Nonatomic game with general preferences over returns

Abstract

We study nonatomic games in which players’ choices are guided by general preferences. Rather than ones over actions while also under influences of player-action profiles, we let the preferences be over returns received by individual players and let the returns be then linked to all players’ actions. Our modeling choice has rendered otherwise standard analysis quite fruitful. Not only can we establish equilibrium existence results, but we can also derive the upper hemi-continuity of equilibrium-environment sets with respect to the return function and players’ preference profile. Advances concerning pure equilibria can also be made on a framework involving a rich set of players, cruder traits, and an externality midway between semi-anonymity and anonymity.

Appendix

1.1 Technical details of Sect. 4

Proof of Proposition 1

For any player $t\in T$ and closed action subset $A'\subseteq A$. we have from (11) that

$$\begin{aligned} \left[ \mathbb B(\delta |\rho ,\psi )\right] (t)\cap A'=\left\{ a\in A':\;\rho (t,A,\delta )\times \rho (t,\{a\},\delta )\subseteq \psi (t)\right\} . \end{aligned}$$

(28)

Fix some closed $A'$. Suppose a sequence $t_n$ in T converges to t and $[{\mathbb {B}}(\delta |\rho ,\psi )](t_n)\cap A'\ne \emptyset$ for each n. Then by (28), for each n there is some $a_n\in A'$ so that

$$\begin{aligned} \rho (t_n,A,\delta )\times \rho (t_n,\{a_n\},\delta )\subseteq \psi (t_n). \end{aligned}$$

Since $A'$ is closed in A while the latter is compact, $A'$ is compact as well. We might as well suppose that $a_n\rightarrow a$ for some $a\in A'$. Also, $t_n\rightarrow t$ and that $\rho \in \mathcal{R}\equiv {\mathscr {C}}(T\times A\times \mathcal{D},R)$ is uniformly continuous in (t, a) because T and A are compact. Hence, for any $\epsilon >0$,

$$\begin{aligned} \rho (t,A,\delta )\times \rho (t,\{a\},\delta )\subseteq (\psi (t_n))^{\epsilon /2}, \end{aligned}$$

as long as n is large enough. Since $\psi \in \mathcal{P}\equiv {\mathscr {C}}(T,\Psi )$, we have $\psi (t_n)\in (\psi (t))^{\epsilon /2}$ when n is large enough. Therefore, $\rho (t,A,\delta )\times \rho (t,\{a\},\delta )\subseteq (\psi (t))^{\epsilon }$. Because $\psi (t)$ is closed,

$$\begin{aligned} \rho (t,A,\delta )\times \rho (t,\{a\},\delta )\subseteq \psi (t). \end{aligned}$$

By (28), the limit point t would satisfy $[\mathbb B(\delta |\rho ,\psi )](t)\cap A'\ne \emptyset$ as well. $\square$

Proof of Proposition 2:

Let $(t_n,a_n,\delta _n,\rho _n,\psi _n)_{n=1,2,\ldots }$ be a sequence in $T\times A\times \mathcal{D}\times \mathcal{R}\times \mathcal{P}$ and let $(t,a,\delta ,\rho ,\psi )$ be a point of the aforementioned product space. Suppose that $(t_n,a_n)\in {\mathbb {B}}(\delta _n|\rho _n,\psi _n)$ for each $n=1,2,\ldots$ and also that $t_n\rightarrow t$, $a_n\rightarrow a$, $\delta _n\rightarrow \delta$, $\rho _n\rightarrow \rho$, and $\psi _n\rightarrow \psi$.

Fix an $\epsilon >0$. By $d_\mathcal{R}(\rho _n,\rho )\rightarrow 0^+$ and $d_\mathcal{R}$’s definition by (14),

$$\begin{aligned} d_R(\rho _n(t_m,a',\delta _m),\rho (t_m,a',\delta _m))\vee d_R(\rho _n(t_m,a_m,\delta _m),\rho (t_m,a_m,\delta _m))<\frac{\epsilon }{4} \end{aligned}$$

for any $a'$ and m would occur when n is large enough. By $\rho$’s membership to $\mathcal{R}\equiv {\mathscr {C}}(T\times A\times \mathcal{D},R)$ whose members are uniformly continuous due to the compactness of $T\times A\times \mathcal{D}$ and by the convergences of $t_n$ to t, $a_n$ to a, and $\delta _n$ to $\delta$, on the other hand,

$$\begin{aligned} d_R(\rho (t_n,a',\delta _n),\rho (t,a',\delta ))\vee d_R(\rho (t_n,a_n,\delta _n),\rho (t,a,\delta ))<\frac{\epsilon }{4} \end{aligned}$$

would occur when n is large enough. Hence for n sufficiently large,

$$\begin{aligned} \rho (t,A,\delta )\subseteq \left( \rho _n(t_n,A,\delta _n)\right) ^{\epsilon /2}, \end{aligned}$$

and

$$\begin{aligned} d_R(\rho (t,a,\delta ),\rho _n(t_n,a_n,\delta _n))<\frac{\epsilon }{2}. \end{aligned}$$

However, by (11) and the hypothesis that $(t_n,a_n)\in {\mathbb {B}}(\delta _n|\rho _n,\psi _n)$ for any n,

$$\begin{aligned} \rho _n(t_n,A,\delta _n)\times \rho _n(t_n,\{a_n\},\delta _n)\subseteq \psi _n(t_n),\quad \forall \,\,n=1,2,\ldots . \end{aligned}$$

In view of (3), we have from the above three that, for a large enough n,

$$\begin{aligned} \rho (t,A,\delta )\times \rho (t,\{a\},\delta )\subseteq (\psi _n(t_n))^{\epsilon /2}. \end{aligned}$$

By $d_\mathcal{P}(\psi _n,\psi )\rightarrow 0^+$ and $d_\mathcal{P}$’s definition by (15), $d_{\Phi }(\psi _n(t_m),\psi (t_m))<\epsilon /4$ for any m would occur when n is large enough. By $\psi$’s membership to $\mathcal{P}$ and the convergence of $t_n$ to t, on the other hand, $d_{\Phi }(\psi (t_n),\psi (t))<\epsilon /4$ would occur when n is large enough. Hence, $d_{\Phi }(\psi _n(t_n),\psi (t))<\epsilon /2$ would occur when n is large enough. By (2), this implies that

$$\begin{aligned} \psi _n(t_n)\subseteq (\psi (t))^{\epsilon /2}, \end{aligned}$$

when n is large enough. When we combine the above two, we would certainly get

$$\begin{aligned} \rho (t,A,\delta )\times \rho (t,\{a\},\delta )\subseteq (\psi (t))^\epsilon . \end{aligned}$$

Since $\psi (t)$ is closed and $\epsilon$ is arbitrary, we can conclude from the above that $\rho (t,A,\delta )\times \rho (t,\{a\},\delta )\subseteq \psi (t)$ and hence by (11), $(t,a)\in {\mathbb {B}}(\delta |\rho ,\psi )$. Thus, $\mathbb B(\cdot |\cdot ,\cdot )$ is closed and hence closed-valued. It is also upper hemi-continuous and compact-valued as $T\times A$ is compact. $\square$

Proof of Proposition 3:

Recall that ${\mathbb {B}}(\delta |\rho ,\psi )$ has a nonempty intersection with any $\{t\}\times A$. It can thus be treated as a correspondence from T to A. By Proposition 1, for any closed subset $A'$ of A,

$$\begin{aligned} \left\{ t\in T:\;\left[ {\mathbb {B}}(\delta |\rho ,\psi )\right] (t)\cap A'\ne \emptyset \right\} \text{ is } \text{ closed } \text{ and } \text{ hence } \text{ a } \text{ member } \text{ of } {\mathscr {B}}(T). \end{aligned}$$

Proposition 2 has shown ${\mathbb {B}}(\delta |\rho ,\psi )$ to be closed in $T\times A$. Hence, each section $\left[ \mathbb B(\delta |\rho ,\psi )\right] (t)$ is closed and therefore compact in the compact set A. By Lemma 18.2 of Aliprantis and Border (1999), the above would guarantee that, for any open subset $A'$ of A,

$$\begin{aligned} \left\{ t\in T:\;\left[ {\mathbb {B}}(\delta |\rho ,\psi )\right] (t)\cap A'\ne \emptyset \right\} \in {\mathscr {B}}(T). \end{aligned}$$

By the Kuratowski–Ryll-Nardzewski measurable selection theorem—see, e.g., 18.13 of Aliprantis and Border (1999), there would be a measurable function $b^*(\delta |\rho ,\psi ):T\rightarrow A$ so that $[b^*(\delta |\rho ,\psi )](t)\in [{\mathbb {B}}(\delta |\rho ,\psi )](t)$ for any $t\in T$.

Meanwhile, $b^*(\delta |\rho ,\psi )$’s graph $B^*(\delta |\rho ,\psi )$ would belong to ${\mathscr {B}}(T\times A)$ which is also the product $\sigma$-field ${\mathscr {B}}(T)\otimes {\mathscr {B}}(A)$; see, e.g., Buckley (1974). We can then define a probability measure $\delta '(\delta |\rho ,\psi )\in {\mathscr {P}}(T\times A)$ so that for any $T'\in {\mathscr {B}}(T)$ and $A'\in {\mathscr {B}}(A)$,

$$\begin{aligned} \left[ \delta '(\delta |\rho ,\psi )\right] (T'\times A')= \theta \left( \left\{ t\in T':\;[b^*(\delta |\rho ,\psi )](t)\in A'\right\} \right) . \end{aligned}$$

Clearly, $\delta '(\delta |\rho ,\psi )|_T=\theta$ and hence $\delta '(\delta |\rho ,\psi )\in \mathcal{D}|_T(\theta )$ by (9). Also,

$$\begin{aligned}{}[\delta '(\delta |\rho ,\psi )](B^*(\delta |\rho ,\psi ))=1. \end{aligned}$$

By Proposition 2, ${\mathbb {B}}(\delta |\rho ,\psi )$ is closed and hence a member of ${\mathscr {B}}(T\times A)$. As $B^*(\delta |\rho ,\psi )\subseteq {\mathbb {B}}(\delta |\rho ,\psi )$, we have $[\delta '(\delta |\rho ,\psi )]({\mathbb {B}}(\delta |\rho ,\psi ))=1$ as well. In view of (12), these are tantamount to the nonemptiness of ${\mathbb {F}}(\delta |\rho ,\psi )$. As the convex combination of two $\delta '$’s satisfying the requirement of (12) would still be in compliance, the latter is also convex. $\square$

Proof of Proposition 4:

Let $(\delta '_n,\delta _n,\rho _n,\psi _n)_{n=1,2,\ldots }$ be a sequence in $\mathcal{D}|_T(\theta )\times \mathcal{D}|_T(\theta )\times \mathcal{R}\times \mathcal{P}$ and let $(\delta ',\delta ,\rho ,\psi )$ be a point of the aforementioned product space. Suppose that $\delta '_n\in {\mathbb {F}}(\delta _n|\rho _n,\psi _n)$ for each $n=1,2,\ldots$ and also that $\delta '_n\rightarrow \delta '$, $\delta _n\rightarrow \delta$, $\rho _n\rightarrow \rho$, and $\psi _n\rightarrow \psi$.

Fix an $\epsilon >0$. The convergences of $\delta _n$ to $\delta$, $\rho _n$ to $\rho$, and $\psi _n$ to $\psi$, as well as Proposition 2 on the upper hemi-continuity of $\mathbb B(\cdot |\cdot ,\cdot )$ would together lead to that, when n is large enough,

$$\begin{aligned} {\mathbb {B}}(\delta _n|\rho _n,\psi _n)\subseteq \left( \mathbb B(\delta |\rho ,\psi )\right) ^{\epsilon /2}. \end{aligned}$$

By (12) and the hypothesis that $\delta '_n\in \mathbb F(\delta _n|\rho _n,\psi _n)$ for any n,

$$\begin{aligned} \delta '_n({\mathbb {B}}(\delta _n|\rho _n,\psi _n))=1.\forall n=1,2,\ldots . \end{aligned}$$

Putting the above two together, we can get that $\delta '_n((\mathbb B(\delta |\rho ,\psi ))^{\epsilon /2})=1$ for n sufficiently large. But the nature of the Prokhorov metric and the fact that $d_\mathcal{D}(\delta '_n,\delta ')\equiv \pi _{T\times A}(\delta '_n,\delta ')\rightarrow 0^+$ would dictate that, as long as n is large enough,

$$\begin{aligned} \delta '(({\mathbb {B}}(\delta |\rho ,\psi ))^\epsilon )\ge \delta '_n((\mathbb B(\delta |\rho ,\psi ))^{\epsilon /2})-\epsilon =1-\epsilon . \end{aligned}$$

For any $k=1,2,\ldots$, the above would mean that

$$\begin{aligned} \delta '\left( \bigcap _{l=k}^{+\infty }(\mathbb B(\delta |\rho ,\psi ))^{1/l}\right) \ge 1-\frac{1}{k}. \end{aligned}$$

Since ${\mathbb {B}}(\delta |\rho ,\psi )$ is closed by Proposition 2, it is equal to $\bigcap _{l=k}^{+\infty }({\mathbb {B}}(\delta |\rho ,\psi ))^{1/l}$. Therefore, $\delta '({\mathbb {B}}(\delta |\rho ,\psi ))=1$. By (12), this implies $\delta '\in \mathbb F(\delta |\rho ,\psi )$. Thus, ${\mathbb {F}}(\cdot |\cdot ,\cdot )$ is closed and hence closed-valued. It is also upper hemi-continuous and compact-valued as $\mathcal{D}|_T(\theta )$ is compact. $\square$

1.2 Technical details of Sect. 5

Proof of Proposition 5:

Let $(\tilde{\delta }'_n,t_n,\tilde{\delta }_n,{\tilde{\rho }}_n,\psi _n)_{n=1,2,\ldots }$ be a sequence in $\tilde{\mathcal{D}}^{\text{ e }}\times T\times \tilde{\mathcal{D}}\times \tilde{\mathcal{R}}\times \mathcal{P}$ and $({\tilde{\delta }}',t,{\tilde{\delta }},{\tilde{\rho }},\psi )$ a point of the aforementioned product space. Suppose $\tilde{\delta }'_n\in \left[ \tilde{\mathbb J}({\tilde{\delta }}_n|{\tilde{\rho }}_n,\psi _n)\right] (t_n)$ for each $n=1,2,\ldots$ and also $\tilde{\delta }'_n\rightarrow \tilde{\delta }_n$, $t_n\rightarrow t$, $\tilde{\delta }_n\rightarrow \tilde{\delta }$, ${\tilde{\rho }}_n\rightarrow {\tilde{\rho }}$, and $\psi _n\rightarrow \psi$. As $\tilde{\mathcal{D}}^{\text{ e }}$ is finite, we may suppose that all the $\tilde{\delta }'_n$’s and $\tilde{\delta }'$ are some common $e_i$.

Fix an $\epsilon >0$. Using similar arguments as the ones in the proof of Proposition 2, but this time using $d_{\tilde{\mathcal{R}}}$’s definition by (16), we can obtain that, as long as n is large enough,

$$\begin{aligned} {\tilde{\rho }}(t,A,{\tilde{\delta }})\subseteq \left( {\tilde{\rho }}_n(t_n,A,{\tilde{\delta }}_n)\right) ^{\epsilon /2}, \end{aligned}$$

and

$$\begin{aligned} d_R\left( {\tilde{\rho }}(t,a_i,{\tilde{\delta }}),{\tilde{\rho }}_n(t_n,a_i,{\tilde{\delta }}_n)\right) <\frac{\epsilon }{2}. \end{aligned}$$

However, by (17) and the hypothesis that $e_i\in \left[ \tilde{\mathbb J}({\tilde{\delta }}_n|{\tilde{\rho }}_n,\psi _n)\right] (t_n)$ for any n,

$$\begin{aligned} {\tilde{\rho }}_n(t_n,A,{\tilde{\delta }}_n)\times {\tilde{\rho }}_n(t_n,\{a_i\},{\tilde{\delta }}_n)\subseteq \psi _n(t_n),\quad \forall n=1,2,\ldots . \end{aligned}$$

In view of (3), we have from the above three that, for a large enough n,

$$\begin{aligned} {\tilde{\rho }}(t,A,{\tilde{\delta }})\times {\tilde{\rho }}(t,\{a_i\},{\tilde{\delta }})\subseteq (\psi _n(t_n))^{\epsilon /2}. \end{aligned}$$

Using the same arguments as in the proof of Proposition 2, we can also obtain

$$\begin{aligned} \psi _n(t_n)\subseteq (\psi (t))^{\epsilon /2}, \end{aligned}$$

when n is large enough. When we combine the above two, it would happen that

$$\begin{aligned} {\tilde{\rho }}(t,A,{\tilde{\delta }})\times {\tilde{\rho }}(t,\{a_i\},{\tilde{\delta }})\subseteq (\psi (t))^\epsilon . \end{aligned}$$

Since $\psi (t)$ is closed and $\epsilon$ is arbitrary, we can conclude from the above that ${\tilde{\rho }}(t,A,{\tilde{\delta }})\times {\tilde{\rho }}(t,\{a_i\},{\tilde{\delta }})\subseteq \psi (t)$ and hence by (17), $e_i\in \left[ \tilde{\mathbb J}({\tilde{\delta }}|{\tilde{\rho }},\psi )\right] (t)$. Thus, $\left[ \tilde{{\mathbb {J}}}(\cdot |\cdot ,\cdot )\right] (\cdot )$ is closed and hence closed-valued. It is also upper hemi-continuous and compact-valued as $\tilde{\mathcal{D}}^{\text{ e }}$ is finite. $\square$

Proof of Proposition 6:

Because $\left[ \tilde{\mathbb J}({\tilde{\delta }}|{\tilde{\rho }},\psi )\right] (t)\ne \emptyset$, we know $\tilde{{\mathbb {F}}}(\tilde{\delta }|\tilde{\rho },\psi )$ of (18) is nonempty. As $\theta$ is atomless, Theorem D.II.3 of Hildenbrand (1974) would guarantee $\tilde{\mathbb F}(\tilde{\delta }|\tilde{\rho },\psi )$’s convexity.

Let $({\tilde{\delta }}_n,{\tilde{\rho }}_n,\psi _n)_{n=1,2,\ldots }$ be a sequence in $\tilde{\mathcal{D}}\times \tilde{\mathcal{R}}\times \mathcal{P}$ that converges to some $({\tilde{\delta }},{\tilde{\rho }},\psi )$ in the aforementioned product space. Then, Proposition 5 would lead to

$$\begin{aligned} \tilde{{\mathbb {H}}}(t)\equiv \text{ Ls }\left( \left( \left[ \tilde{\mathbb J}({\tilde{\delta }}_n|{\tilde{\rho }}_n,\psi _n)\right] (t)\right) _{n=1,2,\ldots }\right) \subseteq \left[ \tilde{\mathbb J}({\tilde{\delta }}|{\tilde{\rho }},\psi )\right] (t),\forall t\in T. \end{aligned}$$

On the other hand, Theorem D.II.6 of Hildenbrand (1974) and (18) would result in

$$\begin{aligned} \text{ Ls }\left( \left( \tilde{\mathbb F}({\tilde{\delta }}_n|{\tilde{\rho }}_n,\psi _n)\right) _{n=1,2,\ldots }\right) \subseteq \int \limits _T \tilde{{\mathbb {H}}}(t)\cdot \theta (dt). \end{aligned}$$

Thus, by combining (18) with the above two, we would obtain

$$\begin{aligned} \text{ Ls }\left( \left( \tilde{\mathbb F}({\tilde{\delta }}_n|{\tilde{\rho }}_n,\psi _n)\right) _{n=1,2,\ldots }\right) \subseteq \int \limits _T \tilde{{\mathbb {H}}}(t)\cdot \theta (dt)\subseteq \int \limits _T\left[ \tilde{\mathbb J}({\tilde{\delta }}|{\tilde{\rho }},\psi )\right] (t)\cdot \theta (dt)=\tilde{\mathbb F}({\tilde{\delta }}|{\tilde{\rho }},\psi ). \end{aligned}$$

This amounts to asserting that $\tilde{\mathbb F}(\cdot |\cdot ,\cdot )$ is closed and hence closed-valued. It is also upper hemi-continuous and compact-valued as $\tilde{\mathcal{D}}$ is compact. $\square$

1.3 Technical details of Sect. 6

Proof of Proposition 7:

Define a subset of $R\times R\times \Psi$:

$$\begin{aligned} \Sigma \equiv \left\{ (r,r',\psi )\in R\times R\times \Psi :\;(r,r')\in \psi \right\} . \end{aligned}$$

(29)

Note $\mathcal{I}(a,a',\delta )\equiv \left[ {\hat{\rho }}(\cdot ,a,\delta ),{\hat{\rho }}(\cdot ,a',\delta ),{\hat{\psi }}(\cdot )\right] ^{-1}(\Sigma )$. By the measurabilities of ${\hat{\rho }}(\cdot ,a,\delta )$, ${\hat{\rho }}(\cdot ,a',\delta )$, and ${\hat{\psi }}(\cdot )$, all we have to show is that $\Sigma$ of (29) belongs to $\mathscr {B}(R\times R\times \Psi )$.

Indeed, $\Sigma$ is a closed subset of $R\times R\times \Psi$. Suppose $(r,r',\psi )\in (R\times R\times \Psi ){\setminus }\Sigma$. Then by (29) it would follow that $(r,r')\in (R\times R){\setminus }\psi$. Since $\psi$ in $\Psi$ of (1) is a closed subset of $R\times R$, there would be some $\epsilon$-neighborhood of $(r,r')$ so that within it any $(r_0,r'_0)$ is still outside $\psi$. Then, by (2) and (3), any $(r_0,r'_0)$ in the $(\epsilon /2)$-neighborhood of $(r,r')$ would be outside any $\psi _0$ in the $(\epsilon /2)$-neighborhood of $\psi$. Therefore, $(R\times R\times \Psi ){\setminus }\Sigma$ is open and hence $\Sigma$ is closed. $\square$

Proof of Proposition 8:

Due to $\hat{\rho }(i,\cdot ,\delta )$’s continuity and A’s compactness, $\hat{\rho }(i,A,\delta )$ would be compact. Also, ${\hat{\psi }}(i)$ as an irreflexive and transitive preference. Then, by Lemma 2 of Schmeidler (1969) or Proposition 3 of Yang (2018), $\hat{{\mathbb {B}}}^A(i,\delta |{\hat{\Gamma }})$ of (25) would be nonempty.

For upper-continuity, let $(a_n,\delta _n)_{n=1,2,\ldots }$ be a sequence in $A\times \mathcal{D}$ and let $(a,\delta )$ be a point of the aforementioned product space. Suppose that $a_n\in \hat{\mathbb B}^A(i,\delta _n|{\hat{\Gamma }})$ for each $n=1,2,\ldots$ and also that $a_n\rightarrow a$ and $\delta _n\rightarrow \delta$. Fix an $\epsilon >0$. By ${\hat{\rho }}(i)$’s membership to $\hat{\mathcal{R}}\equiv {\mathscr {C}}(A\times \mathcal{D},R)$,

$$\begin{aligned} d_R({\hat{\rho }}(i,a',\delta _n),{\hat{\rho }}(i,a',\delta ))\vee d_R({\hat{\rho }}(i,a_n,\delta _n),{\hat{\rho }}(i,a,\delta ))<\epsilon \end{aligned}$$

would occur when n is large enough. Then, for n sufficiently large,

$$\begin{aligned} {\hat{\rho }}(i,A,\delta )\subseteq \left( {\hat{\rho }}(i,A,\delta _n)\right) ^\epsilon . \end{aligned}$$

However, by (25) and the hypothesis that $a_n\in \hat{{\mathbb {B}}}^A(i,\delta _n|{\hat{\Gamma }})$ for any n,

$$\begin{aligned} {\hat{\rho }}(i,A,\delta _n)\times {\hat{\rho }}(i,\{a_n\},\delta _n)\subseteq {\hat{\psi }}(i),\quad \forall n=1,2,\ldots . \end{aligned}$$

In view of (3), we have from the above that, for a large enough n,

$$\begin{aligned} {\hat{\rho }}(i,A,\delta )\times {\hat{\rho }}(i,\{a\},\delta )\subseteq \left( {\hat{\psi }}(i)\right) ^\epsilon . \end{aligned}$$

Since ${\hat{\psi }}(i)$ is closed and $\epsilon$ is arbitrary, we can conclude from the above that ${\hat{\rho }}(i,A,\delta )\times {\hat{\rho }}(i,\{a\},\delta )\subseteq {\hat{\psi }}(i)$ and hence by (25), $a\in \hat{\mathbb B}^A(i,\delta |{\hat{\Gamma }})$ as well. Thus, $\hat{\mathbb B}^A(i,\cdot |{\hat{\Gamma }})$ is closed and hence closed-valued. It is also upper hemi-continuous and compact-valued as A is compact. $\square$

Proof of Proposition 9:

Since A is compact and hence separable, we can suppose some $\{a_n:\;n=1,2,\ldots \}$ to be dense in it. As $A'$ is a closed subset of A, we can further suppose some $\{a'_m:\;m=1,2,\ldots \}$ to be dense in the former. Let

$$\begin{aligned} {\hat{I}}''_{nm}(\delta |\hat{\Gamma })\equiv \left\{ i\in I:\;\left( \hat{\rho }(i,a_n,\delta ),\hat{\rho }(i,a'_m,\delta )\right) \in \hat{\psi }(i)\right\} , \end{aligned}$$

which is a member of ${\mathscr {I}}$ due to Proposition 7. Define

$$\begin{aligned} {\hat{I}}'_m(\delta |\hat{\Gamma })\equiv \left\{ i\in I:\;\hat{\rho }(i,A,\delta )\times \hat{\rho }(i,\{a'_m\},\delta )\subseteq \hat{\psi }(i)\right\} . \end{aligned}$$

(30)

Just because each ${\hat{\rho }}(i,\cdot ,\delta )$ is continuous, each $\hat{\psi }(i)$ is closed, and the $a_n$’s form a dense subset of A, we have from the above two definitions that

$$\begin{aligned} {\hat{I}}'_m(\delta |\hat{\Gamma })=\bigcap _{n=1}^{+\infty }{\hat{I}}''_{nm}(\delta |\hat{\Gamma })\in {\mathscr {I}}. \end{aligned}$$

By (25), we have from its own definition that ${\hat{I}}(A',\delta |\hat{\Gamma })$ is equal to

$$\begin{aligned} \left\{ i\in I:\;\hat{{\mathbb {B}}}^A(i,\delta |\hat{\Gamma })\subseteq (A')^c\right\} ^c=\left\{ i\in I:\;{\hat{\rho }}\left( i,A,\delta \right) \times \hat{\rho }\left( i,\{a'\},\delta \right) \not \subseteq {\hat{\psi }}(i),\;\forall a'\in A'\right\} ^c. \end{aligned}$$

As $\hat{\rho }(i,\cdot ,\delta )$ is continuous, $\{a'_m:\;m=1,2,\ldots \}$ is dense in $A'$, and $\psi$ is closed,

$$\begin{aligned} {\hat{I}}(A',\delta |\hat{\Gamma })=\left\{ i\in I:\;{\hat{\rho }}\left( i,A,\delta \right) \times \hat{\rho }\left( i,\{a'_m\},\delta \right) \not \subseteq {\hat{\psi }}(i),\;\forall m=1,2,\ldots \right\} ^c. \end{aligned}$$

According to (30), ${\hat{I}}(A',\delta |\hat{\Gamma })$ can be further equated to

$$\begin{aligned} \left( \bigcap _{m-1}^{+\infty }\left( {\hat{I}}'_m(\delta |\hat{\Gamma })\right) ^c\right) ^c=\bigcup _{m-1}^{+\infty }{\hat{I}}'_m(\delta |\hat{\Gamma }). \end{aligned}$$

Combining the above three, we may see that ${\hat{I}}(A',\delta |\hat{\Gamma })\in {\mathscr {I}}$. $\square$

Proof of Proposition 10:

By Proposition 9 and Lemma 18.2 of Aliprantis and Border (1999) which allows the closed $A'$ in the former to be replaced by an open one, we can apply the Kuratowski–Ryll-Nardzewski measurable selection theorem (18.13 of Aliprantis and Border (1999)) to demonstrate the existence of some ${\hat{b}}^A(\cdot ,\delta |\hat{\Gamma })\in \hat{\mathcal{A}}\equiv \mathscr {M}(I,A)$ such that each ${\hat{b}}^A(i,\delta |\hat{\Gamma })\in \hat{{\mathbb {B}}}(i,\delta |{\hat{\Gamma }})$. Let each ${\hat{b}}(i,\delta |\hat{\Gamma })$ be $\left( {\hat{\theta }}(i),{\hat{b}}^A(i,\delta |\hat{\Gamma })\right)$. Then, ${\hat{b}}(\cdot ,\delta |\hat{\Gamma })$ would be a measurable selection of the correspondence $\hat{{\mathbb {B}}}(\cdot ,\delta |{\hat{\Gamma }})$ defined by (26). Therefore, $\hat{\mathbb F}(\delta |{\hat{\Gamma }})$ of (27) is nonempty. According to Proposition D(P1) of Khan et al. (2013a), the latter is also convex.

By (26) and Proposition 8, we also have $\hat{{\mathbb {B}}}(i,\cdot |{\hat{\Gamma }})$ being upper hemi-continuous and hence closed-valued at each $i\in I$. In addition, (26), Proposition 9, and Lemma 18.2 of Aliprantis and Border (1999) would together lead to the measurability of the correspondence $\hat{{\mathbb {B}}}(\cdot ,\delta |\hat{\Gamma })$ in both the normal and weak senses. Then, Propositions D(P2) and D(P3) of Khan et al. (2013a) would result in the upper hemi-continuity of $\hat{{\mathbb {F}}}(\cdot |{\hat{\Gamma }})$ defined by (27). At every $\delta \in \mathcal{D}|_T(\lambda \circ {\hat{\theta }}^{-1})$, the set $\hat{{\mathbb {F}}}(\delta |{\hat{\Gamma }})$ is thus closed. It is also compact since $\mathcal{D}|_T(\lambda \circ {\hat{\theta }}^{-1})$ is. $\square$