$\displaystyle\lesssim$ $\displaystyle\sum_{N_{1}\sim N_{2}\gtrsim N\atop
N1,\cdots,N_{6}}\frac{N_{1}^{3}}{m(N_{3})\cdots
m(N_{6})}\|u_{1}\|_{X^{-2,\frac{1}{2}+}_{\delta}}\|u_{2}\|_{X^{-2,\frac{1}{2}+}_{\delta}}\prod^{6}_{j=3}\|u_{j}\|_{X^{\frac{1}{4},\frac{1}{2}+}_{\delta}}$
$\displaystyle\lesssim$ $\displaystyle\sum_{N_{1}\sim N_{2}\gtrsim N\atop
N1,\cdots,N_{6}}N_{2}^{-3}(N_{3}N_{4}N_{5}N_{6})^{-\frac{1}{2}}\prod^{6}_{j=1}\|u_{j}\|_{X^{1,\frac{1}{2}+}_{\delta}}\lesssim
N^{-3}\prod_{j=1}^{6}\|u_{j}\|_{X_{\delta}^{1,\frac{1}{2}+}}.$ (5.47)

If $N_{2}\sim N_{3}\gtrsim N$, $N_{3}\geq N_{1}$, we can get similarly that

$\displaystyle I_{1}\lesssim$ $\displaystyle\sum_{N_{2}\sim N_{3}\gtrsim
N\atop N1,\cdots,N_{6}}\frac{N_{1}^{3}m(N_{1})}{m(N_{2})\cdots
m(N_{6})}\|u_{2}\|_{L^{\infty}_{x}L^{2}_{t}}\|u_{3}\|_{L^{\infty}_{x}L^{2}_{t}}\|u_{1}\|_{L^{4}_{x}L^{\infty}_{t}}\prod^{6}_{j=4}\|u_{j}\|_{L^{4}_{x}L^{\infty}_{t}}$
$\displaystyle\lesssim$ $\displaystyle N^{1-s}\sum_{N_{2}\sim N_{3}\gtrsim
N\atop N1,\cdots,N_{6}}\frac{N_{1}^{s+2}}{m(N_{2})\cdots
m(N_{6})}\|u_{1}\|_{X^{\frac{1}{4},\frac{1}{2}+}_{\delta}}\|u_{2}\|_{X^{-2,\frac{1}{2}+}_{\delta}}\|u_{3}\|_{X^{-2,\frac{1}{2}+}_{\delta}}\prod^{6}_{j=4}\|u_{j}\|_{X^{\frac{1}{4},\frac{1}{2}+}_{\delta}}$
$\displaystyle\lesssim$ $\displaystyle N^{1-s}\sum_{N_{2}\sim N_{3}\gtrsim
N\atop
N1,\cdots,N_{6}}N_{1}^{s+\frac{5}{4}}N_{2}^{-\frac{11}{4}}N_{3}^{-\frac{11}{4}}(N_{4}N_{5}N_{6})^{-\frac{1}{2}}\prod^{6}_{j=1}\|u_{j}\|_{X^{1,\frac{1}{2}+}_{\delta}}$
$\displaystyle\lesssim$ $\displaystyle
N^{-3}\prod_{j=1}^{6}\|u_{j}\|_{X_{\delta}^{1,\frac{1}{2}+}}.$ (5.48)

For term $I_{2}$, the same argument as before, we may assume that

$\displaystyle N_{1}\geq N_{5}\geq N_{6},\quad\quad N_{2}\geq N_{3}\geq
N_{4},\quad\quad N_{2}\gtrsim N.$ (5.49)

If $N_{1}\sim N_{2}\gtrsim N$, the symbol is controlled by
$N_{1}^{3}m(N_{1})/m(N_{2})m(N_{3})m(N_{4})$, which is smaller than
$N_{1}^{3}m(N_{1})/m(N_{2})\cdots m(N_{6})$, therefore, we can get the same
bound as (5.3). If $N_{2}\sim N_{3}\gtrsim N$, using the same techniques as
(5.3) yields the estimates. If $N_{1}\sim N_{5}\gtrsim N$, this case is
better, thus we can obtain the conclusion as before. $\hfill\Box$

###### Proposition 5.5

Let $3/4\leq s<1$ and $T_{I4}$ be as in (5.51), then

$\displaystyle\bigg{|}\int_{0}^{\delta}T_{I4}\bigg{|}\lesssim
N^{-\frac{1}{2}}\|Iu\|^{4}_{X_{\delta}^{1,\frac{1}{2}+}}.$ (5.50)

Proof.As the above proposition, inserting the zero term (5.41) into $T_{I4}$,
it suffices to control

$\displaystyle\sum_{N_{1},\cdots,N_{4}}\bigg{|}\int_{0}^{\delta}\int_{\Gamma_{4}}\xi^{3}_{1}(\xi_{2}+\xi_{3})^{2}\cdot\Big{(}\frac{m(\xi_{2}+\xi_{3}+\xi_{4})}{m(\xi_{2})m(\xi_{3})m(\xi_{4})}-1\Big{)}\widehat{u_{1}}(\xi_{1})\cdots\widehat{u_{4}}(\xi_{4})\bigg{|},$
(5.51)

for any function $u_{j}$, $j=1,2,3,4$ with the frequency supported on
$|\xi_{j}|\sim N_{j}$. A glance at the symbol, we may assume that

$\displaystyle N_{2}\geq N_{3}\geq N_{4},\quad\quad\quad N_{2}\gtrsim N,$
(5.52)

since that if $N_{2}\leq N/8$, the symbol vanishes. The other orders of
$N_{2}$, $N_{3}$ and $N_{4}$ are similar and much easier. Then we control the
above summation by considering the following three cases:

  * 1.

the non-resonant case, where $N_{2}\gg N_{3}\geq N_{4}$;

  * 2.

the semi-resonant case, where $N_{2}\sim N_{3}\gg N_{4}$;

  * 3.

the resonant case, where $N_{2}\sim N_{3}\sim N_{4}$.

1\. In the non-resonant case, from the constraint condition
$\Gamma_{4}=\big{\\{}\sum_{j=1}^{4}\xi_{j}=0\big{\\}}$, it follows immediately
that $N_{1}\sim N_{2}\gg N_{3}\geq N_{4}$. Utilizing bilinear estimates (5.5)
can efficiently control (5.51) by

$\displaystyle\sum_{N_{1}\sim N_{2}\gtrsim N\atop N_{1}\sim N_{2}\gg N_{3}\geq
N_{4}}\frac{N_{1}^{3}N_{2}^{2}m(N_{1})}{m(N_{2})m(N_{3})m(N_{4})}\|u_{1}u_{3}\|_{L^{2}_{x,t}}\|u_{2}u_{4}\|_{L^{2}_{x,t}}$
$\displaystyle\lesssim$ $\displaystyle\sum_{N_{1}\sim N_{2}\gtrsim N\atop
N_{1}\sim N_{2}\gg N_{3}\geq
N_{4}}\frac{N_{1}}{m(N_{3})m(N_{4})}\prod^{4}_{j=1}\|u_{j}\|_{X_{\delta}^{0,\frac{1}{2}+}}$
$\displaystyle\lesssim$ $\displaystyle\sum_{N_{1}\sim N_{2}\gtrsim N\atop
N_{1}\sim N_{2}\gg N_{3}\geq
N_{4}}N_{2}^{-1}(N_{3}N_{4})^{-\frac{3}{4}}\prod^{4}_{j=1}\|u_{j}\|_{X_{\delta}^{1,\frac{1}{2}+}}\lesssim
N^{-1}\prod^{4}_{j=1}\|u_{j}\|_{X_{\delta}^{1,\frac{1}{2}+}}.$ (5.53)

2\. In the semi-resonant case, it is easy to see that $N_{2}\sim N_{3}\gg
N_{4}\sim N_{1}$. Bilinear estimates (5.5) imply that (5.51) is controlled by

$\displaystyle\sum_{N_{2}\sim N_{3}\gtrsim N\atop N_{2}\sim N_{3}\gg N_{4}\sim
N_{1}}\frac{N_{1}^{3}N_{2}^{2}m(N_{1})}{m(N_{2})m(N_{3})m(N_{4})}\|u_{1}u_{3}\|_{L^{2}_{x,t}}\|u_{2}u_{4}\|_{L^{2}_{x,t}}$
$\displaystyle\lesssim$ $\displaystyle\sum_{N_{2}\sim N_{3}\gtrsim N\atop
N_{2}\sim N_{3}\gg N_{4}\sim
N_{1}}\frac{N_{1}^{3}N_{3}^{-2}}{m(N_{2})m(N_{3})}\prod^{4}_{j=1}\|u_{j}\|_{X_{\delta}^{0,\frac{1}{2}+}}$
$\displaystyle\lesssim$ $\displaystyle\sum_{N_{2}\sim N_{3}\gtrsim N\atop
N_{2}\sim N_{3}\gg N_{4}\sim
N_{1}}N_{1}N_{2}^{-\frac{3}{4}}N_{3}^{-\frac{11}{4}}\prod^{4}_{j=1}\|u_{j}\|_{X_{\delta}^{1,\frac{1}{2}+}}\lesssim
N^{-\frac{5}{2}}\prod^{4}_{j=1}\|u_{j}\|_{X_{\delta}^{1,\frac{1}{2}+}}.$
(5.54)

3\. In the resonant case, we have $N_{1}\sim N_{2}\sim N_{3}\sim N_{4}$.
Strichartz estimates (5.2) yield that

$\displaystyle\eqref{I4}\lesssim$ $\displaystyle\
\delta^{\frac{1}{2}}\sum_{N_{1}\sim N_{2}\sim N_{3}\sim N_{4}\gtrsim
N}\frac{N_{1}^{3}N_{2}^{2}m(N_{1})}{m(N_{2})m(N_{3})m(N_{4})}\prod^{4}_{j=1}\|u_{j}\|_{L^{8}_{t}L^{4}_{x}}$
$\displaystyle\lesssim$ $\displaystyle\ \delta^{\frac{1}{2}}\sum_{N_{1}\sim
N_{2}\sim N_{3}\sim N_{4}\gtrsim
N}\frac{N_{1}^{3}N_{2}^{2}}{m(N_{3})m(N_{4})}\prod^{4}_{j=1}\|u_{j}\|_{X_{\delta}^{-\frac{3}{8},\frac{1}{2}+}}$
$\displaystyle\lesssim$ $\displaystyle\
\delta^{\frac{1}{2}}N^{2s-2}\sum_{N_{1}\gtrsim
N}N_{1}^{\frac{3}{2}-2s}\prod^{4}_{j=1}\|u_{j}\|_{X_{\delta}^{1,\frac{1}{2}+}}\lesssim\delta^{\frac{1}{2}}N^{-\frac{1}{2}}\prod^{4}_{j=1}\|u_{j}\|_{X_{\delta}^{1,\frac{1}{2}+}}.$
(5.55)

We now complete the proof. $\hfill\Box$

###  5.4 Global approximation in low regularity spaces

In this subsection, we shall extend the long-time asymptotic behavior to the
rough data. At the beginning, we show that the Beals-Coifman solution is equal
to the strong solution given by (1.29).

###### Lemma 5.4

Let $u_{0}\in H^{4,1}(\mathbb{R})$, the Beals-Coifman solution and the strong
solution are the same (up to a measure zero set)

$\displaystyle
u=\frac{1}{\pi}\left(\int\mu\big{(}w^{+}+w^{-}\big{)}\right)_{12}=\text{e}^{-t\partial_{x}^{5}}{u_{0}}-\int^{t}_{0}\text{e}^{-(t-t^{\prime})\partial_{x}^{5}}{F(u)}(t^{\prime})\,\text{d}t^{\prime}$

in $[-T,T]$ where $T$ is given as in Theorem 5.1.

Proof.For $u_{0}\in H^{4,1}(\mathbb{R})$, there exists a Cauchy sequence
$u_{0,k}\in\mathscr{S}(\mathbb{R})$, such that

$\displaystyle\lim_{k\rightarrow\infty}\|u_{0,k}-u_{0}\|_{H^{4,1}}=0,\quad\text{and}\quad\|u_{0,k}\|_{H^{4}}\leq\|u_{0,k}\|_{H^{4,1}}\leq
C.$

On the one hand, from Theorem 5.1, there exist a strong solution $u_{k}$ with
initial data $u_{0,k}$ such that for $b>1/2$,

$\displaystyle\|u_{k}-u_{l}\|_{X^{4,b}}\lesssim\|u_{0,k}-u_{0,l}\|_{H^{4}}\rightarrow
0\quad\text{as}\quad k,l\rightarrow\infty,$

Therefore, there exists $u_{\infty}$ such that

$\displaystyle\sup_{t\in[-T,T]}\|u_{k}-u_{\infty}\|_{L^{\infty}}\lesssim\|u_{k}-u_{\infty}\|_{X^{4,b}}\rightarrow
0,\quad\text{as}\quad k\rightarrow\infty,$ (5.56)

On the other hand, from the inverse scattering transform, we have the Beals-
Coifman solutions $\tilde{u}_{k}$ with initial data $u_{0,k}$. By utilizing
the bijectivity of the transformation and bi-Lipschitz continuity, we know
reflection coefficients satisfy

$\displaystyle\|r_{k}-r_{l}\|_{H^{1,4}}\lesssim\|u_{0,k}-u_{0,l}\|_{H^{4,1}}\rightarrow
0\quad as\quad k,l\rightarrow\infty.$

By the resolvent estimates,

$\displaystyle\|\tilde{u}_{k}-\tilde{u}_{l}\|_{L^{\infty}}\lesssim\|r_{k}-r_{l}\|_{H^{1}}\rightarrow
0\quad as\quad k,l\rightarrow\infty.$

Therefore, there exists $\tilde{u}_{\infty}$ such that

$\displaystyle\tilde{u}_{\infty}=\lim_{k\rightarrow\infty}\tilde{u}_{k}.$
(5.57)

Since $u_{0,k}\in\mathscr{S}(\mathbb{R})$, $u_{k}$ and $\tilde{u}_{k}$ are
also Schwartz functions, then $u_{k}=\tilde{u}_{k}$. Therefore, we can get the
conclusion from (5.56) and (5.57). $\hfill\Box$

In the end, we divide the proof of Theorem 1.3 into the following two
theorems. We first consider the long-time behavior in the Sobolev space
$H^{1,1}(\mathbb{R})$. It is well known that $u(\cdot,t)$ is uniformly bounded
in $H^{1}(\mathbb{R})$. Indeed, from the mass conservation and energy
conservation, we know that

$\displaystyle\|u\|_{H^{1}(\mathbb{R})}=\|u\|_{L^{2}}+\|u\|_{\dot{H}^{1}}\leq\|u_{0}\|_{L^{2}}+\|\partial_{x}u_{0}\|_{L^{2}}+\|u_{0}\|^{2}_{L^{4}}.$

From Gagliardo-Nirenberg’s inequality and Young’s inequality,

$\displaystyle\|u_{0}\|^{2}_{L^{4}}\lesssim\|\partial_{x}u_{0}\|^{1/2}_{L^{2}}\|u_{0}\|^{3/2}_{L^{2}}\lesssim\frac{1}{\epsilon}\|\partial_{x}u_{0}\|_{L^{2}}+\epsilon\|u_{0}\|^{3}_{L^{2}}.$

We may choose $\epsilon$ satisfying $\epsilon\|u_{0}\|^{2}_{L^{2}}\leq 1$, and
get that

$\displaystyle\|u(\cdot,t)\|_{H^{1}}\lesssim\|u_{0}\|_{H^{1}}.$ (5.58)

###### Theorem 5.2

Let $u_{0}\in H^{1,1}(\mathbb{R})$, the strong solution given by the integral
form (1.29) has the same asymptotic behavior as in Theorem 1.1.

Proof.For $u_{0}\in H^{1,1}(\mathbb{R})$, there exists a sequence $u_{0,k}\in
H^{4,1}(\mathbb{R})$, such that

$\displaystyle\lim_{k\rightarrow\infty}\|u_{0,k}-u_{0}\|_{H^{1,1}}=0,\quad
and\quad\|u_{0,k}\|_{H^{1}}\leq\|u_{0,k}\|_{H^{1,1}}\leq C.$

Following the same strategy as in the proof of Lemma 5.4, we know that there
exist a pointwise limit $u_{\infty}$ of the strong solutions $u_{k}$, and a
pointwise limit $\tilde{u}_{\infty}$ of Beals-Coifman solutions
$\tilde{u}_{k}$, with initial data $u_{0,k}$. Moreover, one has
$u_{\infty}=\tilde{u}_{\infty}$ pointwise for $t\in[-T,T]$. Since $u(\cdot,t)$
is uniformly bounded in $H^{1}(\mathbb{R})$ as (5.58), one can repeat the
above argument many times to extend the time interval to whole line
$\mathbb{R}$. Because $\tilde{u}_{\infty}$ has asymptotic behavior in Theorem
1.1, $u_{\infty}$ also does. $\hfill\Box$

###### Theorem 5.3

Let $u_{0}\in H^{s,1}(\mathbb{R})$, $19/22<s<1$, the strong solution given by
the integral form (1.29) has the same asymptotic behavior as in Theorem 1.1.

Proof.For $u_{0}\in H^{s,1}(\mathbb{R})$, $19/22<s<1$, there exists a sequence
$u_{0,k}\in H^{4,1}(\mathbb{R})$, such that

$\displaystyle\lim_{k\rightarrow\infty}\|u_{0,k}-u_{0}\|_{H^{s,1}}=0,\quad
and\quad\|u_{0,k}\|_{H^{s}}\leq\|u_{0,k}\|_{H^{s,1}}\leq C.$

Denote $u_{\infty}$ and $\tilde{u}_{\infty}$ the pointwise limit of the strong
solutions $u_{k}$ and Beals-Coifman solutions $\tilde{u}_{k}$ with initial
data $u_{0,k}$, respectively. As Lemma 5.4, we know that
$u_{\infty}=\tilde{u}_{\infty}$ pointwise for $t\in[-T,T]$.

From the global well-posedness theory Theorem 1.2, $u_{\infty}$ exists in
$H^{s}$ globally. One can also construct $\tilde{u}_{\infty}$ globally.
Suppose that $u_{\infty}=\tilde{u}_{\infty}$ does not hold for all $t$. We may
assume

$\displaystyle T_{*}=\inf\big{\\{}t\geq
0;u_{\infty}(t)\neq\tilde{u}_{\infty}(t)\big{\\}}.$ (5.59)

From (5.14), we know that

$\displaystyle\|u(\cdot,T_{*})\|_{H^{s}}\leq C_{1}T_{*}^{M}+C_{2},$ (5.60)

then as in the proof of Theorem 5.1, there exists $t_{0}>0$ such that

$\displaystyle\sup_{t\in[0,T_{*}+t_{0}]}\|u_{k}-u_{\infty}\|_{L^{\infty}}\lesssim\|u_{k}-u_{\infty}\|_{X^{s,b}}\rightarrow
0,\quad as\quad k\rightarrow\infty,$

Combining

$\displaystyle\sup_{t\in[0,T_{*}+t_{0}]}\|\tilde{u}_{k}-\tilde{u}_{\infty}\|_{L^{\infty}}\rightarrow
0,\quad as\quad k\rightarrow\infty,\quad and\quad u_{k}=\tilde{u}_{k},$

we can conclude that $u_{\infty}(t)=\tilde{u}_{\infty}(t)$ pointwise for
$t\in[0,T_{*}+t_{0}]$, this is a contradiction with (5.59). Hence we can
obtain that $u_{\infty}(t)=\tilde{u}_{\infty}(t)$ pointwise for all $t\geq 0$.
Since $\tilde{u}_{\infty}$ has asymptotic behavior in Theorem 1.1,
$u_{\infty}$ also does. $\hfill\Box$

Acknowledgments.

N. Liu is supported by the China Postdoctoral Science Foundation under Grant
Nos. 2019TQ0041, 2019M660553; M. Chen is supported in part by the China
Postdoctoral Science Foundation under Grant No. 2019M650019.

##  Appendix A Fourth order Painlevé II RH problem

Fig. 10: The oriented contour $P$.

Let $P$ denote the contour $P=P_{1}\cup P_{2}$ oriented as in Fig. 10, where

$\displaystyle P_{1}=\\{l\text{e}^{\frac{\pi\text{i}}{6}}|l\geq
0\\}\cup\\{l\text{e}^{\frac{5\pi\text{i}}{6}}|l\geq 0\\},\quad
P_{2}=\\{l\text{e}^{-\frac{\pi\text{i}}{6}}|l\geq
0\\}\cup\\{l\text{e}^{-\frac{5\pi\text{i}}{6}}|l\geq 0\\}.$

Fourth order Painlevé II RH problem A.1. Let $s\in{\mathbb{R}}$ be a real
number. Find an analytic function $M^{P}(y;z)$ in ${\mathbb{C}}\setminus P$
parametrized by $y\in{\mathbb{R}},s\in{\mathbb{R}}$ such that:

1\. For $z\in P$, the continuous boundary values $M^{P}_{\pm}(y;z)$ satisfy

$M^{P}_{+}(y;z)=M^{P}_{-}(y;z)v^{P}(s,y;z),~{}~{}z\in P,$ (A.1)

where the jump matrix $v^{P}(s,y;z)$ is defined by

$v^{P}(s,y;z)=\left\\{\begin{aligned} &\begin{pmatrix}1&~{}0\\\
s\text{e}^{2\text{i}(\frac{4}{5}z^{5}-yz)}&~{}1\end{pmatrix},\qquad~{}z\in
P_{1},\\\ &\begin{pmatrix}1&~{}-s\text{e}^{-2\text{i}(\frac{4}{5}z^{5}-yz)}\\\
0&~{}1\end{pmatrix},\quad z\in P_{2}.\end{aligned}\right.$ (A.2)

2\. $M^{P}(y;z)=I+O(\frac{1}{z})$, as $z\rightarrow\infty$.

###### Lemma A.1

The RH problem A.1 has a unique solution $M^{P}(y;z)$ for each
$y\in{\mathbb{R}}$. Moreover, there exists smooth functions
$\\{M_{j}^{P}(y)\\}^{4}$ of $y\in{\mathbb{R}}$ with decay as
$y\rightarrow-\infty$ such that

$M^{P}(y;z)=I+\sum_{j=1}^{4}\frac{M_{j}^{P}(y)}{z^{j}}+O(z^{-5}),\quad
z\rightarrow\infty,$ (A.3)

uniformly for $y$ in compact subsets of ${\mathbb{R}}$ and for $\arg
z\in[0,2\pi]$. The leading coefficient $M_{1}^{Y}$ is given by

$M_{1}^{P}(y)=\text{i}\begin{pmatrix}-2\int^{y}_{-\infty}u_{p}^{2}(y^{\prime})\text{d}y^{\prime}~{}&u_{p}(y)\\\
-u_{p}(y)~{}&2\int^{y}_{-\infty}u_{p}^{2}(y^{\prime})\text{d}y^{\prime}\end{pmatrix},$
(A.4)

where the real-valued function $u_{p}(y)$ satisfies the following fourth order
Painlevé II equation (see [35])

$u_{p}^{{}^{\prime\prime\prime\prime}}(y)-40u_{p}^{2}(y)u_{p}^{{}^{\prime\prime}}(y)-40u_{p}(y)u_{p}^{\prime
2}(y)+96u_{p}^{5}(y)-4yu_{p}(y)=0.$ (A.5)

##  Appendix B  Model RH problem in Region III

Fig. 11: The oriented contour $Z$.

Given a number $z_{0}\geq 0$, let $Z$ denote the contour $Z=Z_{1}\cup
Z_{1}\cup Z_{3}$, where the line segments

$\displaystyle Z_{1}$
$\displaystyle=\\{z_{0}+l\text{e}^{\frac{\pi\text{i}}{6}}|l\geq
0\\}\cup\\{-z_{0}+l\text{e}^{\frac{5\pi\text{i}}{6}}|l\geq 0\\},$ (B.1)
$\displaystyle Z_{2}$
$\displaystyle=\\{z_{0}+l\text{e}^{-\frac{\pi\text{i}}{6}}|l\geq
0\\}\cup\\{-z_{0}+l\text{e}^{-\frac{5\pi\text{i}}{6}}|l\geq 0\\},$
$\displaystyle Z_{3}$ $\displaystyle=\\{l|-z_{0}\leq l\leq z_{0}\\}$

are oriented as in Fig. 11.
Model RH problem B.1. Find an analytic function $M^{Z}(y;z,z_{0})$ in
${\mathbb{C}}\setminus Z$ parametrized by $y>0,s\in{\mathbb{R}},z_{0}\geq 0$
such that:

1\. The continuous boundary values $M^{Z}_{\pm}(y;z,z_{0})$ satisfy the jump
condition

$M^{Z}_{+}(y;z,z_{0})=M^{Z}_{-}(y;z,z_{0})v^{Z}(s,y;z,z_{0}),~{}~{}~{}z\in Z,$
(B.2)

where the jump matrix $v^{Z}(s,y,z;z_{0})$ is defined by

$v^{Z}(s,y;z,z_{0})=\left\\{\begin{aligned} &\begin{pmatrix}1&~{}0\\\
s\text{e}^{2\text{i}(\frac{4}{5}z^{5}-yz)}&~{}1\end{pmatrix},\qquad\qquad\qquad\qquad~{}~{}~{}~{}~{}~{}~{}z\in
Z_{1},\\\ &\begin{pmatrix}1&~{}-s\text{e}^{-2\text{i}(\frac{4}{5}z^{5}-yz)}\\\
0&~{}1\end{pmatrix},~{}~{}\quad\qquad\qquad\qquad~{}~{}~{}z\in Z_{2},\\\
&\begin{pmatrix}1&~{}-s\text{e}^{-2\text{i}(\frac{4}{5}z^{5}-yz)}\\\
0&~{}1\end{pmatrix}\begin{pmatrix}1&~{}0\\\
s\text{e}^{2\text{i}(\frac{4}{5}z^{5}-yz)}&~{}1\end{pmatrix},~{}~{}z\in
Z_{3}.\end{aligned}\right.$ (B.3)

2\. As $z\rightarrow\infty$, $M^{Z}(y;z,z_{0})=I+O(\frac{1}{z})$.

###### Lemma B.2

Define the parameter subset

$\mathbb{P}=\\{(y,t,z_{0})\in{\mathbb{R}}^{3}|0<y<C_{1},t\geq
1,\sqrt[4]{y}/\sqrt{2}\leq z_{0}\leq C_{2}\\},$ (B.4)

where $C_{1},C_{2}>0$ are constants. Then for $(y,t,z_{0})\in\mathbb{P}$, the
RH problem B.1 has a unique solution $M^{Z}(s,y;z,z_{0})$ which satisfies

$M^{Z}(y;z,z_{0})=I+\frac{\text{i}}{z}\begin{pmatrix}-2\int^{y}_{\infty}u_{p}^{2}(y^{\prime})\text{d}y^{\prime}~{}&-u_{p}(y)\\\
u_{p}(y)~{}&2\int^{y}_{\infty}u_{p}^{2}(y^{\prime})\text{d}y^{\prime}\end{pmatrix}+O\bigg{(}\frac{1}{z^{2}}\bigg{)},\quad
z\rightarrow\infty,$ (B.5)

where $u_{p}(y)$ denotes the solution of the fourth order Painlevé II equation
(A.5) and $M^{Z}(y,z,z_{0})$ is uniformly bounded for
$z\in{\mathbb{C}}\setminus Z$. Furthermore, $M^{Z}$ obeys the symmetries

$M^{Z}(y;-z,z_{0})=\overline{M^{Z}(y;\bar{z},z_{0})}=\sigma_{1}M^{Z}(y;z,z_{0})\sigma_{1}.$
(B.6)

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# Approximations of expectations under infinite product measures††thanks:
Solan acknowledges the support of the Israel Science Foundation, grant
#211/22. We are grateful to John (Yehuda) Levy for a stimulating discussion.

Galit Ashkenazi-Golan111London School of Economics and Political Science,
Houghton Street, London WC2A 2AE, UK, E-mail<EMAIL_ADDRESS>János
Flesch222Department of Quantitative Economics, Maastricht University, P.O.Box
616, 6200 MD, The Netherlands. E-mail<EMAIL_ADDRESS>Arkadi Predtetchinski333Department of Economics, Maastricht University,
P.O.Box 616, 6200 MD, The Netherlands. E-mail:
<EMAIL_ADDRESS>Eilon Solan444School of Mathematical
Sciences, Tel-Aviv University, Tel-Aviv, Israel, 6997800, E-mail:
<EMAIL_ADDRESS>
###### Abstract

We are given a bounded Borel-measurable real-valued function on a product of
countably many Polish spaces, and a product probability measure. We are
interested in points in the product space that can be used to approximate the
expected value of this function. We define two notions. A point is called a
_weak $\epsilon$-approximation_, where $\epsilon\geq 0$, if the Dirac measure
on this point, except in finitely many coordinates where another measure can
be taken, gives an expected value that is $\epsilon$-close to the original
expected value. A point is called a _strong $\epsilon$-approximation_ if the
same holds under the restriction that in those finitely many coordinates the
measure is equal to the original one. We prove that both the set of weak
0-approximation points and the set of strong $\epsilon$-approximation points,
for any $\epsilon>0$, have measure 1 under the original measure. Finally, we
provide two applications: (i) in Game Theory on the minmax guarantee levels of
the players in games with infinitely many players, and (ii) in Decision Theory
on the set of feasible expected payoffs in infinite duration problems.

Keywords: Infinite product space, product measure, universally measurable set,
strategic-form game, minmax value.

MSC 2020 Classification: 28A35, 28C20, 91A07.

##  1 Introduction

Approximations of expected payoffs. Suppose that
$\sigma=\otimes_{i\in\mathbb{N}}\sigma_{i}$ is a product probability measure
on a product space $X=\times_{i\in\mathbb{N}}X_{i}$, where each $X_{i}$ is a
Polish space, and $f:X\to\mathbb{R}$ is a bounded and Borel-measurable
function. In this paper, we ask whether $\mathbb{E}_{\sigma}[f]$ can be
approximated by an expectation $\mathbb{E}_{\tau}[f]$, for some “simple”
product measure $\tau$.

If $f$ is a tail function, then there necessarily exists $x\in X$ such that
$f(x)=\mathbb{E}_{\sigma}[f]$; in fact, almost every (under $\sigma$) point of
$X$ has the stated property. In general, no such point $x$ need to exist.

We are interested in two types of approximating measures. We say that
$x=(x_{i})_{i\in\mathbb{N}}\in X$ is a _weak $\epsilon$-approximation_ of the
expectation of $f$ under $\sigma$ if there is some product probability measure
$\tau=\otimes_{i\in\mathbb{N}}\tau_{i}$, such that (i)
$|\mathbb{E}_{\tau}[f]-\mathbb{E}_{\sigma}[f]|\leq\epsilon$, and (ii)
$\tau_{i}$ is the Dirac measure on $x_{i}$ for all but finitely many $i$’s. We
will say that $x\in X$ is a _strong $\epsilon$-approximation_ of the
expectation of $f$ under $\sigma$ if the same holds, where in addition (iii)
$\tau_{i}$ is either $\sigma_{i}$ or a Dirac measure, for every
$i\in\mathbb{N}$.

Our main results are that (i) the set of weak 0-approximations have
$\sigma$-measure 1 (cf. Theorem 4), and (ii) the set of strong
$\epsilon$-approximations have $\sigma$-measure 1 for every $\epsilon>0$ (cf.
Theorem 6). As an example demonstrates (cf. Example 7), a strong
0-approximation does not always exist.

Application in Game Theory. The first motivation for our study comes from Game
Theory, and more specifically, from the study of strategic-form games with
countably many players. Games with countably many players have been studied,
e.g., by Peleg [7], Voorneveld [11], Rachmilevitch [8], and Ashkenazi-Golan,
Flesch, Predtetchinski and Solan [2]. These studies have mainly revolved
around the question of (non)existence of Nash ($\epsilon$-)equilibrium. In
contrast, here we focus on the notion of the minmax value.

The minmax value of a player is one of the central concepts in game theory. It
is defined as the maximal payoff that the player can guarantee to obtain, when
all other players try to decrease her payoff. The underlying assumption on the
behaviour of the punishing players is that they cannot correlate their
lotteries, and thus their joint strategy is a product measure.

Ashkenazi-Golan, Flesch, Predtetchinski and Solan [2] introduced a notion of
the _finitistic minmax value_. This is a version of the minmax where only
finitely many of the punishing players are allowed to randomize; the others
are required to choose a Dirac measure. It is clear that the finitistic minmax
value is not lower than the minmax value. Whether the two values are in fact
equal is an open problem. Here we answer this question in the affirmative (cf.
Theorem 8). This is achieved using the result (ii) on strong
$\epsilon$-approximations.

Application in Decision Theory. The second motivation comes from decision
theory. In dynamic decision problems on the infinite horizon, _Markov
strategies_ play an important role (e.g., Hill [5], Kiefer, Mayr,
Shirmohammadi, Totzke [6], Sudderth [10]). These are strategies in which, at
each stage, the choice of action may depend on the current stage and the
current state, but not on past events. Such strategies are desirable because
of their simplicity, as the decision maker’s behavior depends only on the
current payoff relevant parameters. A strategy is called _eventually pure_ if
after some stage it does not use randomization (cf. Flesch, Herings, Maes,
Predtetchinski [4]).

Using our result (i) on weak 0-approximations, we show that in a class of
decision problems on the infinite horizon, the set of expected payoffs that
the decision maker can obtain by Markov strategies is equal to the set of
expected payoffs under eventually pure Markov strategies. In other words,
using Markov strategies that involve a randomization at infinitely many stages
does not generate additional expected payoffs.

Structure of the paper. Definitions appear in Section 2, the main results and
their proofs are provided in Section 3, and the applications to Game Theory
and Decision Theory are discussed in Section 4. A few concluding remarks are
given, including an open problem, in Section 5.

##  2 Approximations of expectations

Let $\mathbb{N}=\\{1,2,\ldots\\}$. For each $i\in\mathbb{N}$, let $X_{i}$ be a
Polish space. Let $X=\times_{i\in\mathbb{N}}X_{i}$ denote the product space,
which we endow with the product topology.

For each $i\in\mathbb{N}$, let $\Sigma_{i}$ be the set of Borel probability
measures on $X_{i}$, and let $\Sigma$ be the set of all product Borel
probability measures on $X$, i.e., the measures
$\sigma=\otimes_{i\in\mathbb{N}}\sigma_{i}$ where $\sigma_{i}\in\Sigma_{i}$
for each $i\in\mathbb{N}$. With a slight abuse of notation, we write $x_{i}$
to denote both a point of $X_{i}$ and the Dirac measure concentrated on
$x_{i}$, and we use the same symbol $\sigma$ to denote both a measure in
$\Sigma$ and its unique extension to the sigma-algebra of all universally
measurable sets of $X$.

We define two notions for the approximations of expectations under the
measures in $\Sigma$.

###### Definition 1

Let $\epsilon\geq 0$. For a product measure $\sigma\in\Sigma$ and a bounded
Borel-measurable function $f:X\to\mathbb{R}$, a point $x\in X$ is said to be a
_weak $\epsilon$-approximation of the expectation of $f$ under $\sigma$_ if
there is $n\in\mathbb{N}$ and a Borel probability measure $\tau_{i}$ on
$X_{i}$ for each $i<n$, such that:

$\big{|}\mathbb{E}_{\tau_{1}\otimes\cdots\otimes\tau_{n-1}\otimes x_{n}\otimes
x_{n+1}\otimes\cdots}[f]-\mathbb{E}_{\sigma}[f]\big{|}\,\leq\,\epsilon.$ (1)

Intuitively, Eq. (1) means that the expectation changes by at most $\epsilon$
if we replace $\sigma$ by the measure that is the Dirac measure on $x_{i}$ in
each coordinate $i\geq n$ and $\tau_{i}$ in each coordinate $i<n$.

###### Definition 2

Let $\epsilon\geq 0$. For a product measure $\sigma\in\Sigma$ and a bounded
Borel-measurable function $f:X\to\mathbb{R}$, a point $x\in X$ is said to be a
_strong $\epsilon$-approximation of the expectation of $f$ under $\sigma$_ if
there is $n\in\mathbb{N}$ such that:

$\big{|}\mathbb{E}_{\sigma_{1}\otimes\cdots\otimes\sigma_{n-1}\otimes
x_{n}\otimes
x_{n+1}\otimes\cdots}[f]-\mathbb{E}_{\sigma}[f]\big{|}\,\leq\,\epsilon.$ (2)

Thus, Eq. (2) means that the expectation changes by at most $\epsilon$ if we
replace $\sigma$ by the Dirac measure on $x_{i}$ in each coordinate $i\geq n$.
Obviously, if $x\in X$ is a strong $\epsilon$-approximation of the expectation
of $f$ under $\sigma$, then $x$ is also a weak $\epsilon$-approximation.

Let $W_{\epsilon}(\sigma,f)\subseteq X$ and $S_{\epsilon}(\sigma,f)\subseteq
X$ denote the sets of weak and respectively strong $\epsilon$-approximations
of the expectation of $f$ under $\sigma$. Thus,
$W_{\epsilon}(\sigma,f)\subseteq S_{\epsilon}(\sigma,f)$. It follows from the
following lemma that both $W_{\epsilon}(\sigma,f)$ and
$S_{\epsilon}(\sigma,f)$ are universally measurable, and in particular, both
sets are measurable with respect to the measure $\sigma$.

###### Lemma 3

For each $\epsilon\geq 0$, each product measure $\sigma\in\Sigma$, and each
bounded Borel-measurable function $f:X\to\mathbb{R}$, the set
$W_{\epsilon}(\sigma,f)\subseteq X$ is analytic and the set
$S_{\epsilon}(\sigma,f)\subseteq X$ is Borel.

Proof: For each $n\in\mathbb{N}$, we equip $\Sigma_{n}$ with the topology of
weak convergence, and $\Sigma^{n}:=\times_{i=1}^{n}\Sigma_{i}$ with the
product topology. Note that $\Sigma^{n}$ is a Polish space. For each
$n\in\mathbb{N}$, the function $X\times\Sigma^{n}\to\mathbb{R}$ given by
$(x,\tau_{1},\ldots,\tau_{n})\mapsto\mathbb{E}_{\tau_{1}\otimes\cdots\otimes\tau_{n-1}\otimes\tau_{n}\otimes
x_{n+1}\otimes x_{n+2}\otimes\cdots}[f]$ is Borel-measurable (Aliprantis and
Border [1, Theorem 15.13]). Hence the set $B^{n}\subseteq X\times\Sigma^{n}$
consisting of the points $(x,\tau_{1},\ldots,\tau_{n})$ satisfying Eq. (1) is
a Borel set. Let $A^{n}$ be the projection of $B^{n}$ onto the first
coordinate. Then $A^{n}$ is analytic. Since
$W_{\epsilon}(\sigma,f)=\cup_{n\in\mathbb{N}}A^{n}$, the result follows.

The fact that $S_{\epsilon}(\sigma,f)$ is a Borel set follows since the map
$X\to\mathbb{R}$ given by
$x\mapsto\linebreak\mathbb{E}_{\sigma_{1}\otimes\cdots\otimes\sigma_{n-1}\otimes
x_{n}\otimes x_{n+1}\otimes\cdots}[f]$ is Borel-measurable for each
$n\in\mathbb{N}$ (Aliprantis and Border [1, Theorem 15.13]). $\Box$

##  3 Results

###### Theorem 4

For each product measure $\sigma\in\Sigma$ and each bounded Borel-measurable
function $f:X\to\mathbb{R}$, the set $W_{0}(\sigma,f)$ has $\sigma$-measure
$1$.

Proof: Fix a product measure $\sigma\in\Sigma$ and a bounded Borel-measurable
function $f:X\to\mathbb{R}$.

Let $\sim$ be the following tail equivalence relation: for $x,y\in X$ let
$x\sim y$ if there is some $n\in\mathbb{N}$ such that $x_{i}=y_{i}$ for all
$i\geq n$. Let $\mathscr{Z}$ denote the collection of all equivalence classes
of $\sim$. For $Z\in\mathscr{Z}$, let $I_{Z}$ be the convex hull of
$f(Z)=\\{f(x):x\in Z\\}$; in particular, $I_{Z}$ is an interval, with or
without its endpoints.

Let $r=\mathbb{E}_{\sigma}[f]$. Define

$\displaystyle Z_{-}$
$\displaystyle=\bigcup\\{Z\in\mathscr{Z}:I_{Z}\subseteq(-\infty,r)\\},$
$\displaystyle Z_{0}\ $ $\displaystyle=\bigcup\\{Z\in\mathscr{Z}:I_{Z}\ni
r\\},$ $\displaystyle Z_{+}$
$\displaystyle=\bigcup\\{Z\in\mathscr{Z}:I_{Z}\subseteq(r,+\infty)\\}.$

The statement of Theorem 4 follows from Claims 1 and 2 below.

Claim 0: $Z_{-}$ and $Z_{+}$ are co-analytic subsets of $X$.

It follows from Claim 0 that $Z_{0}$ is analytic, and hence $Z_{-}$, $Z_{0}$,
and $Z_{+}$ are measurable with respect to $\sigma$.

Proof of claim 0: Let us write $\sim$ to denote the set $\\{(x,y)\in X\times
X:x\sim y\\}$. Then $\sim$ is a Borel subset of $X\times X$, as can be seen by
rewriting it as

$\bigcup_{n\in\mathbb{N}}\bigcap_{i\geq n}\\{(x,y)\in X\times
X:x_{i}=y_{i}\\}.$

The set $X\setminus Z_{-}$ is a projection of the Borel set

$\sim\bigcap\\{(x,y)\in X\times X:f(y)\geq r\\}$

onto its first coordinate, and therefore analytic. The argument that
$X\setminus Z_{+}$ is analytic is similar.

Claim 1: $Z_{0}\subseteq W_{0}(\sigma,f)$.

Proof of Claim 1: Take any $Z\in Z_{0}$. We show that $Z\subseteq
W_{0}(\sigma,f)$.

As $Z\in Z_{0}$, we have $r\in I_{Z}$, and there exist points $x,y\in Z$ such
that $f(x)\leq r\leq f(y)$. Let $n\in\mathbb{N}$ be such that $x_{i}=y_{i}$
for each $i>n$.

For each $k=1,\ldots,n+1$, let
$z_{k}=(y_{1},\ldots,y_{k-1},x_{k},x_{k+1},\ldots)$. In particular, $z_{1}=x$
and $z_{n+1}=y$, and $z_{k}\in Z$ for each $k=1,\ldots,n+1$. Because $z_{1}=x$
and $z_{n+1}=y$, the interval $[f(x),f(y)]$ is covered by the union of the
intervals $[f(z_{k}),f(z_{k+1})]$ with $k=1,\ldots,n$. Consequently, there is
$k\in\\{1,\ldots,n\\}$ such that $r\in[f(z_{k}),f(z_{k+1})]$. Note that
$z_{k}$ and $z_{k+1}$ only differ in coordinate $k$.

Let $\alpha\in[0,1]$ be such that $\alpha\cdot
f(z_{k})+(1-\alpha)f(z_{k+1})=r$. Define the measure $\tau_{k}$ on $X_{k}$ as
follows: $\tau_{k}$ places probability $\alpha$ on $x_{k}$ and probability
$1-\alpha$ on $y_{k}$. Then,

$\mathbb{E}_{y_{1}\otimes\cdots\otimes y_{k-1}\otimes\tau_{k}\otimes
x_{k+1}\otimes x_{k+2}\otimes\cdots}[f]\,=\,\alpha\cdot
f(z_{k})+(1-\alpha)f(z_{k+1})\,=\,r\,=\,\mathbb{E}_{\sigma}[f].$

This means that $z_{k}=(y_{1},\ldots,y_{k-1},x_{k},x_{k+1},\ldots)$ is a weak
$0$-approximation of the expectation of $f$ under $\sigma$, so that $z_{k}\in
W_{0}(\sigma,f)$. Hence, so is each point in the equivalence class $Z$ of
$z_{k}$.

Claim 2: $Z_{0}$ has $\sigma$-measure 1.

Proof of Claim 2: The sets $Z_{-}$, $Z_{0}$, and $Z_{+}$ are tail. Since
$\sigma$ is a product measure, the $\sigma$-measure of each set is either $0$
or $1$ (see Lemma 10 in the Appendix). Since $Z_{-}$, $Z_{0}$, and $Z_{+}$
partition $X$, exactly one of them has $\sigma$-measure 1. It cannot be the
case that $\sigma(Z_{-})=1$, since then, as $Z_{-}\subseteq
f^{-1}(-\infty,r)$, we would obtain $\mathbb{E}_{\sigma}[f]<r$, a
contradiction. One similarly excludes the possibility that $\sigma(Z_{+})=1$.
Hence $\sigma(Z_{0})=1$. $\Box$

###### Remark 5

The proof of Claim 1 in the proof of Theorem 4 reveals the following
statement: For each product measure $\sigma\in\Sigma$ and each bounded Borel-
measurable function $f:X\to\mathbb{R}$, there exists a weak 0-approximation
$x\in W_{0}(\sigma,f)$ with the following property: there is a coordinate $n$
and a Borel probability measure $\tau_{n}$ on $X_{n}$ such that

$\mathbb{E}_{x_{1}\otimes\cdots\otimes x_{n-1}\otimes\tau_{n}\otimes
x_{n+1}\otimes x_{n+2}\otimes\cdots}[f]\,=\,\mathbb{E}_{\sigma}[f].$

That is, we only need to replace $x$ in a single coordinate by a Borel
probability measure.

###### Theorem 6

For each product measure $\sigma\in\Sigma$, each bounded Borel-measurable
function $f:X\to\mathbb{R}$, and each $\epsilon>0$, the set
$S_{\epsilon}(\sigma,f)$ has $\sigma$-measure $1$.

Proof: Let $\mathscr{G}_{n}$ be the sigma-algebra on $X$ generated by the
coordinate functions $x_{k}$, $k\geq n$. Let
$\mathscr{G}=\bigcap_{n\in\mathbb{N}}\mathscr{G}_{n}$. This is the tail sigma-
algebra on $X$. Consider $g_{n}=\mathbb{E}_{\sigma}[f|\mathscr{G}_{n}]$.

The process $\\{g_{n}\\}_{n\in\mathbb{N}}$ is a reverse martingale with
respect to the sequence $\\{\mathscr{G}_{n}\\}_{n\in\mathbb{N}}$ and the
measure $\sigma$. That is,
$\mathbb{E}_{\sigma}[g_{n}\mid\mathscr{G}_{n+1}]=g_{n+1}$. By Durrett [3,
Theorems 5.6.1 and 5.6.2], the sequence $g_{n}$ converges to a
$\mathscr{G}$-measurable limit $g_{\infty}$, almost surely with respect to
$\sigma$, and
$\mathbb{E}_{\sigma}[g_{\infty}]=\mathbb{E}_{\sigma}[g_{0}]=\mathbb{E}_{\sigma}[f]$.
Since $\sigma$ is a product measure, Kolmogorov’s 0-1 law implies that
$g_{\infty}$ is constant $\sigma$-almost surely. Hence $\sigma$-almost surely,
$g_{n}$ converges to $\mathbb{E}_{\sigma}[f]$. This means that $\sigma$-almost
surely, Eq. (2) holds for large $n\in\mathbb{N}$, as desired. $\Box$

As the following example shows, there are product measures $\sigma\in\Sigma$
and bounded Borel-measurable functions $f$ such that there is no strong
$0$-approximation of the expectation of $f$ under $\sigma$, i.e.,
$S_{0}(\sigma,f)=\emptyset$.

###### Example 7

Let $X_{i}=\\{0,1\\}$ for each $i\in\mathbb{N}$, and let $f(x)$ be 1 if
$x_{i}=1$ for each $i\in\mathbb{N}$ and $0$ otherwise. We identify
$\sigma_{i}\in\Sigma_{i}$ with $\sigma_{i}(\\{1\\})$, the probability of 1.
Take any $\sigma\in\Sigma$ such that (a) $0<\mathbb{E}_{\sigma}[f]$ and (b)
there are infinitely many $i\in\mathbb{N}$ such that $\sigma_{i}<1$ (for
example $\sigma_{i}=1-2^{-i}$ for each $i\in\mathbb{N}$). Then,
$S_{0}(\sigma,f)=\emptyset$. Indeed, consider an $x\in X$ and an
$n\in\mathbb{N}$. If $x_{i}=0$ for some $i\geq n$, then
$\mathbb{E}_{\sigma_{1}\otimes\cdots\otimes\sigma_{n-1}\otimes x_{n}\otimes
x_{n+1}\otimes\cdots}[f]=0$. If, on the other hand, $x_{i}=1$ for each $i\geq
n$, then

$\mathbb{E}_{\sigma}[f]=\prod_{i=1}^{\infty}\sigma_{i}<\prod_{i=1}^{n-1}\sigma_{i}=\mathbb{E}_{\sigma_{1}\otimes\cdots\otimes\sigma_{n-1}\otimes
x_{n}\otimes x_{n+1}\otimes\cdots}[f].$

##  4 Applications in Decision Theory and Game Theory

In this section, we provide two applications of our results: one in Game
Theory and one in Decision Theory.

Application in Game Theory. We apply Theorem 6 on strong
$\epsilon$-approximations of expectations to prove a result on the minmax
values of the players in strategic-form games with an infinite set of players.

We consider a strategic-form game in which the set of players is
$\\{0\\}\cup\mathbb{N}$. Player $0$’s action set is a nonempty finite set $A$,
whereas the action set of each player $i\in\mathbb{N}$ is a Polish space
$X_{i}$. As before, $X=\times_{i\in\mathbb{N}}X_{i}$. Player $0$’s payoff
function is $u:A\times X\to\mathbb{R}$, a bounded Borel-measurable function.
As before, $\Sigma$ denotes the set of product probability measures on $X$,
which is the set of strategy profiles of player $0$’s opponents. Let
$\Sigma^{*}\subseteq\Sigma$ denote the set of strategy profiles
$\sigma=\otimes_{i\in\mathbb{N}}\sigma_{i}$ of player $0$’s opponents such
that $\sigma_{i}$ is pure (i.e., a Dirac measure on $X_{i}$) for all but
finitely many $i\in\mathbb{N}$.

The minmax value and the finitistic minmax value of player $0$ are defined
respectively as

$\displaystyle v=$ $\displaystyle\inf_{\sigma\in\Sigma}\max_{a\in
A}\mathbb{E}_{a\otimes\sigma}[u],$ $\displaystyle v^{*}=$
$\displaystyle\inf_{\sigma\in\Sigma^{*}}\max_{a\in
A}\mathbb{E}_{a\otimes\sigma}[u].$

Intuitively, the minmax value of player 0 is the maximal expected payoff that
player 0 can guarantee to obtain, when all other players try to lower her
payoff. The minmax value is a fundamental concept, and the expected payoff of
a player in a Nash equilibrium can never be below her minmax value.

The finitistic minmax value is a variant where only finitely many opponents
can randomize over their actions. As mention in the introduction, the concept
of the finitistic minmax value played a crucial role in Ashkenazi-Golan,
Flesch, Predtetchinski and Solan [2] in their study of games with infinitely
many players.

It follows directly from the definitions that $v\leq v^{*}$. By using Theorem
6, it turns out, perhaps surprisingly, that in fact equality holds.

###### Theorem 8

$v=v^{*}$.

Proof: Since $\Sigma^{*}\subseteq\Sigma$, we always have $v\leq v^{*}$. To
prove Theorem 8 we therefore need to show that $v^{*}\leq v$. Fix
$\epsilon>0$. We show that $v^{*}\leq v+2\epsilon$.

For each $a\in A$, let $f_{a}=u(a,\cdot):X\to\mathbb{R}$. Let
$\sigma\in\Sigma$ be the $\epsilon$-minmax strategy profile of player $0$’s
opponents; that is, a strategy profile such that
$\mathbb{E}_{a\otimes\sigma}[u]=\mathbb{E}_{\sigma}[f_{a}]\leq v+\epsilon$ for
each $a\in A$.

Take any $x\in\bigcap_{a\in A}S_{\epsilon}(f_{a},\sigma)$, which is nonempty
by Theorem 6 and since $A$ is finite. Using once again the finiteness of $A$,
there exists $n\in\mathbb{N}$ such that Eq. (2) holds for each of the
functions $f_{a}$, and thus
$\mathbb{E}_{\sigma_{1}\otimes\cdots\otimes\sigma_{n-1}\otimes x_{n}\otimes
x_{n+1}\otimes\cdots}[f_{a}]\,\leq\,\mathbb{E}_{\sigma}[f_{a}]+\epsilon$ for
each $a\in A$. This implies that the strategy profile
$\tau=\sigma_{1}\otimes\cdots\otimes\sigma_{n-1}\otimes x_{n}\otimes
x_{n+1}\otimes\cdots\in\Sigma^{*}$ of player $0$’s opponents is a
$2\epsilon$-minmax strategy profile:
$\mathbb{E}_{a\otimes\tau}[u]=\mathbb{E}_{\tau}[f_{a}]\leq\mathbb{E}_{\sigma}[f_{a}]+\epsilon\leq
v+2\epsilon$ for each $a\in A$. Therefore, $v^{*}\leq v+2\epsilon$, as
desired. $\Box$

As the following example shows, the statement of Theorem 8 is not true if
player $0$’s action set $A$ is infinite.

###### Example 9

Suppose that $A=\mathbb{N}\times\\{0,1\\}$ and $X_{i}=\\{0,1\\}$ for each
$i\in\mathbb{N}$. Define player $0$’s payoff function as follows: for
$(n,j)\in A$ and $x\in X$, let $u((n,j),x)$ be $1$ if $j=x_{n}$ and $0$
otherwise. Intuitively, player 0’s action $(n,j)$ can be thought of naming an
opponent $n$ and an action $j$, and if opponent $n$’s action is indeed $j$,
then player 0 obtains payoff 1.

Then $v=\tfrac{1}{2}$, which follows by noticing that if $\sigma_{i}$ places
equal probability on $0$ and $1$ for each $i\in\mathbb{N}$ then
$\mathbb{E}_{(n,j)\otimes\sigma}[u]=\tfrac{1}{2}$ for each $(n,j)\in A$. On
the other hand, $v^{*}=1$, as in any strategy profile $\sigma\in\Sigma^{*}$
there is a player $n\in\mathbb{N}$ who places probability 1 on one of his
actions $j\in\\{0,1\\}$, and then $\mathbb{E}_{(n,j)\otimes\sigma}[u]=1$.

Application in Decision Theory. We apply Theorem 4 to the the following class
of decision problems on the infinite horizon: At each stage $t\in\mathbb{N}$,
a decision maker chooses an action $x_{t}$ from a Polish space $X_{t}$. If the
sequence of chosen actions is $(x_{1},x_{2},\ldots)\in
X=\times_{t\in\mathbb{N}}X_{t}$, then her payoff is $u(x_{1},x_{2},\ldots)$,
where $u:X\to\mathbb{R}$ is a bounded Borel-measurable function.

A _Markov strategy_ is an element of $\Sigma$, which is, as before, the set of
product probability measures on $X$. Intuitively, a Markov strategy
$\sigma=(\sigma_{t})_{t\in\mathbb{N}}\in\Sigma$ chooses an action at each
stage $t$ only depending on the current stage $t$, and thus not depending on
the past actions. Such a strategy is called _eventually pure_ , if
$\sigma_{t}$ is a Dirac measure for all but finitely many stages
$t\in\mathbb{N}$.

It follows from Theorem 4 that the set of expected payoffs that the decision
maker can obtain by Markov strategies is equal to the set of expected payoffs
under eventually pure Markov strategies. In other words, randomizing
infinitely many times in a Markov strategy does not give additional expected
payoffs to the decision maker.

##  5 Concluding Remarks

We do not know whether Theorem 4 could be deduced from Theorem 6 or vice
versa. The two results appear to be logically unrelated, especially as the
proofs use different approaches.

We conclude with an open problem: Can one generalize Theorem 4 to vector-
valued (rather than just real-valued) functions? Consider a bounded Borel-
measurable function $f:X\to\mathbb{R}^{d}$, and a product measure
$\sigma\in\Sigma$. Is there a point $x\in X$, an $n\in\mathbb{N}$, and
probability measures $\tau_{i}\in\Sigma_{i}$ for $i<n$, such that

$\mathbb{E}_{\tau_{1}\otimes\cdots\otimes\tau_{n-1}\otimes x_{n}\otimes
x_{n+1}\otimes\cdots}[f_{i}]=\mathbb{E}_{\sigma}[f_{i}]$

for each $i\in\\{1,\ldots,d\\}$? Theorem 4 gives the affirmative answer when
$d=1$. The answer is also a “yes” when $f$ is a tail function (i.e., when $f$
is measurable with respect to the tail sigma-algebra on $X$). Beyond these
special cases, the problem is open. In particular, it is also open if $f$ is
continuous.

## Appendix

We have applied Kolmogorov’s zero-one law to analytic and co-analytic sets.
Here we justify this application.

###### Lemma 10

If $U$ is a universally measurable tail subset of $X$ and $\sigma\in\Sigma$,
then the $\sigma$-measure of $U$ can only have the values 0 and 1.

Proof: We adapt the proof of the Kolmogorov’s zero-one law in Shiryaev [9,
p.381].

For $n\in\mathbb{N}$ let $\mathscr{F}_{n}$ be the sigma-algebra on $X$
generated by the first $n$ coordinate functions $x_{1},\ldots,x_{n}$. The
Borel sigma-algebra $\mathscr{F}$ on $X$ is the sigma-algebra generated by the
set $\cup_{n\in\mathbb{N}}\mathscr{F}_{n}$. As $U$ is universally measurable,
there exists a Borel set $B$ of $X$ such that $\sigma(B\triangle U)=0$, where
$\triangle$ denotes the symmetric difference.

By Shiryaev [9, Problem 8, p.69], for every $n\in\mathbb{N}$ there exists a
set $B_{n}\in\mathscr{F}_{n}$ such that $\sigma(B_{n}\triangle B)\to 0$. Since
$\sigma$ is a product measure and $U$ is tail, $B_{n}$ and $U$ are
independent. Therefore,

$\sigma(B_{n})\to\sigma(B)=\sigma(U)\ \,\text{ and}$
$\sigma(B_{n})\cdot\sigma(U)=\sigma(B_{n}\cap U)=\sigma(B_{n}\cap
B)\to\sigma(B)=\sigma(U).$

It follows that $\sigma(U)=\sigma(U)\cdot\sigma(U)$, implying the result.
$\Box$

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