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

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.

Appendix: Proofs

1.1 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. $\square$

1.2 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 $g'$. $\square$

1.3 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 ${\mathcal {G}}_s$. $\square$