t(m,a)} u_t(x,m,a),\quad \forall t \in T''_1, \end{aligned}$$

or for some $t\in T_2$,

$$\begin{aligned} u_t(m, a_{t}, a_{-t}) \,+ \, \delta \, < \sup _{x \in \varGamma _t(m,a_{-t})} u_t(m,x,a_{-t}). \end{aligned}$$

This last condition implies that

$$\begin{aligned} \sum _{t \in T_2} u_t(m, a_{t}, a_{-t}) \,+ \, \delta \, < \sum _{t \in T_2} \,\sup _{x \in \varGamma _t(m,a_{-t})} u_t(m,x,a_{-t}). \end{aligned}$$

Since ${\mathcal {G}}$ is upper semicontinuous and $(m(f_n),a_n) \rightarrow _n (m,a)$, it follows from the definition of $f$ that for $n\in {\mathbb {N}}$ large enough, we have that either for all $t\in T'_1 \cap T''_1$,

$$\begin{aligned} u_t(f_n(t), m(f_n), a_n) + \, 0.5\,\delta \, < \sup _{x \in \varGamma _t(m,a)} u_t(x,m,a), \end{aligned}$$

(1)

or

$$\begin{aligned} \sum _{t \in T_2}\,\,u_t(m(f_n), a_{n,t}, a_{n,-t}) + \, 0.5\, \delta \,< \sum _{t \in T_2}\,\,\sup _{x \in \varGamma _t(m,a_{-t})} u_t(m,x,a_{-t}). \end{aligned}$$

The later inequality implies that there is $t \in T_2$ such that

$$\begin{aligned} u_t(m(f_n), a_{n,t}, a_{n,-t}) + \, \frac{0.5\, \delta }{\# T_2}\,< \sup _{x \in \varGamma _t(m,a_{-t})} u_t(m,x,a_{-t}). \end{aligned}$$

(2)

On the other hand, it follows from conditions (d)–(e) above and Definition 11(ii) that for $n\in {\mathbb {N}}$ large enough, there exists $T_n\subseteq T_1$ with $\mu (T_n)\ge \mu (T_1)-\frac{1}{n}$ such that, for any $t \in T_n$,

$$\begin{aligned} u_t(f_n(t), m(f_n), a_n) \ge \sup _{x \in \varGamma _t(m,a)} u_t(x,m,a) - \frac{1}{n}, \end{aligned}$$

and for every atomic player $t\in T_2$,

$$\begin{aligned} u_t(m(f_n), a_{n,t}, a_{n,-t}) \ge \sup _{x \in \varGamma _t(m,a_{-t})} u_t(m,x,a_{-t})- \frac{1}{n}. \end{aligned}$$

Thus, $\underline{\lim }_n u_t(f_n(t), m(f_n), a_n) \ge \sup _{x \in \varGamma _t(m,a)} u_t(x,m,a)$ for almost all non-atomic player $t \in T_1$. Also, for each atomic player $t\in T_2$, we have that $\underline{\lim }_n u_t(m(f_n), a_{n,t}, a_{n,-t}) \ge \sup _{x \in \varGamma _t(m,a_{-t})} u_t(m,x,a_{-t})$. Hence, taking the lower limit in (1) and (2), we obtain a contradiction. $\square$

Proposition 2

$({\mathbb {G}}_d, \rho )$ is a complete metric space.

Proof

Since $(K_t, \varGamma _t)_{t \in T_1 \cup T_2}$ does not change, $({\mathbb {G}}_d, \rho )$ can be considered as a subset of the space of bounded functions ${\mathcal {B}}:={\mathcal {U}}(T_1 \times \widehat{K}\times \widehat{ M}\times \widehat{{\mathcal {F}}}^2) \times \prod _{t \in T_2} {\mathcal {U}}_t(\widehat{M}\times \widehat{{\mathcal {F} }}^2)$, where for any $t\in T_2$, the set ${\mathcal {U}}_t(\widehat{M}\times \widehat{{\mathcal {F} }}^2)$ is the collection of bounded functions $(m,a_t,a_{-t}) \rightarrow u_t(m,a_t,a_{-t})$ which are quasi-concave on $a_t$. Note that $({\mathcal {B}}, \rho )$ is a complete metric space and, therefore, it is sufficient to ensure that ${\mathbb {G}}_d$ is a closed subset of ${\mathcal {B}}$.

Fix a sequence $\{{\mathcal {G}}_n\}_{n \in {\mathbb {N}}} \subset {\mathbb {G}}_d$, with ${\mathcal {G}}_n = {\mathcal {G}}_n( (u^n_{t})_{t\in T_1 \cup T_2})$ for any $n \in {\mathbb {N}}$, which converges to $\overline{\mathcal {G}}={\overline{\mathcal {G}}} ( (\overline{u}_{t})_{t\in T_1 \cup T_2}) \in {\mathcal {B}}.$ We want to prove that $\overline{\mathcal {G}} \in {\mathbb {G}}_d$.

Claim. $\overline{\mathcal {G}}$ is generalized payoff secure.

Given $(m,a)\in \widehat{M}\times \widehat{\mathcal {F}}^2$ and $\epsilon >0$, generalized payoff security of ${\mathcal {G}}_n$ at $((m,a), 0.5\, \epsilon )$ implies that there exists $(U^{n} , (\varphi ^{n}_t)_{t \in T_1\cup T_2}, \alpha ^{n})$ satisfying the requirements of Definition 11.

Thus, for $n$ large enough, for almost all $t \in T_1$, for all $k \in T_2$, and for every $(m',a')\in U^n$, we have that,

$$\begin{aligned} \overline{u}_t(x,m',a')&> \,\,u^n_t(x,m',a') - 0.25\, \epsilon&\ge \,\,\alpha ^{n}(t) - \,0.25\,\epsilon ,\,\,\,\,\,\forall x\in \varphi ^{n}_t(m',a');\\ \overline{u}_k(m',x, a'_{-t})&> \,\,u^n_k (m'. x, a'_{-t})- 0.25\, \epsilon&\ge \,\, \alpha ^{n}(k)-\,0.25\, \epsilon , \,\,\,\,\, \forall x\in \varphi ^{n}_k(m',a'). \end{aligned}$$

Furthermore, Definition 11(ii) ensures that, for $n$ large enough,

$$\begin{aligned} \left( \alpha ^{n}(k)-\,0.25\,\epsilon \right) + \epsilon&= (\alpha ^{n}(k) + 0.5 \epsilon ) + 0,25\,\epsilon \\&\ge \sup _{x \in \varGamma _t(m,a_{-k})}\, u^n_t(m,x,a_{-k}) + 0.25\, \epsilon \\&> \sup _{x \in \varGamma _t(m,a_{-k})}\, \overline{u}_t(m,x,a_{-k}),\quad \forall k \in T_2. \end{aligned}$$

$$\begin{aligned} \mu \left\{ t \in T_1: \left( \alpha ^{n}(t)-\,0.25\,\epsilon \right) + \epsilon \ge \sup _{x \in \varGamma _t(m,a)}\, \overline{u}_t(x,m,a)\right\} \ge \mu (T_1)-0.5\,\epsilon . \end{aligned}$$

Therefore, taking $n$ large enough and choosing $(U^{n} , (\varphi ^{n}_t)_{t \in T_1\cup T_2}, \alpha ^{n}-\,0.25\,\epsilon )$, we ensure that $\overline{\mathcal {G}}$ is generalized payoff secure at $((m,a),\epsilon )$. $\square$

It is a direct consequence of Carbonell-Nicolau (2010, Lemma 1, page 425) that $\overline{\mathcal {G}}$ is upper semicontinuous and atomic players’ objective functions $(\overline{u}_t)_{t \in T_2}$ are quasi-concave. In addition, the same arguments made in the proof of Proposition 1 guarantee that, for every $(m,a)\in \widehat{M}\times \widehat{\mathcal {F}}^2$, the map $(t,x)\in T_1 \times \widehat{K} \rightarrow \overline{u}_t(x,m,a)$ is measurable. This concludes the proof. $\square$

Proof of Theorem 5

Since $({\mathbb {G}}_d,\rho )$ is complete, it follows from the proofs of Theorems 1–4 that it is sufficient to ensure that $\varLambda$ still has a closed graph when its domain is extended to ${\mathbb {G}}_d$.

Let $\{(\mathcal {G}_n,(m_n,a_n))\}_{n \in {\mathbb {N}}} \subset \text{ Graph }(\varLambda )$ such that $(\mathcal {G}_n,(m_n,a_n))\rightarrow (\overline{\mathcal {G}},(\overline{m},\overline{a}))$, where ${\mathcal {G}}_n= {\mathcal {G}}_n((u^n_{t})_{t\in T_1 \cup T_2})$ and $\overline{\mathcal {G}}= \overline{\mathcal {G}}((\overline{u}_{t})_{t\in T_1 \cup T_2})\in \mathbb {G}_d$. We want to prove that $(\overline{m},\overline{a}) \in \varLambda ({\overline{\mathcal {G}}})$.

Since $(m_n,a_n)\in \varLambda ({\mathcal {G}_n})$, there is $f_n\in \widehat{\mathcal {F}}^1$ such that

• (a) the function $g_n: T_1 \rightarrow {\mathbb {R}}^m$ given by $g_n(t)=H(t, f_n(t))$ is measurable and $m_n=m(f_n)$;

• (b) for almost all $t \in T_1$ both $f_n(t)\in \varGamma _t(m_n,a_n)$ and

  $$\begin{aligned} u_t^n(f_n(t), m_n,a_{n})=\sup \limits _{x\in \varGamma _t(m_n, a_n)}{u_t^n(x, m_n,a_{n})}. \end{aligned}$$

Claim A

There exists $\overline{f}\in \widehat{\mathcal {F}}^1$ such that $\overline{m}=\int _{T_1}{H(t,\overline{f}(t))d\mu }$.

Proof

By analogous arguments to those in the proof of Theorem 1 (Claim A), we can find a strategy profile $\overline{f}\in \widehat{\mathcal {F}}^1$ and a full measure set $T^*_1\subseteq T_1$ such that $\overline{m}=m(\overline{f}), \overline{f}(t)\in L_S(f_n(t)),\,\forall t \in T^*_1$. $\square$

Claim B

For almost all $t\in T_1, \overline{f}(t)\in \varGamma _t(\overline{m},\overline{a})$. In addition, for any $t \in T_2, \overline{a}_t \in \varGamma _t(\overline{m},\overline{a}_{-t})$.

Proof

It follows from the proof of Claim A that there is a full measure set $T^*_1\subseteq T_1$ such that $\overline{f}(t)\in L_S(f_n(t)),\,\forall t \in T^*_1$. Thus, the closed graph property of correspondences of admissible strategies ensures that (i) for all $t \in T^*_1, \overline{f}(t)\in \varGamma _t(\overline{m},\overline{a})$; and (ii) for all $t\in T_2, \overline{a}_t \in \varGamma _t(\overline{m},\overline{a}_{-t}).$ $\square$

Claim C

The following properties hold

• (i) For almost all $t \in T_1, \overline{f}(t) \in \mathop {\hbox {argmax}}\nolimits _{x \in \varGamma _t(\overline{m},\overline{a})}\,\,\overline{u}_t(x,\overline{m},\overline{a}).$

• (ii) For any $t\in T_2, \overline{a}_t\in \mathop {\hbox {argmax}}\nolimits _{x\in \varGamma _t(\overline{m},\overline{a}_{-t})}\overline{u}_t(\overline{m},x,\overline{a}_{-t}).$

Proof

Suppose that at least one of the properties (i) and (ii) does not hold, i.e., there is a non-negligible set of agents that are suboptimizing. Since ${\overline{\mathcal {G}}}$ is upper semicontinuous, identical arguments to those made in the proof of Lemma 2 to obtain conditions (1) and (2) imply that there is $\xi >0$ such that,$^{36}$ for $n$ large enough, either

$$\begin{aligned} \overline{u}_t(f_n(t), m_n, a_n) \, + \, \xi \, < \sup _{x \in \varGamma _t(\overline{m},\overline{a})} \overline{u}_t(x,\overline{m},\overline{a}),\quad \forall t \in T^{**}_1\subseteq T_1, \end{aligned}$$

where $T^{**}_1$ is a positive measure set, or there exists $t \in T_2$ such that³⁶

$$\begin{aligned} \overline{u}_t(m_n, a_{n,t}, a_{n,-t}) \, + \, \xi \, < \sup _{x \in \varGamma _t(\overline{m},\overline{a}_{-t})} \overline{u}_t(\overline{m},x,\overline{a}_{-t}). \end{aligned}$$

Thus, as $\rho ({\mathcal {G}}_n, \overline{\mathcal {G}})\rightarrow _n 0$, for $n$ large enough at least one of the following conditions hold:

$$\begin{aligned}&\displaystyle u^n_t(f_n(t), m_n, a_n) \, + \, 0.5\,\xi <\sup _{x \in \varGamma _t(\overline{m},\overline{a})} \overline{u}_t(x,\overline{m},\overline{a}),\quad \forall t \in T^{*}_1 \cap T^{**}_1;&\end{aligned}$$

(3)

$$\begin{aligned}&\displaystyle u^n_t(m_n, a_{n,t}, a_{n,-t}) \, + \, 0.5\,\xi < \sup _{x \in \varGamma _t(\overline{m},\overline{a}_{-t})} \overline{u}_t(\overline{m},x,\overline{a}_{-t}).&\end{aligned}$$

(4)

On the other hand, since $\overline{\mathcal {G}}$ is a generalized payoff secure large game, for every $\epsilon >0$, there exists $(U^\epsilon , (\varphi ^\epsilon _t)_{t \in T_1\cup T_2}, \alpha ^\epsilon )$ satisfying Definition 11. In particular, as $(m_n,a_n)\rightarrow _n (\overline{m},\overline{a})$, there exists a set $T_\epsilon \subseteq T_1$ with $\mu (T_\epsilon )\ge \mu (T_1)-\epsilon$ such that, for any $n$ large enough, we have $(m_n,a_n) \in U^\epsilon$ and the following properties hold for every $(t,k)\in T_\epsilon \times T_2$:

$$\begin{aligned} \sup _{x \in \varGamma _t(m_n,a_n)} \overline{u}_t (x,m_n,a_n)&\ge \sup _{x \in \varphi ^\epsilon _t(m_n,a_n)} \overline{u}_t (x,m_n,a_n)\nonumber \\&\ge \alpha ^\epsilon (t) \,\,\ge \,\, \sup _{x \in \varGamma _t(\overline{m},\overline{a})} \overline{u}_t(x,\overline{m},\overline{a}) - \epsilon ; \end{aligned}$$

(5)

$$\begin{aligned} \sup _{x \in \varGamma _k(m_n,a_{n,-k})} \overline{u}_k (m_n,x, a_{n,-k})&\ge \sup _{x \in \varphi ^\epsilon _k(m_n,a_{n,-k})} \overline{u}_k (m_n,x, a_{n,-k}) \nonumber \\&\ge \alpha ^\epsilon (k) \,\,\ge \,\, \sup _{x \in \varGamma _k(\overline{m},\overline{a}_{-k})} \overline{u}_k(\overline{m},x, \overline{a}_{-k}) - \epsilon . \end{aligned}$$

(6)

As objective functions are bounded, the uniform convergence of ${\mathcal {G}}_n$ to $\overline{\mathcal {G}}$ ensures that, for $n$ large enough and for each $(t,k) \in T^*_1\times T_2$,

$$\begin{aligned} u^n_t (f_n(t), m_n, a_n) + \epsilon&= \sup _{x \in \varGamma _t(m_n,a_n)} u^n_t (x,m_n,a_n) +\epsilon \nonumber \\&\ge \sup _{x \in \varGamma _t(m_n,a_n)} \overline{u}_t (x,m_n,a_n); \end{aligned}$$

(7)

$$\begin{aligned} u^n_k (m_n, a_{n,k}, a_{n,-k}) + \epsilon&= \sup _{x \in \varGamma _k(m_n,a_{n,-k})} u^n_k (m_n,x, a_{n,-k}) +\epsilon \nonumber \\&\ge \sup _{x \in \varGamma _k(m_n,a_{n,-k})} \overline{u}_k (m_n,x, a_{n,-k}). \end{aligned}$$

(8)

Therefore, for every $\epsilon >0$ and for each $(t,k) \in (T_1^* \cap T_\epsilon ) \times T_2$, taking the lower limit as $n$ goes to infinity on inequalities (7)–(8) it follows that,

$$\begin{aligned} \underline{\lim }_n\, u^n_t (f_n(t), m_n, a_n) + \epsilon&\ge \sup _{x \in \varGamma _t(\overline{m},\overline{a})} \overline{u}_t(x,\overline{m},\overline{a}) - \epsilon , \end{aligned}$$

(9)

$$\begin{aligned} \underline{\lim }_n\, u^n_k (m_n, a_{n,k}, a_{n,-k}) + \epsilon&\ge \sup _{x \in \varGamma _k(\overline{m},\overline{a}_{-k})} \overline{u}_k(\overline{m},x, \overline{a}_{-k}) - \epsilon . \end{aligned}$$

(10)

Taking the limit as $\epsilon$ goes to zero on (9–10) and taking the lower limit as $n$ goes to infinity on (3–4), we obtain a contradiction.

It follows from Claims A and C that $(\overline{m},\overline{a}) \in \varLambda ({\overline{\mathcal {G}}})$. $\square$