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Pure-strategy Nash equilibria in nonatomic games with infinite-dimensional action spaces

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Abstract

This paper studies the existence of pure-strategy Nash equilibria for nonatomic games where players take actions in infinite-dimensional Banach spaces. For any infinite-dimensional Banach space, if the player space is modeled by the Lebesgue unit interval, we construct a nonatomic game which has no pure-strategy Nash equilibrium. But if the player space is modeled by a saturated probability space, there is a pure-strategy Nash equilibrium in every nonatomic game. Finally, if every game with a fixed nonatomic player space and a fixed infinite-dimensional action space has a pure-strategy Nash equilibrium, the underlying player space must be saturated.

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Notes

  1. 1.

    The supermodular game is an example; see Topkis (1979). Here is another one, when each player’s payoff function is quasi-concave and continuous, and the strategy space is a convex and compact subset of a finite-dimensional Euclidean space, there exists a pure-strategy Nash equilibrium by Kakutani’s fixed-point theorem.

  2. 2.

    The concept of saturated probability space is first introduced in Hoover and Keisler (1984) where a different but equivalent definition is used; see Remark 4 for more discussion.

  3. 3.

    Similar sufficient and necessary results hold for finite-player games with nonatomic information spaces, see Khan and Zhang (2012b). See Fu (2007) for the relation between the games with nonatomic player spaces and finite-player games with nonatomic information spaces.

  4. 4.

    When players’ payoff functions are linear in their own actions, see Khan (1986) for the existence of the approximate pure-strategy Nash equilibria for nonatomic games with infinite-dimensional action spaces; see also Rustichini and Yannelis (1991) for the existence of exact pure-strategy Nash equilibria for such nonatomic games with the condition “many more” players than actions.

  5. 5.

    See, for example, Bewley (1972), Yannelis (2009) and the books Khan and Yannelis (1991) and Stokey et al. (1989).

  6. 6.

    For nonatomic games with a compact action set in a finite-dimensional space, the existence of pure-strategy Nash equilibria is shown in Rath (1992). See also Khan et al. (1997) for the positive results on pure-strategy Nash equilibria for games with countable actions in an infinite-dimensional space.

  7. 7.

    See Remark 4 below.

  8. 8.

    See Proposition 1.f.3 in Lindenstrauss and Tzafriri (1977).

  9. 9.

    For more details, see Chapter 2 of Diestel and Uhl (1977).

  10. 10.

    This is Mazur’s Theorem; see Theorem 12 in Diestel and Uhl (1977).

  11. 11.

    In the case that a number has two binary representations, e.g., \(\frac{1}{2}\) has two binary representations, either \(\frac{1}{2} =\frac{1}{2}+\frac{0}{2^2}+\frac{0}{2^3}+\cdots \) or \(\frac{1}{2} =\frac{0}{2}+\frac{1}{2^2} +\frac{1}{2^3} + \cdots \), we choose the one with finitely many \(1\)s.

  12. 12.

    See Walsh (1923).

  13. 13.

    Take \(s_m(t) = \sum \nolimits _{n=0}^m \frac{x_n}{2^n\Vert x_n\Vert } W_n(t)\). Then \(\{s_m(t)\}_{m\in {\mathbb {N}}}\) is a sequence of Bochner integrable simple functions, and \(\lim _{m\rightarrow \infty } \int _0^1 \Vert \psi (t)-s_m(t)\Vert {{\mathrm{d\!}}}\eta (t) = 0\).

  14. 14.

    For more discussion on the Lyapunov example on the range of vector measures, see e.g., Khan and Zhang (2012a).

  15. 15.

    See p. 33 in Khan et al. (1997), “by an appealing to Corollary 6 (p. 265) in Diestel and Uhl (1977), one should hopefully be able to set a version of the counterexample in any arbitrary infinite-dimensional Banach space.”

  16. 16.

    In Sun and Zhang (2009), the construction of a saturated extension of the Lebesgue unit interval is not an issue, while the key is to construct a rich Fubini extension based on this extended Lebesgue interval.

  17. 17.

    Here, \(1_{A_{\gamma }} \) is the indicator function of \(A_{\gamma }\).

  18. 18.

    A measure space is called super-atomless if for any subset \(S\in {\mathcal {T}}\) with \(\mu (S)>0\), the least cardinal number of any subset \(\mathcal{A} \subseteq {\mathcal {T}}^S\) which (completely) generates \({\mathcal {T}}^S\) (modulo the null sets) is not countable; please see Podczeck (2008) for more discussion.

  19. 19.

    A result similar to Proposition 1 is also obtained independently by Yu (2013), see Lemma 2 therein.

  20. 20.

    For more discussion on this “transferring” technique, see Sect. 5 of Keisler and Sun (2009).

  21. 21.

    For example, see the sufficiency part of Theorem 4.6 and Proposition 5.3(1) in Keisler and Sun (2009), or Corollary 4(4) in Carmona and Podczeck (2009), or Theorem 2 of Noguchi (2009).

  22. 22.

    Based on the compactness result, Yannelis (1990) points out “preservation of upper hemi-continuity” for the Bochner integral of correspondences can be proved directly; see Theorem 3.1 and Remark 3.1 therein. We are grateful to a referee for pointing out this fact.

  23. 23.

    Measure-preserving map here is also called “inverse measure-preserving map” in some literature.

  24. 24.

    In Khan and Zhang (2012b), “sufficient and necessary” results on saturated probability spaces are established in the context of finite-player games with diffused private information.

  25. 25.

    This remark is motivated by the comments of a referee.

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Author information

Correspondence to Yongchao Zhang.

Additional information

The authors are very grateful to Yi-Chun Chen, Wei He, M. Ali Khan, Xiao Luo, Kali P. Rath, Yeneng Sun, Nicholas C. Yannelis, and Haomiao Yu. We are also indebted to two anonymous referees for many constructive comments. The results reported in this paper constitute Chapter 4 of Sun’s PhD dissertation submitted to Department of Mathematics, National University of Singapore, and Sun gratefully acknowledges NUS for support. Zhang is supported by Shanghai Leading Academic Discipline Project (B801), and Sun by the NUS grant (R-146-000-170-112). This project is also supported by National Natural Science Foundation of China (No. 11201283). This paper was presented by Zhang at the 12th Conference of the Society for the Advancement of Economic Theory held at Brisbane, Australia, June 29–July 2, 2012.

Appendix: Proofs

Appendix: Proofs

Proofs for results in Sect. 3

Proof of Lemma 1

We prove this result by contradiction. Suppose that \(e/2 \in \int _I \Psi (t){{\mathrm{d\!}}}\eta (t)\), that is, there is a selection \(f :I \rightarrow X\) of \(\Psi \) such that \(\int _I f {{\mathrm{d\!}}}\eta =~e/2.\) Here \(f\) is a \({\mathcal {L}}\)-measurable function and for any \(t \in I, f(t)\) is either \(\mathbf 0 \) or \(\psi (t)\). Let \(E = \{t \in I :f(t) \ne \mathbf 0 \}\). It is clear that \(E \in {\mathcal {L}}\) and \(\int _I f {{\mathrm{d\!}}}\eta = \int _E \psi {{\mathrm{d\!}}}\eta .\) As a result \(e/2 = \int _E\psi (t){{\mathrm{d\!}}}\eta (t)\), that is,

$$\begin{aligned} \frac{1}{2}\frac{x_0}{\Vert x_0\Vert } = \sum _{n=0}^\infty \frac{x_n}{2^n\Vert x_n\Vert }\int \limits _EW_n(t){{\mathrm{d\!}}}\eta (t). \end{aligned}$$
(9)

Apply \(x_0^*\) on the both sides of Eq. (9), we will have \(\eta (E)=1/2\). Then for each integer \(n\ge 1\), apply \(x_n^*\) on the both sides of Eq. (9), we will have

$$\begin{aligned} \eta (E\cap E_n)=\eta (E\cap E_n^c), \end{aligned}$$

where \(E_n^c\) is the complement of \(E_n\).

Let \(g=1_E-1_{E^c}\), where \(1_E\) is the indicator function of \(E\). Then

$$\begin{aligned} \int \limits _I W_n(t)g(t) {{\mathrm{d\!}}}\eta (t)=0 \end{aligned}$$

for each integer \(n\ge 0\). Because that the Walsh system is a complete orthogonal basis of the space of all square-integrable functions on the Lebesgue unit interval, \(g\equiv 0\); hence, it is a contradiction to the definition of \(g\). Therefore, \(e/2\notin \int _I \Psi {{\mathrm{d\!}}}\eta \). \(\square \)

Before we offer a proof of Theorem 1, we shall need three additional inequalities.

Lemma 2

For \(\alpha >0\) and any real numbers \(b_1, b_2\) and \(c\), with \(b_1<b_2\),

$$\begin{aligned} \left| \int \limits _{b_1}^{b_2} (-1)^{ \left[ \frac{t+c}{\ell } \right] } {{\mathrm{d\!}}}\eta (t) \right| \le \ell , \end{aligned}$$

where for any real number \(x, [x]\) denotes the largest integer not greater than \(x\).

Proof

Routine. \(\square \)

Lemma 3

For \(\ell >0\), and any integer \(n\ge 0\),

$$\begin{aligned} \left| \int \limits _0^1 (-1)^{\left[ \frac{t}{\ell }\right] } W_n(t) {{\mathrm{d\!}}}\eta (t) \right| \le \min \{1,2n\ell \}. \end{aligned}$$

Proof

It is obvious that

$$\begin{aligned} \left| \int \limits _0^1 (-1)^{\left[ \frac{t}{\ell }\right] } W_n(t) {{\mathrm{d\!}}}\eta (t) \right| \le 1. \end{aligned}$$

We next show the other part. Consider the binary representation of \(n\),

$$\begin{aligned} n = n_0 + 2n_1 + 2^2n_2 + \cdots + 2^{m-1}n_{m-1}, \end{aligned}$$

where \(n_{m-1} =1\) and \(n_0,n_1,\ldots , n_{m-2}\) are either \(0\) or \(1\). For any \(t \in \left( \frac{s-1}{2^m}, \frac{s}{2^m} \right) \), where \(s\) is an integer from \(\{1,2,\ldots ,2^m\}\), it is clear that the first \(m\) numbers in the binary representation of \(t\) are fixed. By definition, \(W_n(t) = (-1)^{c_s}\) over \(\left( \frac{s-1}{2^m}, \frac{s}{2^m} \right) \), where \(c_s\) is a constant integer. Then we have

$$\begin{aligned} \left| \int \limits _0^1 (-1)^{\left[ \frac{t}{\ell }\right] } W_n(t) {{\mathrm{d\!}}}\eta (t) \right| \le&\sum _{s=1}^{2^m} \left| \int \limits _{\frac{s-1}{2^m}}^{\frac{s}{2^m}} (-1)^{\left[ \frac{t}{\ell }\right] } W_n(t) {{\mathrm{d\!}}}\eta (t) \right| \\ =&\sum _{s=1}^{2^m} \left| \int \limits _{\frac{s-1}{2^m}}^{\frac{s}{2^m}} (-1)^{\left[ \frac{t+c_s\ell }{\ell }\right] } {{\mathrm{d\!}}}\eta (t) \right| . \end{aligned}$$

It follows from Lemma 2 that

$$\begin{aligned} \left| \int \limits _0^1 (-1)^{\left[ \frac{t}{\ell }\right] } W_n(t) {{\mathrm{d\!}}}\eta (t) \right| \le \sum _{s=1}^{2^m}\ell = 2^m\ell \le 2n\ell , \end{aligned}$$

where the last equality holds because of the binary representation of \(n\). \(\square \)

Lemma 4

For \(\ell >0\),

$$\begin{aligned} \sum _{n=0}^\infty \frac{1}{2^{n+1}}\left| \int \limits _0^1(-1)^{\left[ \frac{t}{\ell }\right] } W_n(t) {{\mathrm{d\!}}}\eta (t)\right| < 2\ell . \end{aligned}$$

Proof

Take \(r=\left[ 1/(2\ell )\right] \), then \(r\le {1/\ell }\) and \(r+1>{1/(2\ell )}\). By Lemma 3,

$$\begin{aligned} \sum _{n=0}^\infty \frac{1}{2^{n+1}} \left| \int \limits _0^1(-1)^{\left[ \frac{t}{\ell }\right] } W_n(t) {{\mathrm{d\!}}}\eta (t)\right|&\le \sum _{n=0}^\infty \frac{1}{2^{n+1}} \min \{1,2n\ell \} \\&\le \sum _{n=0}^r \frac{1}{2^{n+1}} 2n\ell + \sum _{n=r+1}^\infty \frac{1}{2^{n+1}}\\&= \ell \left( 2-\frac{r+2}{2^r} \right) + \frac{1}{2^{r+1}}\\&= 2\ell - \frac{2(r+2)\ell -1}{2^{r+1}} < 2\ell . \end{aligned}$$

\(\square \)

Proof of Theorem 1

We will prove this result by contradiction. Suppose that there exists a pure-strategy Nash equilibrium in the game \({\mathcal {G}}_1\), and assume that the \({\mathcal {L}}\)-measurable function \(f :I \rightarrow A\) is a pure-strategy Nash equilibrium. We next prove that the following two cases cannot happen, (1) \(\int _I f(t){{\mathrm{d\!}}}\eta (t) = e/2\) and (2) \(\int _I f(t){{\mathrm{d\!}}}\eta (t) \ne e/2\), where \(\int _I f(t){{\mathrm{d\!}}}\eta (t)\) is the societal response when players take actions as in the Nash equilibrium \(f\) and \(e = x_0 /\Vert x_0\Vert \) as in Eq. (3). Therefore, we obtain a contradiction. Next, we discuss these two cases separately by dividing our arguments into the following two parts.

Part 1. Suppose that \(\int _If(t){{\mathrm{d\!}}}\eta (t)=e/2\). Then for any player \(t\in I\), the payoff function of player \(t\) reduces to the following form, for any \(a \in A\),

$$\begin{aligned} u_t\left( a, \int \limits _I f {{\mathrm{d\!}}}\eta \right) = u_t\left( a,{e/2}\right) =-\Vert a\Vert \cdot \Vert a-\psi (t)\Vert , \end{aligned}$$
(10)

As a result, the best response of player \(t\) is the doubleton set \(\{0,\psi (t)\}\) for any \(t \in I\). Hence, the pure-strategy Nash equilibrium \(f\), as a function from the Lebesgue unit interval to \(A\), is a measurable selection of the correspondence \(\Psi \) in Eq. (4). However, Lemma 1 ensures that there does not exist a Lebesgue measurable selection from this correspondence \(\Psi \) whose Bochner integral is \({e/2}\). Hence, it is a contradiction that \(\int _If(t){{\mathrm{d\!}}}\eta (t) = e/2\).

Part 2. Suppose that \(\int _If(t){{\mathrm{d\!}}}\eta (t) \ne e/2\). That is, in the pure-strategy Nash equilibrium \(f :I \rightarrow A\), the societal response is no longer \(e/2\). As a result, in the first item \(h\), see Eq. (6), of the payoff function for any player \(t \in I\), the fourth argument is no longer zero. Let

$$\begin{aligned} \ell _0=\beta d\left( \int \limits _If(t){{\mathrm{d\!}}}\eta (t),\frac{e}{2}\right) . \end{aligned}$$

Certainly, \(0<\ell _0\le 1\). Divide the unit interval \(I\) into intervals of length \(\ell _0\), so that we obtain \(I\cap \big (\cup _{n\in {\mathbb {N}}}[n\ell _0,(n+1)\ell _0)\big )\).

We first characterize the set of best responses for any player when other players take actions in the Nash equilibrium \(f\), i.e., the societal response is \(\int _I f {{\mathrm{d\!}}}\eta \). On the one hand, we first fix a player \(t \in (n\ell _0,(n+1)\ell _0)\). If \(n\) is even, then the integer part of \(t /\ell _0, \left[ t /\ell _0\right] \), is even, as a result the payoff function of this player \(t\) (see Eq. (7)) reduces to the following form, for any \(a\in A\),

$$\begin{aligned} u_t\left( a, \int \limits _I f {{\mathrm{d\!}}}\eta \right) = -\ell _0 \left| \sin \frac{t}{\ell _0}\pi \right| \cdot \Vert a\Vert \cdot (\Vert a-\psi (t)\Vert +2) - \Vert a\Vert \cdot \Vert a-\psi (t)\Vert . \end{aligned}$$

Note that the value of \(u_t(a, \int _I f {{\mathrm{d\!}}}\eta ) \le 0\) for any \(a \in A\), and the value is \(0\) if and only if \(a = \mathbf 0 \). As a result, \(u_t (a, \int _I f {{\mathrm{d\!}}}\eta )\) takes the maximum value only at \(a=\mathbf 0 \). That is, the best response for this player \(t\), when facing the societal response \(\int _I f {{\mathrm{d\!}}}\eta \), is the singleton set \({\{\mathbf{0}\}}\). Similarly, for player \(t \in \left( n\ell _0,(n+1)\ell _0 \right) \), if \(n\) is an odd natural number, so is the integer part \(\left[ {t}/{\ell _0}\right] \). As a result, the payoff function of this player \(t\) reduces to the following form, for any \(a \in A\),

$$\begin{aligned} u_t\left( a, \int \limits _I f {{\mathrm{d\!}}}\eta \right) = -\ell _0 \left| \sin \frac{t}{\ell _0}\pi \right| \cdot (\Vert a\Vert +2) \cdot \Vert a-\psi (t)\Vert - \Vert a\Vert \cdot \Vert a-\psi (t)\Vert . \end{aligned}$$

Using the similar argument as above as \(n\) is even, we can obtain that the best response for this player \(t\), when facing the societal response \(\int _I f {{\mathrm{d\!}}}\eta \), is the singleton set \({\{\psi (t)\}}\).

On the other hand, we consider the player \(t=n\ell _0\) for some \(n \in {\mathbb {N}}\). In this case, the payoff function of player \(t\) (see Eq. (7)) reduces to the following form, for any \(a\in A\),

$$\begin{aligned} u_t\left( a, \int \limits _I f {{\mathrm{d\!}}}\eta \right) =-\Vert a\Vert \cdot \Vert a-\psi (t)\Vert . \end{aligned}$$

It is clear that the set of best responses for this player \(t\), when facing the societal response \(\int _I f {{\mathrm{d\!}}}\eta \), is a doubleton set \({\{\mathbf{0},\psi (t)\}}\).

To summarize, except for the players in the set \(\{t = n \ell _0 \in I :n \in {\mathbb {N}}\}\), the best response for any other player is a singleton set. Note also that \(\{t = n \ell _0 :n \in {\mathbb {N}}\}\) is a \(\eta \)-null set in the Lebesgue interval. As a result, the pure-strategy Nash equilibrium \(f :I \rightarrow A\) is of the following form, for almost all \(t\in ~I\),

$$\begin{aligned} f(t)=\frac{1-(-1)^{\left[ \frac{t}{\ell _0}\right] }}{2}\psi (t). \end{aligned}$$
(11)

Next, we calculate the distance between the societal response \(\int _I f {{\mathrm{d\!}}}\eta \) and \(e/2\), where the pure-strategy Nash equilibrium \(f\) is determined in Eq. (11) above.

$$\begin{aligned} d\left( \int \limits _If(t){{\mathrm{d\!}}}\eta (t),\frac{e}{2}\right)&=\left\| \int \limits _If(t){{\mathrm{d\!}}}\eta (t)-\frac{e}{2}\right\| \\&=\left\| \int \limits _If(t){{\mathrm{d\!}}}\eta (t)-\frac{1}{2}\int \limits _0^1\psi (t) {{\mathrm{d\!}}}\eta (t)\right\| \\&=\left\| \frac{1}{2}\int \limits _I(-1)^{\left[ \frac{t}{\ell _0}\right] }\psi (t) {{\mathrm{d\!}}}\eta (t)\right\| \\&=\left\| \frac{1}{2}\int \limits _I(-1)^{\left[ \frac{t}{\ell _0}\right] }\sum _{n=0}^\infty \frac{x_n}{2^n\Vert x_n\Vert }W_n(t) {{\mathrm{d\!}}}\eta (t)\right\| \\&=\left\| \sum _{n=0}^\infty \int \limits _I\frac{1}{2}(-1)^{\left[ \frac{t}{\ell _0}\right] }\frac{x_n}{2^n\Vert x_n\Vert }W_n(t) {{\mathrm{d\!}}}\eta (t)\right\| \\&\le \sum _{n=0}^\infty \frac{1}{2}\left\| \int \limits _I(-1)^{\left[ \frac{t}{\ell _0}\right] }\frac{x_n}{2^n\Vert x_n\Vert }W_n(t) {{\mathrm{d\!}}}\eta (t)\right\| \\&\le \sum _{n=0}^\infty \frac{1}{2^{n+1}}\left| \int \limits _I(-1)^{\left[ \frac{t}{\ell _0}\right] }W_n(t) {{\mathrm{d\!}}}\eta (t)\right| . \end{aligned}$$

By virtue of the inequality in Lemma 4, we have

$$\begin{aligned} d\left( \int \limits _If(t){{\mathrm{d\!}}}\eta (t), \frac{e}{2}\right) < \sum _{n=0}^\infty \frac{1}{2^{n+1}} 2 \ell _0 = 2\ell _0. \end{aligned}$$
(12)

Notice that \(\ell _0 = \beta {{\mathrm{d}}}\left( \int f {{\mathrm{d\!}}}\eta ,e/2 \right) \), where \(\beta = 1/(2M)\) and \(M = \max \{\Vert a\Vert :a \in A\}\). Note that \(M \ge 1\), it follows that \(\beta \le 1/2\). It follows from Eq. (12) that \({{\mathrm{d}}}\left( \int f {{\mathrm{d\!}}}\eta ,e/2 \right) <2\ell _0 = 2 \beta {{\mathrm{d}}}\left( \int f {{\mathrm{d\!}}}\eta ,e/2 \right) \). This implies that \(\beta >1/2\). It is a contradiction with \(\beta \le 1/2\). Therefore, we complete the proof of Part 2 that there does not exist a pure-strategy Nash equilibrium \(f :I \rightarrow A\) with \(\int _I f {{\mathrm{d\!}}}\eta \ne e/2\). \(\square \)

Proof of Corollary 1

We show it by contradiction. Suppose that the function \(f :[0,1]\) \( \rightarrow A\) is a pure-strategy Nash equilibrium in the game \({\mathcal {G}}_s\). Note that for any player \(t \in (s, 1]\), the best response is \(\mathbf 0 \). As a result, in this Nash equilibrium \(f\), the actions for players in \((s, 1]\) do not affect the societal response. In particular,

$$\begin{aligned} \frac{1}{s} \int \limits _I f {{\mathrm{d\!}}}\eta = \frac{1}{s}\int \limits _0^{s} f {{\mathrm{d\!}}}\eta = \int \limits _0^{s} f {{\mathrm{d\!}}}\eta ^{[0, s]}, \end{aligned}$$

which is exactly an element in \({{\mathrm{\overline{con}}}}(A)\). Moreover, for any player \(t \in [0, s]\), if the other players follow actions as in this Nash equilibrium \(f\), for any \(a \in A\),

$$\begin{aligned} {\mathcal {G}}_s(t) \left( a, \int \limits _I f {{\mathrm{d\!}}}\eta \right) = {\mathcal {G}}_1 \left( \frac{t}{s} \right) \left( a, \frac{1}{s}\int \limits _0^{s} f {{\mathrm{d\!}}}\eta \right) . \end{aligned}$$

For \(\eta \)-almost all player \(t\in [0,s]\), since \(f\) is a pure-strategy Nash equilibrium, \(f(t)\) is a best response for this player \(t\) given others players follow \(f\). That is, for any \(a\in A\)

$$\begin{aligned} {\mathcal {G}}_1 \left( \frac{t}{s} \right) \left( f(t), \frac{1}{s}\int \limits _0^{s} f {{\mathrm{d\!}}}\eta \right) \ge {\mathcal {G}}_1 \left( \frac{t}{s} \right) \left( a, \frac{1}{s}\int \limits _0^{s} f {{\mathrm{d\!}}}\eta \right) . \end{aligned}$$
(13)

Let \(g:[0,1] \rightarrow A\) defined by \(g(t') = f(t' \cdot s)\) for \(\eta \)-almost all \(t' \in [0,1]\). It follows from substitution of variables that \( \frac{1}{s}\int _0^{s} f(t){{\mathrm{d\!}}}\eta (t) = \int _0^1 g(t') {{\mathrm{d\!}}}\eta (t')\). According to Eq. (13), for all \(t'\in [0,1]\) and \(a \in A\)

$$\begin{aligned} {\mathcal {G}}_1 \left( t' \right) \left( g(t'), \int \limits _0^1 g {{\mathrm{d\!}}}\eta \right) \ge {\mathcal {G}}_1 \left( t' \right) \left( a, \int \limits _0^1 g {{\mathrm{d\!}}}\eta \right) . \end{aligned}$$

Hence, \(g\) is a pure-strategy Nash equilibrium for the nonatomic game \({\mathcal {G}}_1\). It is a contradiction with Theorem 1. \(\square \)

Proof of Theorem 2

Since \((T,{\mathcal {T}},\mu )\) is not a saturated probability space, by Definition 3, there exists a non-negligible subset \(S \in {\mathcal {T}}\) with \(\mu \)-measure \(s \in (0,1]\) such that the restricted probability space \((S,{\mathcal {T}}^{S},\mu ^{S})\) is essentially countably generated. As a result of Maharam’s theorem (see Maharam 1942), the measure algebra of the Lebesgue subinterval, \(([0,s], {\mathcal {L}}^{[0, s]}, \eta ^{[0, s]})\), is isomorphic to that of \((S,{\mathcal {T}}^{S},\mu ^{S})\). It is a known result that this isomorphism can be realized by a measure-preserving map \(q\) from \((S,{\mathcal {T}}^{S},\mu ^{S})\) to \(([0,s], {\mathcal {L}}^{[0, s]}, \eta ^{[0, s]})\); see Theorem 4.12 (p. 937) in Fremlin (1989).

Now, consider the following nonatomic game with the player space \((T,{\mathcal {T}},\mu )\). For any player \(t \in T\), for any action \(a \in A\) of this player and any societal response \(b \in {{\mathrm{\overline{con}}}}A\), the payoff function of player \(t\) is defined as

$$\begin{aligned} {\mathcal {G}}_s'(t)(a, b) =\left\{ \begin{array}{ll} {\mathcal {G}}_s(q(t)) (a, b), &{} \text { if } t \in S,\\ -\Vert a\Vert , &{} \text { if } t \notin S. \end{array}\right. \end{aligned}$$
(14)

where \({\mathcal {G}}_s\) is defined in Eq. (14), Sect. 3.3.

We next show by contradiction that there is no pure-strategy Nash equilibrium in the nonatomic game \({\mathcal {G}}_s'\). Suppose not, let \(g :T \rightarrow A\) be a pure-strategy Nash equilibrium for \({\mathcal {G}}_s'\). Since the space of all Bochner integrable functions from \([0,s]\) to \(A\) is complete, for the Bochner integrable function \(g\) restricted on \(S\), there exists a Lebesgue measurable function \(g':[0,s]\rightarrow A\) such that \(g(t)=g'\circ q(t)\) for \(\mu \)-almost all \(t \in S\). Further, define \(g'(t') = \mathbf 0 \) for any \(t' \in (s, 1]\). Thus, \(g'\) is a Lebesgue measurable function from \([0,1]\) to \(A\).

We complete the proof by proving that \(g'\) is a pure-strategy equilibrium in the nonatomic game \({\mathcal {G}}_s\), which contradicts Corollary 1. First, for any \(t' \in (s, 1]\), by the definition of \({\mathcal {G}}_s\) in Eq. (8), \(g'(t') =\mathbf 0 \) is a best response. Next, since \(g\) is a pure-strategy equilibrium in \({\mathcal {G}}_s'\), according to Eq. (14), for any player \(t \notin S\), the best response is \(\mathbf 0 \); moreover, \( \frac{1}{s} \int _T g {{\mathrm{d\!}}}\mu = \frac{1}{s}\int _S g {{\mathrm{d\!}}}\mu = \int _{S} g {{\mathrm{d\!}}}\mu ^S \in {{\mathrm{\overline{con}}}}(A). \) As a result, the payoff for any player \(t \in S\), for all \(a \in A\), is

$$\begin{aligned} {\mathcal {G}}_s'(t)\left( a, \int \limits _T g {{\mathrm{d\!}}}\mu \right) = {\mathcal {G}}_s(q(t)) \left( a, \int \limits _T g {{\mathrm{d\!}}}\mu \right) . \end{aligned}$$
(15)

It is clear that

$$\begin{aligned} \int \limits _T g {{\mathrm{d\!}}}\mu = \int \limits _{S} g {{\mathrm{d\!}}}\mu = \int \limits _{S} g'(q) {{\mathrm{d\!}}}\mu = \int \limits _0^s g' {{\mathrm{d\!}}}\eta = \int \limits _0^1 g' {{\mathrm{d\!}}}\eta , \end{aligned}$$
(16)

where the third equation follows from the substitution of variables, and the first and last equations from that \(g\) and \(g'\) are \(0\) elsewhere. Therefore, since for \(\mu \)-almost all \(t \in S, g(t)\) is a best response against \(g\), it follows from Eqs. (15) and (16) and the construction of \(q\) that for \(\eta \)-almost all player \(t' \in [0,s], g'(t')\) is a best response against \(g'\). \(\square \)

Proof of Proposition 2

By Lemma 2 of Khan and Zhang (2012a), there exists a subset \(E \in {\mathcal {I}}\) such that

$$\begin{aligned} \lambda ([0,t] \cap E) = t/2, \ \text { for all } t\in [0,1]. \end{aligned}$$
(17)

For any \(s \in (0, 1]\), define a function \(f_s :I \rightarrow X\) as follows, for all \(t \in I, f_s(t) = \psi (t) \cdot 1_{E} (t) \cdot 1_{[0,s]}(t),\) where \(1_S\) is the indicator function on a subset \(S\). Note that for all \(s, f_s\) is an \({\mathcal {I}}\)-measurable function.

We next show that for any \(s \in (0, 1], f_s\) is a pure-strategy Nash equilibrium in the nonatomic game \({\mathcal {G}}_s\). We first calculate \(\int _I f_s {{\mathrm{d\!}}}\lambda \). As in Eq. (2), \(\psi (t) = \sum _{n=0}^\infty \frac{x_n}{2^n\Vert x_n\Vert } W_n(t)\). For \(n = 0\), it is clear that \(\int _I W_0 (t) \cdot 1_{E} (t) \cdot 1_{[0,s]}(t) {{\mathrm{d\!}}}\lambda = \lambda (E \cap [0, s]) = \frac{s}{2}.\) For \(n \ge 1\), note that \(E_n = \{t : W_n(t) = 1\}\) is a union of subintervals, then \(\int _I W_n (t) \cdot 1_{E} (t) \cdot 1_{[0,s]}(t) {{\mathrm{d\!}}}\lambda = \lambda (E \cap E_n \cap [0,s]) - \lambda (E \cap E^{c}_n \cap [0,s]) = 0,\) where the last equation follows from Eq. (17). Hence \(\int _I f_s {{\mathrm{d\!}}}\lambda = \frac{x_0}{||x_0||} \cdot \int _I W_0 (t) \cdot 1_{E} (t) \cdot 1_{[0,s]}(t) {{\mathrm{d\!}}}\lambda = s \cdot e/2\). Finally, it is routine to check that \(f_s\) is a pure-strategy Nash equilibrium for the game \({\mathcal {G}}_s\). \(\square \)

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Sun, X., Zhang, Y. Pure-strategy Nash equilibria in nonatomic games with infinite-dimensional action spaces. Econ Theory 58, 161–182 (2015). https://doi.org/10.1007/s00199-013-0795-6

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Keywords

  • Infinite-dimensional action space
  • Nonatomic game
  • Pure-strategy Nash equilibrium
  • Saturated probability space

JEL Classification

  • C62
  • C72