# Trust AI Regulation? Discerning users are vital to build trust and effective
AI regulation

Zainab Alalawi1 Paolo Bova1 Theodor Cimpeanu2 Alessandro Di Stefano1 Manh Hong
Duong3 Elias Fernández Domingos4,5 The Anh Han1,∗ Marcus Krellner2 Bianca
Ogbo1 Simon T. Powers6 Filippo Zimmaro7,8

1 School Computing, Engineering and Digital Technologies, Teesside University
2 School of Mathematics and Statistics, University of St Andrews
3 School of Mathematics, University of Birmingham
4 Machine Learning Group, Université libre de Bruxelles
5 AI Lab, Vrije Universiteit Brussel
6 School of Computing, Engineering and the Built Environment, Edinburgh Napier
University
7 Department of Mathematics, University of Bologna
8 Department of Computer Science, University of Pisa
⋆ Corresponding author: The Anh Han<EMAIL_ADDRESS>
## abstract

There is general agreement that some form of regulation is necessary both for
AI creators to be incentivised to develop trustworthy systems, and for users
to actually trust those systems. But there is much debate about what form
these regulations should take and how they should be implemented. Most work in
this area has been qualitative, and has not been able to make formal
predictions. Here, we propose that evolutionary game theory can be used to
quantitatively model the dilemmas faced by users, AI creators, and regulators,
and provide insights into the possible effects of different regulatory
regimes. We show that creating trustworthy AI and user trust requires
regulators to be incentivised to regulate effectively. We demonstrate the
effectiveness of two mechanisms that can achieve this. The first is where
governments can recognise and reward regulators that do a good job. In that
case, if the AI system is not too risky for users then some level of
trustworthy development and user trust evolves. We then consider an
alternative solution, where users can condition their trust decision on the
effectiveness of the regulators. This leads to effective regulation, and
consequently the development of trustworthy AI and user trust, provided that
the cost of implementing regulations is not too high. Our findings highlight
the importance of considering the effect of different regulatory regimes from
an evolutionary game theoretic perspective.

Figure 1: Core features. The figure schematically illustrates the core
features of the three-population model of AI governance. Users can either
trust (T) or not trust (N) the AI system, in which case they do not adopt the
system and get zero benefit. Creators can either defect by creating unsafe AI
products (D) or cooperate by creating safe ones (C), which entails additional
costs. Applying regulations (C) also comes with some costs, while punishing
defecting creators requires further costs.

##  I Introduction

Debates are taking place across the world about what kind of regulation should
apply to the development of artificial intelligence (AI) systems. Governments
want _trustworthy_ AI systems to be developed, and for users to actually
_trust_ these systems – this is exemplified by the aims of the EU AI act [1].
Regulation is typically assumed to be the way to achieve this [2]. Economic
history supports this assumption; safety critical systems such as automotive
vehicles and medical devices are indeed regulated in this way. But it is less
clear what forms the regulations should take and who should implement them.
See, for example, proposals in the EU compared to the USA on the amount of
restrictions that should be placed on AI creators [3, 4]. Moreover, there is
the key question of _who_ should create and enforce the regulations? Should
the regulations be _public_ or _private_ [5]? That is, should they be created
by government bodies, or by industrial organisations and accrediting bodies
such as the IEEE? And once they have been created, who will enforce them?
Enforcement could involve actions such as auditing source code, training runs,
training data, and monitoring for consumer complaints. This could be done by
government sector organisations, but it could also be done by third-party
auditors [6].

To help governments choose the most suitable kind of regulatory framework, we
need to be able to predict the effects of different regulatory systems. Most
of the discourse around this at the moment is qualitative and does not lead to
formal predictions [7, 8, 9, 10]. This limits the ability of governments,
technology companies, and citizens to foresee what the effects of different
regulatory systems might be [11]. There is a small amount of literature on AI
race modelling [12, 13, 14], including from the perspective of (evolutionary)
game theory [15, 16, 17, 18], but this has not considered how different
regulatory mechanisms influence both user trust and the compliance of
companies with AI safety considerations. To address this, we propose that
evolutionary game theory [19] can be used to formally model the effects of
different regulatory systems in terms of their incentives on tech companies,
end users, and regulators.

We present a framework for formalising the strategic interaction between
users, AI system creators, and regulators as a game (Fig. 1). This game
captures three key decisions that these actors face:

  1. 1.

Users: do they trust and hence use an AI system or not?

  2. 2.

AI creators: do they follow a safety optimal development path in compliance
with regulations, or do they pursue a competitive development path that
violates regulations in a race to the market?

  3. 3.

Regulators: do they invest in monitoring AI system creators and enforcing the
regulations effectively, or do they cut corners to save the costs of doing
this?

This highlights the dilemmas facing users, creators, and regulators. Users can
benefit from using an AI system, but also run the risk that the system may not
act in their best interest, i.e. may not be trustworthy [20]. This follows
from the fact that AI creators are themselves in competition with each other,
as highlighted by the current “AI race” to develop artificial general
intelligence [12, 21, 8, 22, 23]. Consequently, we cannot assume that creators
will always act in the best interests of their users by complying with
regulations and developing systems worthy of user trust. Finally, regulators
themselves may be self-interested. This may occur when governments delegate
the enforcement of regulations to other actors, such as private audit firms
[24, 6, 9]. This kind of delegation may reduce government costs, but it also
introduces a principal agent problem [25]: regulators are themselves agents
with their own profit maximising goals.

To analyse the model, we use the methods of evolutionary game theory.
Evolutionary game theory is based on the idea that agents can learn behaviours
that benefit them from social learning, i.e. by copying the behaviour of other
agents in their population that are doing better than themselves. This avoids
the need to assume that the agents are fully rational and have complete
information. In our model, we consider three populations corresponding to the
three actors: users, creators, and regulators.

Our analysis demonstrates how incentives for regulators are important.
Governments desire that all creators produce trustworthy AI systems, and all
users trust these systems. Such a state cannot be reached if regulators that
do their job properly cannot be distinguished from regulators that cut
corners. This holds regardless of the severity of punishment for defecting
creators.

We consider two possible institutional solutions to this problem. First, we
show that if governments can provide rewards to regulators that do a good job,
and use of the AI system is not too risky for users, then some level of
trustworthy development and trust by users occurs. We then consider an
alternative solution, where users may condition their trust decision on the
effectiveness of the regulators, for example, where information about the past
performance of the regulators is available. This leads to effective
regulation, and consequently the development of trustworthy AI and user trust,
provided that the cost of implementing regulations is not too high.

##  II Models and Methods

###  II.1 Three population model of AI governance

We create a baseline model of an AI development ecosystem [2]. It is similar
to current models of regulations for information systems, e.g. the General
Data Protection Regulations (GDPR) and the AI Act of the European Union. The
model involves three populations representing the three actors in the
ecosystem: AI users, AI system creators, and regulators. In each population,
individuals can choose different options (also called strategies). A user can
decide to trust (T) or not (N) an AI system: This is a combination of trust in
the regulator and the creator. The creator either complies (C) with the rules
set out by the regulators or not (D). The regulator chooses whether to create
rules and enforce compliance (C) or not (D).

The individual payoff earned in any one encounter (also called a game) depends
on the strategy of the participating individuals. In each game, one user, one
developer, and one regulator participate. If the user trusts and adopts an AI
system when both the developer and the regulator cooperate, the user benefits
significantly from AI adoption, denoted by $b_{U}$. However, if AI adoption
occurs when the developer defects, the user is affected by unsafe AI, gaining
a reduced or even negative benefit, denoted by $\epsilon\times b_{U}$, where
$\epsilon\in[-\infty,1]$. This parameter, $\epsilon$, also represents a risk
factor that users take when adopting the AI system. For the system creator,
since it takes more time and effort to comply with the precautionary
requirements, we assume that playing C requires an extra cost $c_{P}$,
compared to playing D (the cost for this is normalised to 0). Regarding the
regulators, they earn the benefit $b_{R}$ when the user trusts and adopts AI.
Since it is costly to create rules and development technologies to capture
unsafe development, we assume that playing C requires a extra cost $c_{R}$,
compared to playing D (the cost for this is normalised to 0). The model
strategies and parameters are summarised in Table 1.

Table 1: AI Governance model (User Us., Creator Cr and Regulator Re).

Parameters'description |  Symbol
---|---
Users trust (T) or not (N) trust an AI system – this is a combination of trust in regulators and trust in creators |  T, N
Creators comply with the rules (C) or not (D) as set out by the Regulators |  C, D
Regulators create rules and ensure compliance (C) or not (D) |  C, D
Benefit users get from trust and adopt the AI system |  $b_{U}$
Fraction of user benefit when creators play D, where $\varepsilon$ in [-$\infty$,1], also referred to as the (inverse) risk factor users take when adopting an AI |  $\varepsilon$
Benefit the creator gets from selling the product |  $b_{P}$
Additional cost of creating safe AI (the cost of creating unsafe AI is normalised to 0) |  $c_{P}$
Funding for regulators (which is only generated upon AI adoption) |  $b_{R}$
The cost of developing rules and unsafe capture technologies (the cost of not doing this is normalised to 0) |  $c_{R}$
The cost of institutional punishment |  $v$
The impact of institutional punishment |  $u$

Strategies | Payoffs
---|---
Us. |  Cr |  Re |  User |  Creator |  Regulator
T |  C |  C |  $b_{U}$ |  $b_{P}-c_{P}$ |  $b_{R}-c_{R}$
T |  C |  D |  $b_{U}$ |  $b_{P}-c_{P}$ |  $b_{R}$
T |  D |  C |  $\varepsilon b_{U}$ |  $b_{P}-u$ |  $b_{R}-c_{R}-v$
T |  D |  D |  $\varepsilon b_{U}$ |  $b_{P}$ |  $b_{R}$
N |  C |  C |  0 |  $-c_{P}$ |  $-c_{R}$
N |  C |  D |  0 |  $-c_{P}$ |  0
N |  D |  C |  0 |  0 |  -$c_{R}$
N |  D |  D |  0 |  0 |  0

As shown in our analysis below, in this system, regulators are always better
off not cooperating, and trust is rarely encouraged. Thus, below we consider
two extended models that might enable investment of high quality regulation,
safe development, and users’ trust.

  1. 1.

First, we consider that the cooperative regulator is rewarded an amount of
$b_{fo}$ if they catch defective creators (when users trust and adopt AI). The
payoff matrix is summarised in Table 2.

  2. 2.

Second, we assume that regulators’ reputation is publicly available and users
can act conditionally on whether the regulators’ reputation is good or not;
see Table 3. Namely, the CT strategy means users do not trust when the
regulator’s reputation is bad.

In the general case, this leads to a frequency-dependent dynamics in each of
the three populations, where the dynamical process of strategy adoption
assumes that those individuals that are more fit are more often imitated by
their peers, an adaptive scheme akin to social learning (see Methods below).

Figure 2: Reward for regulators. The figure shows the changes to the original model. Regulators gain extra reward for capturing unsafe development. Table 2: AI Governance extended model (User Us., Creator Cr and Regulator Re), where regulators are rewarded for capturing unsafe creators (when users trust and adopt). Strategies | Payoffs
---|---
Us. |  Cr |  Re |  User |  Creator |  Regulator
T |  C |  C |  $b_{U}$ |  $b_{P}-c_{P}$ |  $b_{R}-c_{R}$
T |  C |  D |  $b_{U}$ |  $b_{P}-c_{P}$ |  $b_{R}$
T |  D |  C |  $\varepsilon b_{U}$ |  $b_{P}-u$ |  $b_{R}-c_{R}-v{+b_{fo}}$
T |  D |  D |  $\varepsilon b_{U}$ |  $b_{P}$ |  $b_{R}$
N |  C |  C |  0 |  $-c_{P}$ |  $-c_{R}$
N |  C |  D |  0 |  $-c_{P}$ |  0
N |  D |  C |  0 |  0 |  -$c_{R}$
N |  D |  D |  0 |  0 |  0
Figure 3: Conditional trust. The figure shows how the state of regulation changes the behaviour of users, which impacts the whole system. Table 3: AI Governance extended model—conditional trust (User Us., Creator Cr and Regulator Re). In this model, regulators’ reputation are assumed to be publicly available before the game, and users can act conditionally on it: CT does not trust if regulator has a bad reputation.  Strategies | Payoffs
---|---
Us. |  Cr |  Re |  User |  Creator |  Regulator
CT |  C |  C |  $b_{U}$ |  $b_{P}-c_{P}$ |  $b_{R}-c_{R}$
CT |  C |  D |  $0$ |  $-c_{P}$ |  $0$
CT |  D |  C |  $\varepsilon b_{U}$ |  $b_{P}-u$ |  $b_{R}-c_{R}-v+b_{fo}$
CT |  D |  D |  $0$ |  $0$ |  $0$
N |  C |  C |  0 |  $-c_{P}$ |  $-c_{R}$
N |  C |  D |  0 |  $-c_{P}$ |  0
N |  D |  C |  0 |  0 |  -$c_{R}$
N |  D |  D |  0 |  0 |  0

###  II.2 Methods

####  II.2.1 Stochastic dynamics for finite populations

The baseline model. We consider three different well-mixed populations of
Users (U), Creators (C) and Regulators (R) of sizes, respectively $N_{U}$,
$N_{C}$ and $N_{R}$. Let $x$ be the fraction of users that trust the AI
system. Let $y$ and $z$ be respectively the fraction of Creators and
Regulators that cooperate. Each game involves an individual randomly drawn
from each population. The fitness that a user, regulator and creator obtains
in each game is respectively given by

$\displaystyle f^{U}_{X\in\\{T,N\\}}$
$\displaystyle=yzP^{U}_{XCC}+(1-y)zP^{U}_{XDC}+y(1-z)P^{U}_{XCD}+(1-y)(1-z)P^{U}_{XDD},$
(1) $\displaystyle f^{C}_{Y\in\\{C,D\\}}$
$\displaystyle=xzP^{C}_{TYC}+(1-x)zP^{C}_{NYC}+x(1-z)P^{C}_{TYD}+(1-x)(1-z)P^{C}_{NYD},$
(2) $\displaystyle f^{R}_{Z\in\\{C,D\\}}$
$\displaystyle=xyP^{R}_{TCZ}+(1-x)zP^{R}_{NCZ}+x(1-y)P^{R}_{TDZ}+(1-x)(1-y)P^{R}_{NDZ}.$
(3)

We now derive the fitness/average payoffs for each model. First, for the
baseline model, from (1) and the payoff table 1, we calculate explicitly the
difference of the fitness between two strategies in users:

$\displaystyle f^{U}_{T}-f^{U}_{N}$
$\displaystyle=yz(P^{U}_{TCC}-P^{U}_{NCC})+(1-y)z(P^{U}_{TDC}-P^{U}_{TDC})+y(1-z)(P^{U}_{TCD}-P^{U}_{NCD})+(1-y)(1-z)(P^{U}_{TDD}-P^{U}_{NDD})$
$\displaystyle=yzb_{U}+(1-y)z(\varepsilon
b_{U})+y(1-z)b_{U}+(1-y)(1-z)(\varepsilon b_{U})$
$\displaystyle=b_{U}\Big{(}yz+\varepsilon(1-y)z+y(1-z)+\varepsilon(1-y)(1-z)\Big{)}$
$\displaystyle=b_{U}(y+\varepsilon(1-y)).$ (4)

Similarly, the difference of the fitness between two strategies in creators is

$\displaystyle f^{C}_{C}-f^{C}_{D}$
$\displaystyle=xz(P^{C}_{TCC}-P^{C}_{TDC})+(1-x)z(P^{C}_{NCC}-P^{C}_{NDC})+x(1-z)(P^{C}_{TCD}-P^{C}_{TDD})+(1-x)(1-z)(P^{C}_{NCD}-P^{C}_{NDD})$
$\displaystyle=xz\Big{(}b_{P}-c_{P}-(b_{P}-u)\Big{)}+(1-x)z(-c_{P})+x(1-z)(b_{P}-c_{P}-b_{P})+(1-x)(1-z)(-c_{P})$
$\displaystyle=xz(u-c_{P})+(1-x)z(-c_{P})+x(1-z)(-c_{P})+(1-x)(1-z)(-c_{P})$
$\displaystyle=-c_{P}+uxz.$ (5)

Finally, the difference of the fitness between two strategies in regulators is

$\displaystyle f^{R}_{C}-f^{R}_{D}$
$\displaystyle=xy(P^{R}_{TCC}-P^{R}_{TCD})+(1-x)y(P^{R}_{NCC}-P^{R}_{NCD})+x(1-y)(P^{R}_{TDC}-P^{R}_{TDD})+(1-x)(1-y)(P^{R}_{NDC}-P^{R}_{NDD})$
$\displaystyle=xy\Big{(}(b_{R}-c_{R})-b_{R}\Big{)}+(1-x)y(-c_{R})+x(1-y)\Big{(}b_{R}-c_{R}-v-b_{R}\Big{)}+(1-x)(1-y)(-c_{R})$
$\displaystyle=xy(-c_{R})+(1-x)y(-c_{R})+x(1-y)(-c_{R}-v)+(1-x)(1-y)(-c_{R})$
$\displaystyle=-c_{R}-x(1-y)v.$ (6)

We notice that $f^{R}_{C}-f^{R}_{D}$ is always negative. Thus, it is always
better off for the regulator to defect, that is, to provide regulator rules.

Now, for the extended model where the reward $b_{fo}$ is provided to
regulators for capturing unsafe creators and if they catch defective creators
(when users adopt only). Then only the payoff difference $f^{R}_{C}-f^{R}_{D}$
is changed to

$f^{R}_{C}-f^{R}_{D}=-c_{R}+x(1-y)(b_{fo}-v).$ (7)

Finally, when users can make a conditional decision based on the regulator’s
reputation, we have

$\begin{split}&f^{U}_{T}-f^{U}_{N}=b_{U}z(y+\epsilon(1-y))\\\
&f^{C}_{C}-f^{C}_{D}=-c_{P}+uxz\\\
&f^{R}_{C}-f^{R}_{D}=-c_{R}+b_{R}x+(b_{fo}-v)x(1-y)\end{split}$ (8)

For a finite population setting, at each time step, a randomly selected
individual A, with fitness $f_{A}$, may adopt a different strategy by
imitating a randomly chosen individual B from the same population (with
fitness $f_{B}$) with probability given by the Fermi distribution

$p=[1+e^{-\beta(f_{B}-f_{A})}]^{-1},$

where $\beta\geq 0$ is the strength of selection. $\beta=0$ corresponds to
neutral drift where imitation decisions are random, while for large
$\beta\rightarrow\infty$, the imitation decision becomes increasingly
deterministic.

In the absence of mutations or exploration, the end states of evolution are
inevitably monomorphic: once such a state is reached, it cannot be escaped
through imitation. We thus further assume that with a certain mutation
probability, an agent switches randomly to a different strategy without
imitating another agent. In the limit of small mutation rates, the dynamics
will proceed with, at most, two strategies in the population, such that the
behavioural dynamics can be conveniently described by a Markov chain, where
each state represents a monomorphic population, whereas the transition
probabilities are given by the fixation probability of a single mutant [26,
27, 28]. The resulting Markov chain has a stationary distribution, which
characterises the average time the population spends in each of these
monomorphic end states.

Now, the probability to change the number $k$ of agents using strategy A by
$\pm$ one in each time step can be written as ($Z$ is the population size)
[29]

$T^{\pm}(k)=\frac{Z-k}{Z}\frac{k}{Z}\left[1+e^{\mp\beta[f_{A}(k)-f_{B}(k)]}\right]^{-1}.$
(9)

The fixation probability of a single mutant with a strategy A in a population
of $(Z-1)$ agents using B is given by [29, 27]

$\rho_{B,A}=\left(1+\sum_{i=1}^{Z-1}\prod_{j=1}^{i}\frac{T^{-}(j)}{T^{+}(j)}\right)^{-1}.$
(10)

The transition matrix $\Lambda$ corresponding to the set of
$\left\\{1,\ldots,s\right\\}$ strategies is given by:

$\displaystyle\Lambda_{ij,j\neq i}=\frac{\rho_{ji}}{3}\hskip 2.84526pt\text{
and }\hskip 2.84526pt\Lambda_{ii}=1-\sum_{j=1,j\neq i}^{s}\Lambda_{ij}.$ (11)

Fixation probability $\rho_{ij}$ denotes the likelihood that a population
transitions from a state $i$ to a different state $j$ when a mutant of one of
the populations adopts an alternate strategy s. The fixation probability is
divided by the number of populations (3) representing the interaction of three
players at a time [30, 31].

####  II.2.2 Population dynamics for infinite populations: The multi-
population replicator dynamics

In this section, we recall the framework of the replicator dynamics for multi-
populations [32, 33]. To describe the dynamics, we consider a set of $m$
different populations ($m$ is some positive integer), which are infinitely
large and well-mixed. Each population $i$, $i=1,\ldots m$, consists of $n_{i}$
($n_{i}$ is some positive integer) different strategies (types). Let
$x_{ij},1\leq i\leq m,1\leq j\leq n_{i}$, be the frequency of the strategy $j$
in the population $i$. We denote by $x_{i}=(x_{ij})_{j=1}^{n_{i}}$, which is
the collection of all strategies in the population $i$, and
$x=(x_{1},\ldots,x_{m})$, which is the collection of all strategies in all
populations.

For each $i\in\\{1,\ldots,m\\}$ and $j\in\\{1,\ldots,n_{i}\\}$, let
$f_{ij}(x)$ be the fitness (reproductive rate) of the strategy $j$ in the
population $i$. This fitness is obtained when the strategy $j$ interacts with
all other strategies in all populations; thus, it depends on all the
strategies in the populations. The average fitness of the population $i$ is
defined by

$\bar{f}_{i}(x)=\sum_{j=1}^{n_{i}}x_{ij}f_{ij}(x).$

The multi-population replicator dynamics is then given by

$\dot{x}_{ij}=x_{ij}(f_{ij}(x)-\bar{f}_{i}(x)),\quad 1\leq i\leq m,\quad 1\leq
j\leq n_{i}.$ (12)

This is in general an ODE system of $\sum_{i=1}^{m}n_{i}$ equations. Noting,
however that since $\sum_{j=1}^{n_{i}}x_{ij}=1$ for all $i=1,\ldots,m$, we can
reduce the above system to a system of $\sum_{i=1}^{m}n_{i}-m$ equations.

Now we focus on the case when there are two strategies in each population
(which is the case for our models of AI governance and trust in the present
paper), that is $n_{i}=2$ for all $i=1,\ldots,n_{i}$. Let $\eta_{i}$ be the
frequency of the first strategy in the population $i$, $i=1,\ldots,m$ (thus
$1-\eta_{i}$ will be the frequency of the second strategy in the population
$i$), let $\eta=(\eta_{1},\ldots,\eta_{m})$. Let $f_{1i}(\eta)$ and
$f_{2i}(\eta)$ be the fitness of the first and second strategy in the
population $i$. Since

$\bar{f}_{i}(\eta)=\eta_{i}f_{1i}(\eta)+(1-\eta_{i})f_{2i}(\eta),$

we have

$f_{1i}(\eta)-\bar{f}_{i}(\eta)=f_{1i}(\eta)-(\eta_{i}f_{1i}(\eta)+(1-\eta_{i})f_{2i}(\eta))=(1-\eta_{i})(f_{1i}(\eta)-f_{2i}(\eta)).$

Thus we obtain the following system of equations

$\dot{\eta}_{i}=\eta_{i}(1-\eta_{i})(f_{1i}(\eta)-f_{2i}(\eta)),\quad
i=1,\ldots,m.$ (13)

This is a system of $m$ coupled nonlinear ordinary differential equations for
$m$ variables.

In the subsequent sections, we employ (13) to our models of AI governance
trust, where the fitness are computed from the payoff matrix constructed in
the models, see Tables 1-2-3, where we assume that the payoffs are directly
translated to the biological fitnesses (infinite strength of selection).

##  III Equilibrium analysis in infinite populations

####  III.0.1 The baseline model

We consider three different well-mixed populations of Users (U), Creators (C)
and Regulators (R). Let $x$ be the frequency of users that trust the AI
system. Let $y$ and $z$ be respectively the frequency of Creators and
Regulators that cooperate.

The replicator dynamics is

$\displaystyle\dot{x}$
$\displaystyle=x(1-x)(f^{U}_{T}-f^{U}_{N})=x(1-x)b_{U}(y+\varepsilon(1-y)),\quad$
$\displaystyle x(0)=x_{0}$ (14a) $\displaystyle\dot{y}$
$\displaystyle=y(1-y)(f^{C}_{C}-f^{C}_{D})=y(1-y)(-c_{P}+uxz),\quad$
$\displaystyle y(0)=y_{0}$ (14b) $\displaystyle\dot{z}$
$\displaystyle=z(1-z)(f^{R}_{C}-f^{R}_{D})=z(1-z)(-c_{R}-x(1-y)v),\quad$
$\displaystyle z(0)=z_{0}$ (14c)

where $(x_{0},y_{0},z_{0})\in[0,1]^{3}$ is the initial data. This is a
complex, nonlinear system of three coupled ordinary differential equations
(ODEs) for three variables. In general, the dynamical solutions
$x(t),y(t),z(t)$ may show different behaviour depending on the values of the
parameters and the initial data. We now focus on stationary solutions
(equilibria), which describe the long-time behaviour of the dynamics.

##### Stationary solutions (equilibrium points)

Stationary solutions of (14) are points $(x^{*},y^{*},z^{*})\in[0,1]^{3}$ that
make the right hand side of (14) vanish. It is clear that the vertices of the
unit cube $(x,y,z)\in\\{0,1\\}^{3}$ are stationary solutions (we call them
vertex equilibria).

There may exist other, non-vertex, stationary solutions depending on the sign
of $\varepsilon$ and the relation between $c_{P}$ and $u$.

If $\varepsilon>0$, then $y+\varepsilon(1-y)>0$ for all $y\in[0,1]$. It
implies that there are no other stationary solutions (except in the very
specific case where $u=c_{p}$, in which
$(x^{*},y^{*},z^{*})\in\\{0,1\\}\times[0,1]\times\\{0,1\\}$ is a stationary
solution).

If $\varepsilon<0$, and $0<c_{P}\leq u$, then other, non-vertex, stationary
solutions are given by

$y^{*}=\frac{\varepsilon}{\varepsilon-1},\quad-c_{P}+ux^{*}z^{*}=0,\quad
z^{*}=1,$

which is

$x^{*}=\frac{c_{P}}{u},\quad y^{*}=\frac{\varepsilon}{\varepsilon-1},\quad
z^{*}=1.$ (15)

##### Stability Analysis

Next, we study the stability of the equilibria. To this end, we define

$\displaystyle F(x,y,z)=(F_{1}(x,y,z),F_{2}(x,y,z),F_{3}(x,y,z)),\quad
F_{1}(x,y,z)=x(1-x)b_{U}(y+\varepsilon(1-y)),$ $\displaystyle
F_{2}(x,y,z)=y(1-y)(-c_{P}+uxz),\quad F_{3}(x,y,z)=z(1-z)(-c_{R}-x(1-y)v).$

Then (14) can be reformulated as

$\dot{x}=F_{1}(x,y,z),\quad\dot{y}=F_{2}(x,y,z),\quad\dot{z}=F_{3}(x,y,z).$

The Jacobian matrix is

$\displaystyle DF(x,y,z)$ $\displaystyle=\begin{pmatrix}\frac{\partial
F_{1}}{\partial x}&\frac{\partial F_{1}}{\partial y}&\frac{\partial
F_{1}}{\partial z}\\\ \frac{\partial F_{2}}{\partial x}&\frac{\partial
F_{2}}{\partial y}&\frac{\partial F_{2}}{\partial z}\\\ \frac{\partial
F_{3}}{\partial x}&\frac{\partial F_{3}}{\partial y}&\frac{\partial
F_{3}}{\partial z}\\\ \end{pmatrix}$
$\displaystyle=\begin{pmatrix}b_{U}(1-2x)(y+\varepsilon(1-y))&b_{U}x(1-x)(1-\varepsilon)&0\\\
uy(1-y)z&(1-2y)(-c_{P}+uxz)&uy(1-y)x\\\
-z(1-z)(1-y)v&xvz(1-z)&(1-2z)(-c_{R}-x(1-y)v)\end{pmatrix}.$

We recall that an equilibrium $(x^{*},y^{*},z^{*})$ is asymptotically stable
if all eigenvalues of $DF(x^{*},y^{*},z^{*})$ have negative real parts.

##### Stability Analysis for vertex equilibria

We now determine the stability for vertex-equilibria,
$(z^{*},y^{*},z^{*})\in\\{0,1\\}^{3}$. For these equilibria,
$DF(x^{*},y^{*},z^{*})$ is always a diagonal matrix and is explicitly given by

$DF(x^{*},y^{*},z^{*})=\begin{pmatrix}b_{U}(1-2x^{*})(y^{*}+\varepsilon(1-y^{*}))&0&0\\\
0&(1-2y^{*})(-c_{P}+ux^{*}z^{*})&0\\\
0&0&(1-2z^{*})(-c_{R}-x^{*}(1-y^{*})v)\end{pmatrix}.$

For instance at the origin $(x^{*},y^{*},z^{*})=(0,0,0)$ we have

$DF(x,y,z)=\begin{pmatrix}\varepsilon b_{U}&0&0\\\ 0&-c_{P}&0\\\
0&0&-c_{R}\end{pmatrix}.$

For another example at $(x^{*},y^{*},z^{*})=(1,0,0)$ then

$DF(x,y,z)=\begin{pmatrix}-\varepsilon b_{U}&0&0\\\ 0&-c_{P}&0\\\
0&0&-c_{R}\end{pmatrix}.$

Since for vertex equilibria, $DF(x^{*},y^{*},z^{*})$ is diagonal and its
eigenvalues read

$\lambda_{1}=b_{U}(1-2x^{*})(y^{*}+\varepsilon(1-y^{*})),\quad\lambda_{2}=(1-2y^{*})(-c_{P}+ux^{*}z^{*}),\quad\lambda_{3}=(1-2z^{*})(-c_{R}-x^{*}(1-y^{*})v).$

Thus $(x^{*},y^{*},z^{*})$ is stable only if

$\lambda_{1}<0,\quad\lambda_{2}<0,\quad\lambda_{3}<0.$

Since $(-c_{R}-x^{*}(1-y^{*})v)<0$, it follows that $\lambda_{3}<0$ iff
$z^{*}=0$. For $z^{*}=0$, $\lambda_{2}=-c_{P}(1-2y^{*})<0$ iff $y^{*}=0$. Then
$\lambda_{1}=\varepsilon b_{U}(1-2x^{*})$.
For $\varepsilon>0$, then $\lambda_{1}<0$ iff $x^{*}=1$. On the other hand, if
$\varepsilon<0$, then $\lambda_{1}<0$ iff $x^{*}=0$.
In conclusion, for vertex equilibria, $(x^{*},y^{*},z^{*})\in\\{0,1\\}^{3}$:

  * •

when $\varepsilon>0$, the equilibrium $(x^{*},y^{*},z^{*})=(1,0,0)$ is stable,
thus the users will fully trust (even if the creators and the regulators
defect).

  * •

when $\varepsilon<0$, the origin $(x^{*},y^{*},z^{*})=(0,0,0)$ is stable, both
regulators and creators defect and there is no trust from users.

##### Stability analysis for the non-vertex equilibrium

. For the non-vertex equilibrium (15), the Jacobian matrix is of the form

$DF=\begin{pmatrix}0&b_{U}\frac{c_{P}}{u}(1-\frac{c_{p}}{u})(1-\varepsilon)&0\\\
-u\frac{\varepsilon}{(\varepsilon-1)^{2}}&0&-u\frac{\varepsilon}{(\varepsilon-1)^{2}}\frac{c_{P}}{u}\\\
0&0&c_{R}+\frac{c_{P}}{u}\frac{1}{1-\varepsilon}v\end{pmatrix}.$

Thus it has one positive (recalling that $\varepsilon<0$ in this case) real
root, which is the last diagonal entry. Therefore, the non-vertex equilibrium
is not stable.

####  III.0.2 Extended model: incentive for regulator to coperate

For the extended model, using $f_{C}^{R}-f_{D}^{R}$ calculated in (7) in (14),
the replicator dynamics become

$\displaystyle\dot{x}$
$\displaystyle=x(1-x)(f^{U}_{T}-f^{U}_{N})=x(1-x)b_{U}(y+\varepsilon(1-y)),\quad$
$\displaystyle x(0)=x_{0}$ (16a) $\displaystyle\dot{y}$
$\displaystyle=y(1-y)(f^{C}_{C}-f^{C}_{D})=y(1-y)(-c_{P}+uxz),\quad$
$\displaystyle y(0)=y_{0}$ (16b) $\displaystyle\dot{z}$
$\displaystyle=z(1-z)(f^{R}_{C}-f^{R}_{D})=z(1-z)(-c_{R}+x(1-y)(b_{fo}-v)),\quad$
$\displaystyle z(0)=z_{0}$ (16c)

where $(x_{0},y_{0},z_{0})\in[0,1]^{3}$ is the initial data. Practically,
compared to the baseline model, the only difference is that we allowed $v$ to
be negative. This is again a complex nonlinear coupled system of $3$ ordinary
differential equations. Its solutions $x(t),y(t),z(t)$ may exhibit different
behaviour depending on the values of the parameters and the initial data, see
Figure 4.

The vertex equilibria in the baseline model are obviously still equilibria for
this extended model. In addition, due to the presence of the incentive
$b_{fo}$ there may be another internal equilibrium

$x^{*}=\frac{c_{R}(1-\varepsilon)}{b_{fo}-v},\quad
y^{*}=\frac{\varepsilon}{\varepsilon-1},\quad
z^{*}=\frac{c_{P}(b_{fo}-v)}{u(1-\varepsilon)},$ (17)

if $\varepsilon<0$ and the parameters $c_{P},c_{R},(b_{fo}-v)$ are such that
$0<x^{*},y^{*},z^{*}<1$ (in particular $b_{fo}<v$). The Jacobian matrix is

$\displaystyle
DF(x,y,z)=\begin{pmatrix}b_{U}(1-2x)(y+\varepsilon(1-y))&b_{U}x(1-x)(1-\varepsilon)&0\\\
uy(1-y)z&(1-2y)(-c_{P}+uxz)&uy(1-y)x\\\
-z(1-z)(1-y)(v-b_{fo})&x(v-b_{fo})z(1-z)&(1-2z)(-c_{R}-x(1-y)(v-b_{fo})).\end{pmatrix}$

We are now interested in determining the stability for vertex equilibria,
$(z^{*},y^{*},z^{*})\in\\{0,1\\}^{3}$. For these equilibria,
$DF(x^{*},y^{*},z^{*})$ is always a diagonal matrix and is explicitly given by

$DF(x^{*},y^{*},z^{*})=\begin{pmatrix}b_{U}(1-2x^{*})(y^{*}+\varepsilon(1-y^{*}))&0&0\\\
0&(1-2y^{*})(-c_{P}+ux^{*}z^{*})&0\\\
0&0&(1-2z^{*})(-c_{R}-x^{*}(1-y^{*})(v-b_{fo})).\end{pmatrix}$

$DF(x^{*},y^{*},z^{*})$ still has three real eigenvalues, corresponding to the
diagonal entries:

$\lambda_{1}=b_{U}(1-2x^{*})(y^{*}+\varepsilon(1-y^{*})),\quad\lambda_{2}=(1-2y^{*})(-c_{P}+ux^{*}z^{*}),\quad\lambda_{3}=(1-2z^{*})(-c_{R}-x^{*}(1-y^{*})(v-b_{fo})).$

Thus $(x^{*},y^{*},z^{*})$ is stable only if

$\lambda_{1}<0,\quad\lambda_{2}<0,\quad\lambda_{3}<0.$

If $v>b_{fo}$ then $(-c_{R}-x^{*}(1-y^{*})(v-b_{fo})<0$, the stability for
vertex equilibria is the same as in the case where $b_{fo}=0$.
We consider now $v<b_{fo}$.
When $\varepsilon>0$ then $\lambda_{1}<0$ iff $x^{*}=1$, in which case
$\lambda_{2}$ and $\lambda_{3}$ reduce to

$\lambda_{2}=(1-2y^{*})(-c_{p}+uz^{*}),\lambda_{3}=(1-2z^{*})(-c_{R}-(1-y^{*})(v-b_{fo}).$

  * •

$(y^{*},z^{*})=(0,0)$. Then $\lambda_{2}=-c_{P}<0$,
$\lambda_{3}=-c_{R}-(v-b_{fo})$. Thus the equilibrium $(1,0,0)$ is stable if
$-c_{R}<v-b_{fo}<0$ and unstable otherwise.

  * •

$(y^{*},z^{*})=(0,1)$. Then $\lambda_{2}=-c_{P}+u$,
$\lambda_{3}=c_{R}+(v-b_{fo})$. Thus $(1,0,1)$ is stable only if $u<c_{P}$ and
$c_{R}+(v-b_{fo})<0$ and is unstable otherwise.

  * •

$(y^{*},z^{*})=(1,0)$. Then $\lambda_{2}=c_{P}>0$. Thus the equilibrium
$(1,1,0)$ is unstable.

  * •

$(y^{*},z^{*})=(1,1)$. Then $\lambda_{2}=c_{P}-u$, $\lambda_{3}=c_{R}>0$. Thus
$(1,1,1)$ is unstable.

When $\varepsilon<0$

  * •

$y^{*}=0$. Then $\lambda_{1}=b_{U}(1-2x^{*})\varepsilon$. Hence
$\lambda_{1}<0$ for $x^{*}=0$. Then $\lambda_{2}=-c_{P}<0$ and
$\lambda_{3}=-c_{R}(1-2z^{*})$. Hence $\lambda_{3}<0$ for $z^{*}=0$. In this
case, $(0,0,0)$ is a stable equilibrium.

  * •

$y^{*}=1$. Then $\lambda_{1}=b_{U}(1-2x^{*})$. Hence $\lambda_{1}<0$ for
$x^{*}=1$. Then $\lambda_{2}=c_{P}-uz^{*}$. Thus $\lambda_{2}<0$ for $z^{*}=0$
and $c_{P}<u$. Thus $\lambda_{3}=-c_{R}<0$. If $c_{P}<u$ then $(1,1,0)$ is a
stable equilibrium.

(a) $x_{0}=y_{0}=z_{0}=0.1$, $\epsilon=0.01$

(b) $x_{0}=y_{0}=z_{0}=0.5$, $\epsilon=0.01$

(c) $x_{0}=y_{0}=z_{0}=0.1$, $\epsilon=-0.5$

(d) $x_{0}=y_{0}=z_{0}=0.5$, $\epsilon=-0.5$

(e) $x_{0}=y_{0}=z_{0}=0.1$, $\epsilon=-1$

(f) $x_{0}=y_{0}=z_{0}=0.5$, $\epsilon=-1$

Figure 4: Numerical integration of the evolution equation for the extended
model (16), describing the evolution of the density of trusting users $x(t)$,
cooperating creators $y(t)$ and cooperating regulators $z(t)$. In all these
simulations: $b_{U}=4$, $c_{P}=0.5$, $u=1.5$, $c_{R}=0.5$, $b_{fo}-v=1.5$. For
the initial conditions and the other parameters, see the captions.

####  III.0.3 Model with conditional trust

For the model with conditional trust, the replicator dynamics reads

$\displaystyle\dot{x}$ $\displaystyle=x(1-x)b_{U}z(y+\epsilon(1-y)),\quad$
$\displaystyle x(0)=x_{0}$ (18a) $\displaystyle\dot{y}$
$\displaystyle=y(1-y)[-c_{P}+uxz],\quad$ $\displaystyle y(0)=y_{0}$ (18b)
$\displaystyle\dot{z}$
$\displaystyle=z(1-z)[-c_{R}+b_{R}x+(b_{fo}-v)x(1-y)],\quad$ $\displaystyle
z(0)=z_{0}$ (18c)

The stability analysis for this model is similar to the extended model
(incentive for the regulator to cooperate). Hence we omit it here.

####  III.0.4 Numerical results for replicator dynamics

We present here numerical results for the infinite population setting. First,
Figure 4 shows different outcomes in the evolutionary dynamics when varying
initial conditions (fractions of strategists $x,y,z$ in the population). In
figure 4a), the system first reaches trust and regulation ($TCC$), which leads
to profitability for regulators to stop enforcing safety standards ($TCD$). In
panels 4b) and d) we find a stable limit cycle between regulation and unsafe
development. Users fully trust, but regulators mirror unsafe AI development.
Safe AI reduces regulation, whereas unsafe AI prompts them to reinstate
incentives. Thus, we could simplify the equations (16), taking $x(t)=1$
constantly and explaining mathematically the existence of the cycle.
Qualitatively, when defecting creators emerge, regulators enforce standards
(because $b_{fo}>v+c_{R}$), thus it is profitable to comply when meeting a
defecting creator. Conversely, when cooperating regulators emerge, for
creators it is better to cooperate as well (as $u>c_{P}$); but, if creators
cooperate, the enforcing standards becomes unprofitable and regulators defect.
In the absence of regulation, unsafe developers re-emerge and the limit cycle
restarts. In 4c), e) and f), trust is never established in the users’
population, there is no incentive to regulate as there is no adoption, and
creators ignore safety standards - $NDD$ is reached.

Figure 5: Low regulation cost ($c_{R}=0.5$). Conditional trust can lead to
full trust, cooperative regulation and safe development. Parameters set to:
$b_{U}=b_{R}=b_{P}=4$, $u=1.5$, $v=0.5$, $c_{P}=0.5$, $\beta=0.1$,
$N_{U}=N_{P}=N_{R}=100$.

##  IV Stochastic analysis

We present here numerical results for the finite population setting (see
Methods). Figure 5 shows results for varying $b_{fo}$ for both proposed models
when the cost of regulation $c_{R}$ is sufficiently small. We identify two
regimes, based on the sign of the risk factor $\epsilon$. We assume that
defection in the creator population (i.e., strategies of the form $xDz$) is
the most deleterious outcome for users, but there still exists a potential
benefit when the sign of $\epsilon$ is positive. In such cases, we observe the
evolution of trust regardless of the success of regulators. Given low
incentives for capturing unsafe developers ($b_{fo}$), regulators fail to
adequately enforce safety. Increasing incentives compels regulators to act,
thus giving rise to a mixed outcome between $TCC,TCD$ and $TDC$ (refer to
Figure 7 for an in-depth view of the dominance cycles that govern these
states). Whether the population evolves to $TDD$ (for low $b_{fo}$) or the
mixed state, we note that for this regime, it is always beneficial to adopt
even unsafe AI (see discussion). If $\epsilon<0$, no trust can evolve, and
technology is never adopted. Importantly, When regulation is cheaply
implementable (low $c_{R}$), conditional trust solves these issues for all
incentives and in both risk regimes.

Figure 6 shows results for varying $b_{fo}$ for different models, when the
cost of regulation $c_{R}$ is large compared to the benefit, e.g., when the
technology is highly sophisticated and it is difficult to develop technology
to capture unsafe behaviour. In the regime of low risk $\epsilon<0$, we
observe defection because users do not trust the technology and the
regulators. For very low values of $b_{fo}$, there is no incentive to deviate
from defection, and this remains the main regime even increasing the $b_{fo}$
values. With conditional trust, illustrated in Fig. 6 (b), even in the absence
of trust in regulations, other strategies start emerging, due to cyclic
dominance between $TCC\rightarrow TCD\rightarrow TDD\rightarrow TDC$ (see
Figure 7). This cycle only emerges if regulators are given a high enough
incentive to enforce safety standards ($b_{fo}$). With a high risk factor
$\epsilon<0$, trust, regulation and safety never emerge. Interestingly,
extending this model to conditional trust, we see a decrease in trust and
adoption. Previously, low risk ($\epsilon>0$) provided enough of a benefit for
users to adopt AI even in the case of rapid, risky development, yet regulators
prevent this from occurring. Regulation is too costly, thus regulators
restrict access to AI, deeming it unsafe. If we consider conditional trust in
such cases, users cannot elect to adopt technology unless regulators approve
of creators, which hinges on the cost of regulation.

Figure 6: High regulation cost ($c_{R}=5$). Conditional trust is not
sufficient to enable safe development and AI adoption when the cost of
regulation is high, as this discourages regulators to cooperate. Parameters
set to: $b_{U}=b_{R}=b_{P}=4$, $u=1.5$, $v=0.5$, $c_{P}=0.5$, $\beta=0.1$,
$N_{U}=N_{P}=N_{R}=100$ Figure 7: Stationary distribution and transitions
amongst the strategies (with conditional trust and without conditional trust)
($b_{R}=4$). Only arrows where the transition probability is larger than the
opposite one, are shown. Dashed lines represent neutral transitions.
Parameters set to: $b_{U}=b_{P}=4$, $u=1.5$, $v=0.5$, $c_{P}=0.5$,
$\beta=0.1$. Population sizes: $N_{U}$ = $N_{P}$ = $N_{R}=100$.

##  V Discussion

###  V.1 Key Takewaways

Our conceptual model of an AI governance ecosystem highlights several key
factors relevant to both current and future regulatory landscapes for AI:

  * I.

AI companies are motivated to implement safety protocols only when regulatory
bodies have the capacity to enforce compliance,

  * II.

In scenarios where the potential negative consequences of unsafe AI
development are significant, user trust in the system hinges on the
dependability of regulatory authorities,

  * III.

For regulatory bodies to consistently enforce safety measures, the cost
associated with regulation must be manageable, and their capacity to identify
non-compliant AI creators should be high,

  * IV.

Critical users, namely those whose trust in the regulatory system is
conditional on its perceived effectiveness, are essential to enable a stable
and long-lasting trust in the system, where creators comply with regulations
and regulators continue to enforce them.

We found that trust can only emerge when regulators have an incentive to
regulate. Moreover, the presence of critical users, that is, users that place
their trust conditionally, are able to break the cyclic dynamics that we
observe in Fig. 4a. Figs. 7b and d show how the presence of these users who
trust conditionally breaks the cycle in the dimension of the regulator and
creator populations, by making TCC dominant over TCD. That is, when users
trust conditionally, enforcing regulation risk dominates not enforcing it.

###  V.2 Implications for AI Governance

We focus on the policy implications of our main findings III and IV: high
costs in providing effective regulation pose an obstacle to building trust in
future AI systems, and that it is vital to help users discern which systems
have undergone effective regulation.

Scenarios where regulators face large costs in assessing the risks of AI
systems are plausible. Existing AI evaluators already experience relatively
high costs in providing effective regulation for fronteir AI systems.
Foundation models like Open AI’s GPT-4 require large amounts of computational
hardware and access to the top AI talent [22]. Evaluations of these systems
are likewise expensive and demand teams with expertise in AI. Academia and
public institutions who wish to assess the risks of new AI systems therefore
compete with AI creators in acquiring these resources.

To overcome these barriers to entry, policy makers should consider (i)
investing in building up government capacity to assess the risks of large AI
systems [34], and (ii) subsidising organisations that show promise in
providing services that improve the effectiveness of AI regulation.
Governments may therefore wish to take heed of the UK’s investment in their AI
Safety Institute, k2023institute [35].

Governments and regulators can benefit from communicating information that
allows users to discern which methods that regulators use are reliable (so
that, in turn, they can discern which AI systems are safe and trustworthy).
Doing so requires investing in the expertise needed to follow the
recommendations on how to assess the risks presented by AI systems [7]. This
could mean conducting machine learning model evaluations or other tests as
part of an audit or risk assessment of new frontier AI systems [36, 37]. The
role of academia and the commentariat should also not be underestimated here
[2]. Academia and commentators can also play a key role in communicating to
users information about the effectiveness of regulations. Our model can be
extended to capture this by including academics and commentators as an
additional evolving population of agents.

In the case where the regulator is a public institution with greater powers,
they may wish to monitor the size of training runs for frontier AI systems or
to impose fines on companies whose AI systems cause significant economic harm
[38]. The governance of computing hardware (shortened to compute) has also
been noted as a point of potentially high leverage, given that large amounts
of physical infrastructure are much easier to monitor, track, and limit in
principle than software, data, or model weights [39].

One more avenue regulators can pursue to help demonstrate to users that
trustworthy AI systems will be developed and deployed safely is to encourage
international coordination in the governance of AI systems, perhaps by
proposing designs for international institutions to govern new AI
breakthroughs or AI safety projects [40, 41]. World powers have expressed an
awareness that AI could pose enormous risks when signing the Bletchley
Declaration [42], highlighting that international coordination of some form is
tractable and desired.

Finally, public institutions will be necessary to ensure that companies and
users, particularly large enterprises that serve a wide number of additional
users, can form reliable impressions of the reputation of different regulators
[35]. Model registration is an essential prerequisite for such a reputation
system; Users need to know which regulatory evaluations match which AI systems
[43]. Moreover, governments may wish those providing regulatory services to
preregister a set of experiments they plan to run to test the model.
Regulators could later add details about how they dealt with unforeseen
findings in their experiments. This would help them to demonstrate vigilance
(i.e. that they have a high likelihood of finding early signs of dangerous
capabilities when they in fact exist), while ensuring that their insights can
be shared with other organisations quickly and safely [34, 44]. Keep in mind
that, as our results show, this information is useful whenever we have
multiple organisations evaluating frontier AI systems, including at the
international level [41].

###  V.3 Limitations and Areas for Future Research

Above, we discussed how decision makers can act on the insights from our
results. Now, we turn to the limitations of the present model. An important
simplification we made was to limit each actor in our model to a small set of
strategic choices. Future research could address these limitations by
incorporating partner selection into these models between populations. In a
more realistic setting, users would choose which AI system they want to use.
Companies too may choose to relocate to avoid especially burdensome
regulation.

We also think it is essential to model more explicitly competition between
different regulatory blocs. One appropriate formalism for achieving this would
be to model each regulatory bloc as an island network that is connected
loosely to other islands. These islands of regulation would be in competition
with each other through a process of cultural group selection [45, 46, 47].
For example, creators may move to regulatory regimes more favourable to
themselves, while regulators may imitate the regulatory regimes of other
groups that are more successful than themselves.

Our model also does not fully capture the race dynamics between AI companies.
OpenAI’s initial commercial breakthrough with ChatGPT has sparked interest in
building more capable foundation models among the world’s leading tech
companies [22]. As predicted by other game-theoretic models of tech race
dynamics, remaining at the frontier of AI capabilities appears to take
precedence over pursuing AI safety research agendas [12, 15]. It is noteworthy
that to date, companies have had relatively limited success making generative
AI systems fundamentally safer.

Scaling AI systems quickly encounters a brisk trade-off with safety research.
It allows less time to investigate and understand the impacts of smaller
models, particularly when new algorithms are used to pretrain or fine tune
those models. More talent and computational resources must also be allocated
to these large experiments, allowing fewer resources to pursue research
agendas on AI safety. Therefore, we cannot expect even relatively safety-
minded AI labs to choose the optimal trade-off, given the huge competitive
dynamics at play. For a further discussion of what can be done to directly
kerb race dynamics, consult the following references [38, 21, 48, 44, 14].

While we recognise the limitations of our simpler scenario, we find our
approach useful as a tool for thinking through what assumptions policymakers
must make to assume that different regulatory regimes are likely to achieve
their policy objectives.

Further avenues of research should test the predictions of our model,
providing real-world evidence of strategic interactions between users,
creators, and regulators in the AI domain. Exploring additional incentive
structures for regulators could further enhance our understanding of effective
AI governance. A useful incentive structure to consider is that AI companies
could be mandated to pay for regulatory services that governments consider
essential to demonstrate the safety of the model [6, 9].

Our findings offer valuable insights into the dynamics at play in the AI
regulation landscape and highlight the indispensable role of reliable
regulators, incentivised by either governmental rewards or the maintenance of
a prestigious reputation. They highlight the value of taking a game-theoretic
approach to formally model the effects of different kinds of regulatory system
on the behaviour of both AI creators and users.

##  VI Appendix

(a) $x_{0}=y_{0}=z_{0}=0.1$, $\epsilon=-1.2$, $b_{U}=b_{R}=4$

(b) $x_{0}=y_{0}=z_{0}=0.5$, $\epsilon=-1.2$, $b_{U}=b_{R}=4$

(c) $x_{0}=y_{0}=z_{0}=0.1$, $\epsilon=-0.5$, $b_{U}=b_{R}=4$

(d) $x_{0}=y_{0}=z_{0}=0.5$, $\epsilon=-0.5$, $b_{U}=b_{R}=4$

(e) $x_{0}=y_{0}=z_{0}=0.1$, $\epsilon=-1.2$, $b_{U}=b_{R}=1$

(f) $x_{0}=y_{0}=z_{0}=0.5$, $\epsilon=-1.2$, $b_{U}=b_{R}=1$

Figure 8: Numerical integration of the evolution equation for the model with
conditional trust (18), describing the evolution of the density of trusting
users $x(t)$, cooperating creators $y(t)$ and cooperating regulators $z(t)$.
In all these simulations: $c_{P}=0.5$, $u=1.5$, $c_{R}=0.5$, $b_{fo}-v=5$. For
the initial conditions and the other parameters, see the captions.

## Acknowledgement

This work is produced during the workshop ”AI Governance Modelling”, funded
through the generous support from the Future of Life institute (TAH).

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The electricity grid is expected to require vast investments due to the decarbonization-by-electrification trend, calling for a change in grid tariff design which provides proper incentives for reducing peak loads. However, price signals from grid tariffs could be “distorted” from electricity spot prices which also represents a significant of the total consumer electricity bill. This paper attempts to identify whether there is a price signal conflict between grid tariffs and spot prices. Four different grid tariff designs are compared, using a generic demand response model as part of a cost-minimizing linear program to simulate the reduction in peak load. The method is applied to metered electricity demand from 3608 consumers in Oslo, Norway. Results show that new grid tariff designs reduce peak loads by 1-4%, and that reduction in peak load is smaller when consumers are subject to electricity spot prices.

[I]$T$Set of hourly time steps, index $t$

[I]$D$Set of days, index $d$
[I]$M$Set of months, index $m$

[P]$L_{t}$Original load series [kWh/h]

[P]$C^{red}$Discomfort cost of reducing load [$\frac{ \euro }{kWh}$]

[P]$C^{ET}$Energy term [$\frac{ \euro }{kWh}$]
[P]$C^{h}$Energy term above subscribed capacity [$\frac{ \euro }{kWh}$]
[P]$C^{sub}$Annual subscribed capacity cost[$\frac{\euro }{kW \times year}$]
[P]$C^{spot}$Electricity spot price [$\frac{ \euro }{kWh}$]
[P]$C_t^{TOU}$Static time-of-use network tariff [$\frac{ \euro }{kWh}$]
[P]$C_t^{dynTOU}$Dynamic time-of-use network tariff [$\frac{ \euro }{kWh}$]
[P]$C^{peak}$Demand charge peak cost[$\frac{ \euro }{kWh}$]

[P]$C^{e}$Total electricity cost []
[P]$C^{g}$Total grid tariff cost []
[P]$C^{f}$Total flexibility use cost []

[P]${Q}^{flex}$Share of reducible load in an hour [%]
[P]${E}^{flex}$Share of reducible energy in a day [%]

[V]$x_{t}^{i}$New load series with demand response [kWh/h]
[V]$x_{t}^{l}$Electricity consumption below sub. cap. [kWh/h]
[V]$x_{t}^{h}$Electricity consumption above sub. cap. [kWh/h]
[V]$x_{m}^{max}$Monthly peak demand [kWh/h]
[V]$x^{sub}$Subscribed capacity [kW]
[V]$q_{t}^{red}$Load reduction [kWh/h]


# Construction of A Lattice on the completion space of an algebra and an
isomorphism to its Caratheodory Extension

Jun Tanaka  University of California, Riverside, USA<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>and  Peter F. Mcloughlin
Crafton Hills College, USA<EMAIL_ADDRESS>
(Date: August, 10, 2008)

###### Abstract.

In this paper, we will show how one is able to construct a lattice on the
completion of an algebra and to obtain an isomorphism to its Caratheodory
Extension. In addition, it will be shown that the lattice form a
$\sigma$-algebra and a complete Heyting algebra of countable type.

###### Key words and phrases:

Caratheodory Extension Theorem, Lattice Theory, Measure Theory

###### 2000 Mathematics Subject Classification:

Primary: 28A12, 28B20

##  1\. Introduction

The Caratheodory Extension Theorem is a crucial part of any advanced Real
Analysis course; the process extends an algebra to a $\sigma$-algebra and a
measure on an algebra to a measure on a $\sigma$-algebra. This is a powerful
tool in both measure theory and statistics (see for example, [5],[2]). A
Boolean Algebra structure on the Caratheodory Extension was discussed in
several papers (e.g. Kolmogorov, [8], Coquand and Palmgren [3]). Furthermore,
the Caratheodory Extension is complete with a pseudometric
$d(.,.)=\mu^{\ast}(.\triangle.)$ where $\mu^{\ast}(.)$ is the outer measure on
the power set P(X) (Please refer to [4]). The Caratheodory Extension has
various rich structures which are open to further investigation.

In this paper, we will show how the Caratheodory Extension process is
intimately related to the metric completion process. In particular, it will be
shown how one is able to construct a lattice on the completion of an algebra
and to obtain an isomorphism to its Caratheodory Extension. The author
believes this method to be new, intuitive, and constructive. Especially
noteworthy is that, in Definition 4.14 of this paper, we give a precise
definition of $\vee^{\infty}$ on the completion space
$(\bar{d},\overline{\Omega}$) of an algebra $\Omega$ and show that the lattice
is a complete Heyting algebra of countable type.

Let X be an arbitrary nonempty set, and let $\mu$ be a finite measure on an
algebra $\Omega\subset\mathbf{P}(X)$. Note that
$d(.,.)=\mu^{\ast}(.\triangle.)$ defines a pseudometric on $\Omega$. Denote by
$(\bar{d},\overline{\Omega})$ the completion of $(d,\Omega)$. Also, let $S$ be
the set of all Cauchy sequences in $(d,{\Omega})$. By the metric completion
procedures, we know
$\bar{d}(\\{\overline{B^{\alpha_{1}}_{n}}\\},\\{\overline{B^{\alpha_{2}}_{n}}\\})=\lim
d({B^{\alpha_{1}}_{n}},{B^{\alpha_{2}}_{n}})$ where
$\\{{B^{\alpha_{i}}_{n}}\\}$ is in $S$ with
$\\{\overline{B^{\alpha_{i}}_{n}}\\}=\\{\\{B_{n}\\}\in S:\lim
d(B_{n},B_{n}^{\alpha_{i}})=0\\}$ for $i=1$or 2.

######  Definition 1.1.

Let
$\bar{\mu}(\\{\overline{B^{\alpha}_{n}}\\})=\bar{d}(\\{\overline{B^{\alpha}_{n}}\\},\\{\bar{\phi}\\})=\lim\mu({B^{\alpha}_{n}})$.

######  Definition 1.2.

If $B_{n}\in\Omega$, then $\\{B_{n}\\}$ is called a $\mu$-Cauchy sequence if
$\mu(B_{n}\triangle B_{m})\rightarrow 0$ as n,m $\rightarrow\infty$.

######  Definition 1.3.

For A, B in $\mathbf{P}(X)$, A = B a.e. if $\mu^{\ast}(A\triangle B)=0.$

In Section 4, we will show that a lattice structure can be naturally defined
on $\overline{\Omega}$ which makes it a $\sigma$-algebra. Let
$\widetilde{\mathbf{S}}$ = $\\{S\in\mathbf{P}(X)\ |$ $\exists\ \mu$-Cauchy
sequence $\\{B_{n}\\}$ s.t. $\lim\mu^{\ast}(B_{n}\triangle S)=0\\}$. In [9],
we proved that $\widetilde{\mathbf{S}}$ is a $\sigma$-algebra where, for any
$\mu$-Cauchy sequence $\\{B_{n}\\}$ such that $\lim\mu^{\ast}(B_{n}\triangle
S)=0$, the measure $\widetilde{\mu}(S)$ on $\widetilde{\mathbf{S}}$ is defined
as $\widetilde{\mu}(S)$ = $\lim\mu(B_{n})$. In addition, we proved that
$\widetilde{\mu}$ is a countably additive measure on $\widetilde{\mathbf{S}}$.
Thus, ($\widetilde{\mu}$, $\widetilde{\mathbf{S}}$) is a measure space. We
showed that the Caratheodory Extension of $\Omega$ can be expressed as the set
of limit points of $\mu$-Cauchy sequences under the pseudometric $d(A,B)$ =
$\mu^{\ast}(A\triangle B)$. Moreover, when the measure is a $\sigma$-finite
measure, we obtained an equivalent expression of the Caratheodory Extension,
$\widetilde{\mathbf{S}}$ = $\\{S\in\mathbf{P}(X)$ $|$ $\exists\ \mu$-Cauchy
sequence $\\{B_{n}\\}$ s.t. $\lim\mu^{\ast}(B_{n}\triangle S)=0\\}$. Theorem 2
in [9] shows that $E$ is a measurable set iff $E$ is in
$\widetilde{\mathbf{S}}$. Thus, the measure space ($\widetilde{\mu}$,
$\widetilde{\mathbf{S}}$) agrees with the Caratheodory Extension when $\mu$ is
a finite measure. Moreover, it shows that measurable sets are exactly limit
points of $\mu$-Cauchy sequences. The $\sigma$-finite case follows from the
finite case. In Section 5, we will define a $\sigma$-algebra lattice
isomorphism between two $\sigma$-algebras, and define a map $F:$
$\overline{\Omega}\rightarrow\mathbf{P}(X)$ given by
$F(\overline{\\{B_{n}\\}})=B$ where $\lim\mu^{\ast}(B_{n}\triangle B)=0$. We
will show that $F$ is an isometry and a $\sigma$-algebra lattice isomorphism
between the completion $\overline{\Omega}$ and the Caratheodory Extension of
$\Omega$ under the equivalence relation $\sim$ defined as A $\sim$ B iff
$\mu^{\ast}(A\triangle B)$ =0.

##  2\. Main Definitions and Notations

Our notation agrees with that in [9]. For emphasis, let X be a nonempty set.
Throughout the paper, unless otherwise stated, $\mu$ will be a finite measure
on an algebra $\Omega\subset\mathbf{P}(X)$. Unless otherwise stated,
$\\{B^{\alpha}_{n}\\}$ will be a $\mu$-Cauchy sequence.

In addition, for each $\mu$-Cauchy sequence $\\{B_{n}\\}$, a capital letter
corresponds to the limit point of the $\mu$-Cauchy sequence. For instance,
$\lim\mu^{\ast}(A_{n}\triangle A)=0$, $\lim\mu^{\ast}(B_{n}\triangle B)=0$,
$\lim\mu^{\ast}(Y_{L}\triangle Y)=0$. There exist such limit points in
$\mathbf{P}(X)$ since the Caratheodory Extension is complete under the
pseudometric d $($[4]$)$.

We recall that by a $\sigma$-algebra of subsets of X, we mean a family
$\Omega$ of subsets such that (1) every countable union of sets in $\Omega$ is
in $\Omega$, and (2) $A^{C}$ is in $\Omega$ whenever A is.

######  Definition 2.1.

Define a map $F:$ $\overline{\Omega}\rightarrow\mathbf{P}(X)$ given by
$F(\overline{\\{B_{n}\\}})=B$

where $\lim\mu^{\ast}(B_{n}\triangle B)=0$. Note that such a B always exists
in $\mathbf{P}(X)$ since the Caratheodory Extension is complete $($[4]$)$.

######  Definition 2.2.

$\sigma$-algebra lattice isomorphism

Suppose lattices $($X, $\vee_{X}$,$\wedge_{X}$ $)$ and $($Y,
$\vee_{Y}$,$\wedge_{Y}$ $)$ are $\sigma$-algebras, and H: X $\rightarrow$ Y is
a one-to-one, onto well-defined map. Then H is called a $\sigma$-algebra
lattice isomorphism if

$\displaystyle H(\cdot\vee_{X}\cdot)=$ $\displaystyle
H(\cdot)\vee_{Y}H(\cdot),\ \ \ \ \ \ \ H(\vee^{\infty}_{X}\ \cdot)$
$\displaystyle=$ $\displaystyle\vee^{\infty}_{Y}H(\cdot),$ $\displaystyle
H(\cdot\wedge_{X}\cdot)=$ $\displaystyle H(\cdot)\wedge_{Y}H(\cdot),\ \ \ \ \
\ \ H(\cdot^{\textbf{C}})$ $\displaystyle=$ $\displaystyle
H(\cdot)^{\textbf{C}}.$

We will show that F is a $\sigma$-algebra lattice isomorphism between
$\overline{\Omega}$ and $\widetilde{\mathbf{S}}_{\diagup_{\sim}}$ where $\sim$
is an equivalence relation in the sense of Definition 1.3.

##  3\. Preliminaries

In this section, we shall briefly review the well-known facts about lattice
theory (e.g. Birkhoff [1], Iwamura [7]), propose an extension lattice, and
investigate its properties, as well as some lemmas from [9]. A nonempty set L
is called a lattice if L is closed under the operations of meet ($\wedge$) and
join ($\vee$). It is denoted by (L,$\wedge$,$\vee$) or simply L. If it
satisfies, in addition, the distributive law, then it is called a distributive
lattice. For two lattices L and $L^{\prime}$, a bijection from L to
$L^{\prime}$, which preserves lattice operations is called a lattice
isomorphism, or simply an isomorphism. If there is an isomorphism from L to
$L^{\prime}$, then L is called lattice-isomorphic with $L^{\prime}$, and we
write L $\cong$ $L^{\prime}$. We write x $\leq$ y if x $\wedge$ y = x or,
equivalently, if x $\vee$ y = y. L is called complete if, for any subset A of
L, L contains the supremum $\vee$A and the infimum $\wedge$A, with respect to
the above order. A complete lattice L includes the maximum and minimum
elements, which are denoted by I and O, or 1 and 0, respectively. A
distributive lattice is called a Boolean algebra or a Boolean lattice, if, for
any element x in L, there exists a unique complement $x^{C}\in L$ such that x
$\vee$ $x^{C}$ = 1 and $x\wedge x^{C}$ = 0. Let L be a lattice and
$\cdot^{c}$: L $\rightarrow$ L be an operator. Then $\cdot^{c}$ is called a
lattice complement in L if the following conditions are satisfied:

(1) $\forall x$ $\in$ L, $x^{C}\wedge x$ = 0, $x^{C}\vee x$ = I,

(2) $\forall$ x, y $\in$ L, x $\leq$ y $\Rightarrow$ $x^{C}\geq y^{C}$,

(3) $\forall$ x $\in$ L, $(x^{C})^{C}$ = x.

######  Definition 3.1.

A complete lattice L is called a lattice $\sigma$-algebra or simply a
$\sigma$-algebra if the following conditions are satisfied:

(1) $\forall$ A $\in$ L, $A^{C}\in$L where $\cdot^{C}$ is a lattice
complement,

(2) $\forall$ $A_{i}\in$ L, $\vee^{\infty}A_{i}\in$ L.

######  Definition 3.2.

A complete lattice is called a complete Heyting algebra $($cHa$)$, if

$\displaystyle\vee_{i\in I}\ (x_{i}\wedge y)=(\vee_{i\in I}\ x_{i})\wedge y$

holds for $\forall x_{i},y\in L$ $(i\in I)$; where I is an index set of
arbitrary cardinal number. In the case that the cardinality of I is countable,
it is called a complete Heyting Algebra of countable type.

It is well-known that for a set E, $|P(E)|$= $2^{|E|}$. The set of all subsets
of E is a Boolean algebra.

######  Lemma 3.3.

If $\\{B_{n}\\}$ is a $\mu$-Cauchy sequence, then $\\{\mu(B_{n})\\}$ is a
Cauchy sequence of real numbers.

###### Proof.

Note $B_{n}\subseteq(B_{n}\triangle B_{m})\cup B_{m}$ and
$B_{m}\subseteq(B_{n}\triangle B_{m})\cup B_{n}$ implies
$\mu(B_{n})\leq\mu(B_{n}\triangle B_{m})+\mu(B_{m})$ and
$\mu(B_{m})\leq\mu(B_{n}\triangle B_{m})+\mu(B_{n})$. Hence,
$|\mu(B_{n})-\mu(B_{m})|\leq\mu(B_{n}\triangle B_{m})$. ∎

######  Lemma 3.4.

$d(A_{1}\cup A_{2},A_{3}\cup A_{4})\leq d(A_{1},A_{3})+d(A_{2},A_{4})$ for any
$A_{i}\in\Omega$.

###### Proof.

The proof is in [9] ∎

######  Lemma 3.5.

$\\{B^{\alpha}_{n}\cup B^{\gamma}_{n}\\}$ and $\\{(B^{\alpha}_{n})^{C}\\}$ are
$\mu$-Cauchy sequences.

###### Proof.

By Lemma 3.4,

$d(B^{\alpha}_{n}\cup B^{\gamma}_{n},B^{\alpha}_{m}\cup B^{\gamma}_{m})\leq
d(B^{\alpha}_{n},B^{\alpha}_{m})+d(B^{\gamma}_{n},B^{\gamma}_{m}).$
$\displaystyle\mbox{Moreover},\,\,d((B^{\alpha}_{n})^{C},(B^{\alpha}_{m})^{C})=\mu((B^{\alpha}_{n})^{C}\triangle(B^{\alpha}_{m})^{C})=\mu(B^{\alpha}_{n}\Delta
B^{\alpha}_{m})=d(B^{\alpha}_{n},B^{\alpha}_{m}).$ $\displaystyle\mbox{Hence,
the claim follows}.$

∎

##  4\. The Completion Space $(\bar{d},\overline{\Omega}$) and the Lattice
(L,$\wedge$,$\vee$)

In this section, we introduce the completion of ($d$,$\Omega$) where $\Omega$
is an algebra of subsets of X.

######  Definition 4.1.

Let $E_{\alpha}=\\{\overline{B^{\alpha}_{n}}\\}$ and
$E_{A}=\overline{\\{A\\}}$ when $A\in\Omega$ and $\\{B^{\alpha}_{n}\\}$ is a
sequence in $\Omega$.

######  Definition 4.2.

Define $E_{\alpha}\vee E_{\gamma}=\\{\overline{B^{\alpha}_{n}\cup
B^{\gamma}_{n}}\\}$, $E_{\alpha}\wedge
E_{\gamma}=\\{\overline{B^{\alpha}_{n}\cap B^{\gamma}_{n}}\\}$, and
$E^{C}_{\alpha}=\\{\overline{(B^{\alpha}_{n})^{C}}\\}$.

######  Remark 4.3.

$E_{\alpha}\wedge E_{\gamma}=E_{\phi}$ iff $\bar{\mu}(E_{\alpha}\wedge
E_{\gamma})=0$ iff $\lim\mu(B^{\alpha}_{n}\cap B^{\gamma}_{n})=0$. Moreover,
$E_{\alpha}\subset E_{\gamma}$ iff $\bar{\mu}(E_{\alpha}\wedge
E^{C}_{\gamma})=0$. The containment may be written in the usual lattice sense;
namely, $E_{\alpha}\wedge E_{\gamma}=E_{\alpha}$.

######  Lemma 4.4.

The map
$\vee:\overline{\Omega}\times\overline{\Omega}\rightarrow\overline{\Omega}$
defined by $\vee(E_{\alpha}\times E_{\gamma})=E_{\alpha}\vee
E_{\gamma}=\\{\overline{B^{\alpha}_{n}\cup B^{\gamma}_{n}}\\}$ is well-
defined.

###### Proof.

Suppose $\\{\overline{B^{\alpha}_{n}}\\}=\\{\overline{A^{\alpha}_{n}}\\}$ and
$\\{\overline{A^{\gamma}_{n}}\\}=\\{\overline{B^{\gamma}_{n}}\\}$.

Then $\lim d(B^{\alpha}_{n},A^{\alpha}_{n})=\lim
d(B^{\gamma}_{n},A^{\gamma}_{n})=0$. Also, by Lemma 3, $\\{B^{\alpha}_{n}\cup
B^{\gamma}_{n}\\}$ and $\\{A^{\alpha}_{n}\cup A^{\gamma}_{n}\\}$ are
$\mu$-Cauchy sequences.

Hence, by Lemma 3.4, $d(B^{\alpha}_{n}\cup B^{\gamma}_{n},A^{\alpha}_{n}\cup
A^{\gamma}_{n})\leq
d(B^{\alpha}_{n},A^{\alpha}_{n})+d(B^{\gamma}_{n},A^{\gamma}_{n})$ and the
claim follows. ∎

######  Lemma 4.5.

The map $\cdot{}^{C}:\overline{\Omega}\rightarrow\overline{\Omega}$ defined by

$(\\{\overline{B^{\alpha}_{n}}\\})^{C}=\overline{\\{(B^{\alpha}_{n})^{C}\\}}$

is well-defined.

###### Proof.

Suppose $\\{\overline{B^{\alpha}_{n}}\\}=\\{\overline{A^{\alpha}_{n}}\\}$.
Then $\lim d(B^{\alpha}_{n},A^{\alpha}_{n})=0$. Now,
$d((B^{\alpha}_{n})^{C},(A^{\alpha}_{n})^{C})=\mu((B^{\alpha}_{n})^{C}\triangle(A^{\alpha}_{n})^{C})=\mu(B^{\alpha}_{n}\triangle
A^{\alpha}_{n})=d(B^{\alpha}_{n},A^{\alpha}_{n})$ which implies
$\\{(B^{\alpha}_{n})^{C}\\}\sim\\{(A^{\alpha}_{n})^{C}\\}$. ∎

######  Lemma 4.6.

The map
$\wedge:\overline{\Omega}\times\overline{\Omega}\rightarrow\overline{\Omega}$
defined by $\wedge(E_{\alpha}\times E_{\gamma})=E_{\alpha}\wedge
E_{\gamma}=\\{\overline{B^{\alpha}_{n}\cap B^{\gamma}_{n}}\\}$ is well-
defined.

###### Proof.

The proof follows immediately from the two previous lemmas. ∎

######  Lemma 4.7.

$\cdot{}^{C}$ in Lemma 4.5 is a lattice complement.

###### Proof.

The proof is obvious. ∎

######  Lemma 4.8.

$\bar{\mu}(E_{\alpha}\vee
E_{\gamma})\leq\bar{\mu}(E_{\alpha})+\bar{\mu}(E_{\gamma})$ for
$E_{\alpha},E_{\gamma}\in\overline{\Omega}$.

###### Proof.

$E_{\alpha}=\\{\overline{B^{\alpha}_{n}}\\}$ and
$E_{\gamma}=\\{\overline{B^{\gamma}_{n}}\\}$ implies $E_{\alpha}\vee
E_{\gamma}=\\{\overline{B^{\alpha}_{n}\cup B^{\gamma}_{n}}\\}$. Hence
$\bar{\mu}(E_{\alpha}\vee E_{\gamma})=\lim\mu(B^{\alpha}_{n}\cup
B^{\gamma}_{n})\leq\lim[\mu(B^{\alpha}_{n})+\mu(B^{\gamma}_{n})]=\lim\mu(B^{\alpha}_{n})+\lim\mu(B^{\gamma}_{n})\leq\bar{\mu}(E_{\alpha})+\bar{\mu}(E_{\gamma})$.
∎

######  Lemma 4.9.

If $E_{\alpha_{1}}$ and $E_{\alpha_{2}}$ are disjoint, then
$\bar{\mu}(E_{\alpha_{1}}\vee
E_{\alpha_{2}})=\bar{\mu}(E_{\alpha_{1}})+\bar{\mu}(E_{\alpha_{2}})$.

###### Proof.

By Remark 4.3, we must have $\lim\mu(B^{\alpha_{1}}_{n}\cap
B^{\alpha_{2}}_{n})=0$. Note, $B^{\alpha_{i}}_{n}=(B^{\alpha_{i}}_{n}\cap
B^{\alpha_{j}}_{n})\ \cup\ (B^{\alpha_{i}}_{n}\cap(B^{\alpha_{j}}_{n})^{C})$
implies $\mu(B^{\alpha_{i}}_{n})=\mu(B^{\alpha_{i}}_{n}\cap
B^{\alpha_{j}}_{n})+\mu(B^{\alpha_{i}}_{n}\cap(B^{\alpha_{j}}_{n})^{C})$ and
$\lim\mu(B^{\alpha_{i}}_{n})=\lim\mu(B^{\alpha_{i}}_{n}\cap(B^{\alpha_{j}}_{n})^{C})$
for $\\{i,j\\}=\\{1,2\\}$.

Moreover, $\mu(B^{\alpha_{1}}_{n}\cup B^{\alpha_{2}}_{n})=\mu(\
(B^{\alpha_{1}}_{n}\cap B^{\alpha_{2}}_{n})\ \cup\
(B^{\alpha_{1}}_{n}\cap(B^{\alpha_{2}}_{n})^{C})\ \cup\
(B^{\alpha_{2}}_{n}\cap(B^{\alpha_{1}}_{n})^{C})\ )=\mu(B^{\alpha_{1}}_{n}\cap
B^{\alpha_{2}}_{n})+\mu(B^{\alpha_{1}}_{n}\cap(B^{\alpha_{2}}_{n})^{C})+\mu(B^{\alpha_{2}}_{n}\cap(B^{\alpha_{1}}_{n})^{C})$
implies $\bar{\mu}(E_{\alpha_{1}}\vee
E_{\alpha_{2}})=\lim\mu(B^{\alpha_{1}}_{n}\cup
B^{\alpha_{2}}_{n})=\lim\mu(B^{\alpha_{1}}_{n})+\lim\mu(B^{\alpha_{2}}_{n})=\bar{\mu}(E_{\alpha_{1}})+\bar{\mu}(E_{\alpha_{2}})$.
∎

######  Theorem 4.10.

$\overline{\Omega}$ is an algebra and $\bar{\mu}$ is a measure on
$\overline{\Omega}$.

###### Proof.

By Lemmas 3.5, 4.6 and 4.5, $\overline{\Omega}$ is an algebra. Moreover, the
fact that $\bar{\mu}$ takes on values in $\mathbb{R}^{+}$ along with Lemmas 6
and 7 imply $\bar{\mu}$ is a measure on $\overline{\Omega}$. ∎

######  Remark 4.11.

$\bar{d}(E_{\alpha},E_{\gamma})=\lim
d(B^{\alpha}_{n},B^{\gamma}_{n})=\lim[\mu(B^{\alpha}_{n}\cap(B^{\gamma}_{n})^{C})+\mu(B^{\gamma}_{n}\cap(B^{\alpha}_{n})^{C})]=\lim\mu(B^{\alpha}_{n}\cap(B^{\gamma}_{n})^{C})+\lim\mu(B^{\gamma}_{n}\cap(B^{\alpha}_{n})^{C})=\bar{\mu}(E_{\alpha}\wedge(E_{\gamma})^{C})+\bar{\mu}(E_{\gamma}\wedge(E_{\alpha})^{C})=\bar{\mu}(E_{\alpha}\triangle
E_{\gamma})$.

######  Lemma 4.12.

Let $E_{i}$ = $\overline{\\{B^{i}_{n}\\}}$ $\in\overline{\Omega}$ for i $\geq$
1 and by following the proof of Lemma 8 in [9], construct $Y_{L}$ =
$\cup_{i=1}^{N_{L}}B^{i}_{K_{L}}$ for each $L$ such that

$\mu^{\ast}(\cup_{i=1}^{\infty}S_{i}\triangle\cup_{i=1}^{N_{L}}B^{i}_{K_{L}})\leq\mu^{\ast}(\cup_{i=N_{L}+1}^{\infty}S_{i})+\mu^{\ast}(\cup_{i=1}^{N_{L}}S_{i}\triangle\cup_{i=1}^{N_{L}}B^{i}_{K_{L}})<\frac{1}{L}.$

Then E := $\overline{\\{Y_{L}\\}}$ satisfies the following conditions;

$\displaystyle 1.$ $\displaystyle\vee^{n}_{i=1}E_{i}\leq E\ \ \mbox{for all}\
n,$ $\displaystyle 2.$
$\displaystyle\lim\bar{\mu}(E\wedge(\vee^{n}_{i=1}E_{i})^{C})=0.$

In particular, E is uniquely determined in $\overline{\Omega}$.

###### Proof.

Note that $E_{i}$ = $\overline{\\{B^{i}_{n}\\}}$ =
$\overline{\\{B^{i}_{K_{L}}\\}}$.

$(\vee_{i=1}^{n}E_{i})\wedge\overline{\\{Y_{L}\\}}$ =
$\overline{\\{\cup_{i=1}^{n}B^{i}_{K_{L}}\cap Y_{L}\\}}$ =
$\overline{\\{\cup_{i=1}^{n}B^{i}_{K_{L}}\\}}$ = $\vee_{i=1}^{n}E_{i}$ for any
n.

Let $N_{L}>n$.

$\displaystyle\mu(\cup_{i=1}^{N_{L}}B^{i}_{K_{L}}\cap(\cup_{i=1}^{n}B^{i}_{K_{L}})^{\textbf{C}})=$
$\displaystyle\mu(\cup_{i=1}^{N_{L}}B^{i}_{K_{L}}\triangle\cup_{i=1}^{n}B^{i}_{K_{L}})=\mu^{\ast}(\cup_{i=1}^{N_{L}}B^{i}_{K_{L}}\triangle\cup_{i=1}^{n}B^{i}_{K_{L}})$
$\displaystyle\leq$
$\displaystyle\mu^{\ast}(\cup_{i=1}^{\infty}S_{i}\triangle\cup_{i=1}^{N_{L}}B^{i}_{K_{L}})+\mu^{\ast}(\cup_{i=1}^{\infty}S_{i}\triangle\cup_{i=1}^{n}B^{i}_{K_{L}}).$

This implies that
$\lim\overline{\mu}(\overline{\\{Y_{L}\\}}\wedge(\vee_{i=1}^{n}E_{i})^{\textbf{C}})$
= 0.

Suppose $E,E^{\prime}\in\overline{\Omega}$ both satisfy condition 1 and 2
above. Then $\vee^{n}_{i=1}E_{i}\leq E\wedge E^{\prime}\leq E$ for all n.
Hence, we must have $E=E\wedge E^{\prime}$. Similarly, we must have
$E^{\prime}=E\wedge E^{\prime}$. Therefore E is unique.

∎

######  Lemma 4.13.

Let $E_{i}$ = $\overline{\\{B^{i}_{n}\\}}$ $\in\overline{\Omega}$ for i $\geq$
1 and $E_{i}\wedge E_{j}=E_{\phi}$ for $i\neq j$. Then
$\bar{\mu}(E)=\sum^{\infty}_{i=1}\bar{\mu}(E_{i})$ where E is defined as in
Lemma 4.12.

###### Proof.

Suppose that $E_{i}$ = $\overline{\\{B^{i}_{n}\\}}$ $\in\overline{\Omega}$ for
i $\geq$ 1 and $E_{i}\wedge E_{j}=E_{\phi}$ for $i\neq j$. Then it follows
that
$\bar{\mu}(\vee^{n}_{i=1}E_{i})=\sum^{k}_{i=1}\bar{\mu}(E_{i})\leq\bar{\mu}(E)$
for all $k$.

Hence, we must have

(1) $\sum^{\infty}_{i=1}\bar{\mu}(E_{i})\leq\bar{\mu}(E).$

Now since
$\bar{\mu}(E)=\bar{\mu}(\vee^{n}_{i=1}E_{i})+\bar{\mu}(E\wedge(\vee^{n}_{i=1}E_{i})^{C})$
for all n and $\lim\bar{\mu}(E\wedge(\vee^{n}_{i=1}E_{i})^{C})=0$ , we must
have $\bar{\mu}(E)=\sum^{\infty}_{i=1}\bar{\mu}(E_{i})$ by (1).

∎

######  Definition 4.14.

For ${E}_{i}$ in $\overline{\Omega}$, we define $\vee^{\infty}_{i=1}{E}_{i}=E$
where $E$ is as in Lemma 4.12.

######  Theorem 4.15.

$(\bar{\mu},\overline{\Omega})$ is a measure space.

###### Proof.

This follows by Theorem 4.10, 4.12, and 4.13, and Definition 4.14. ∎

##  5\. The Isomorphism

In this section, we introduce an isomorphism $F:$
$\overline{\Omega}\rightarrow\mathbf{P}(X)$ given by
$F(\overline{\\{B_{n}\\}})=B$.

######  Lemma 5.1.

$F$ : $\overline{\Omega}\rightarrow\widetilde{\mathbf{S}}_{\diagup_{\sim}}$ is
a well-defined map.

###### Proof.

Suppose that $\overline{\\{A_{n}\\}}=\overline{\\{B_{n}\\}}$.

There exist A and B in $\mathbf{P}(X)$ such that
$\lim\mu^{\ast}(A_{n}\triangle A)=0$ and $\lim\mu^{\ast}(B_{n}\triangle B)=0$.

$\mu^{\ast}(A\triangle B)\leq$ $\mu^{\ast}(A\triangle
A_{n})+\mu^{\ast}(A_{n}\triangle B_{n})+\mu^{\ast}(B_{n}\triangle B)$ by the
triangle inequality.

By taking limits on both sides, $\mu^{\ast}(A\triangle B)=0$.

Thus, A = B a.e.. Therefore, F is well-defined.

∎

######  Theorem 5.2.

F is an isometry between $\overline{\Omega}$ and
$\widetilde{\mathbf{S}}_{\diagup_{\sim}}$.

###### Proof.

First, we show F is onto $\widetilde{\mathbf{S}}$. Let
$X\in\widetilde{\mathbf{S}}$. Then there exists a $\mu$-Cauchy sequence
$\\{B_{n}\\}$ such that $\lim\mu^{\ast}(B_{n}\triangle X)=0$. Thus
$F(\overline{\\{B_{n}\\}})=X$ a.e.. Therefore, F is onto.

Second, we will show F preserves the metric. Let $\overline{\\{A_{n}\\}}$,
$\overline{\\{B_{n}\\}}$ $\in$ $\overline{\Omega}$. Letting A and B be as
before, we have

$\lim\mu(A_{n}\triangle B_{n})=\mu^{\ast}(A\triangle B).$

Therefore, $\overline{d}(\overline{\\{A_{n}\\}},\overline{\\{B_{n}\\}})$ =
$\lim\mu(A_{n}\triangle B_{n})$ = $\mu^{\ast}(A\triangle B)$

= $\mu^{\ast}(F(\overline{\\{A_{n}\\}})\triangle F(\overline{\\{B_{n}\\}}))$ =
$d(F(\overline{\\{A_{n}\\}}),F(\overline{\\{B_{n}\\}}))$. Thus, F preserves
the metric.

Lastly, we will show that F is one-to-one. Let
$F(\overline{\\{A_{n}\\}}),F(\overline{\\{B_{n}\\}})\in\widetilde{\mathbf{S}}$
such that $F(\overline{\\{A_{n}\\}})=F(\overline{\\{B_{n}\\}})$ a.e..

Then $A=B$ a.e. implies $\mu^{\ast}(A\triangle B)=0$. Then, as in the proof of
F being onto, $\lim\mu(A_{n}\triangle B_{n})$ = $\mu^{\ast}(A\triangle B)$.
Thus $\overline{\\{A_{n}\\}}$ = $\overline{\\{B_{n}\\}}$ and F is one-to-one.
Therefore, F is an isometry between $\overline{\Omega}$ and
$\widetilde{\mathbf{S}}_{\diagup_{\sim}}$.

∎

######  Theorem 5.3.

F is a $\sigma$-algebra lattice isomorphism between $\overline{\Omega}$ and
$\widetilde{\mathbf{S}}_{\diagup_{\sim}}$.

###### Proof.

We already showed that F is a one-to-one, onto map in Theorem 5.2.

$F(\overline{\\{A_{n}\\}}\vee\overline{\\{B_{n}\\}})$ =
$F(\overline{\\{A_{n}\cup B_{n}\\}})$ = $F(\overline{\\{A_{n}\\}})\cup
F(\overline{\\{B_{n}\\}})$ a.e. since $\lim\mu(A_{n}\cup B_{n}\ \ \triangle\ \
A\cup B)$ = 0.

Let $\overline{\\{B_{n}\\}}\in\overline{\Omega}$.

Then,

$F(\overline{\\{B_{n}\\}}^{\textbf{C}})=F(\overline{\\{(B_{n})^{\textbf{C}}\\}})=B^{\textbf{C}}\
\ a.e..$

Thus, $F(\cdot^{\textbf{C}})$ = $F(\cdot)^{\textbf{C}}$ in
$\widetilde{\mathbf{S}}_{\diagup_{\sim}}$.

Similarly, $F(\cdot\wedge\cdot)$ =
$F(\cdot^{\textbf{C}}\vee\cdot^{\textbf{C}})^{\textbf{C}}$ =
$[F(\cdot^{\textbf{C}})\cup F(\cdot^{\textbf{C}})]^{\textbf{C}}$=
$F(\cdot)\cap F(\cdot)$.

Let $E_{\alpha_{i}}$ $\in\overline{\Omega}$ for i $\geq$ 1 and
$E_{\alpha_{i}}$ = $\overline{\\{B^{\alpha_{i}}_{n}\\}}$.

Then for each i, there exists an $S_{i}\in\widetilde{\mathbf{S}}$ such that
$\lim\mu^{\ast}(B^{\alpha_{i}}_{n}\triangle S_{i})=0$.

Now suppose we have $\\{Y_{L}\\}$ in the same manner as in Lemma 4.12. By
design, $\\{Y_{L}\\}$ converges to $\cup_{i=1}^{\infty}S_{i}$. Then
$\overline{\\{Y_{L}\\}}$ = $\vee_{i=1}^{\infty}E_{\alpha_{i}}$ by Lemma 4.12.
Now we have

$F(\vee_{i=1}^{\infty}E_{\alpha_{i}})=F(\vee_{i=1}^{\infty}\overline{\\{B^{\alpha_{i}}_{n}\\}})=F(\overline{\\{Y_{L}\\}})=Y.$

Since $\lim\mu^{\ast}(Y_{L}\triangle Y)=0$ and
$\lim\mu^{\ast}(Y_{L}\triangle\cup^{\infty}_{i=1}S_{i})=0$, we have $Y$ =
$\cup^{\infty}_{i=1}S_{i}$ a.e.. In addition, $\cup^{\infty}_{i=1}S_{i}$ =
$\cup^{\infty}_{i=1}F(\overline{\\{B^{\alpha_{i}}_{n}\\}})$.

Thus,

$F(\vee_{i=1}^{\infty}E_{\alpha_{i}})=\cup^{\infty}_{i=1}F(E_{\alpha_{i}}).$

Therefore, the claim follows.

∎

######  Theorem 5.4.

$(\overline{\Omega}$,$\wedge$,$\vee)$ is a complete Heyting algebra of
countable type. More precisely, it is the case that the cardinality of I in
Definition 3.2 is countable.

###### Proof.

Since $\widetilde{\mathbf{S}}$ is a complete Heyting algebra of countable
type, the proof follows from Theorem 5.3. ∎

##  6\. Conclusion

Theorems 5.2 and 5.3 show that the completion of $\Omega$ is isometric and
$\sigma$-algebra lattice isomorphic to
$\widetilde{\mathbf{S}}_{\diagup_{\sim}}$. Thus, by the conclusion in [9] the
completion of $\Omega$ is isometric to, as well as $\sigma$-algebra isomorphic
to, the Caratheodory Extension under the equivalence relation $\sim$ . An
important note is that the isomorphism shows that Caratheodory Extension
Theorem and the Metric Completion Process are essentially the same, differing
only on measure zero sets. In addition, the completion $\overline{\Omega}$ of
$\Omega$ is a complete Heyting algebra of countable type by Theorem 5.4. The
$\sigma$-finite case follows from the finite case.

##  7\. Acknowledgement

The first author would like to thank Professors who gave him very professional
advice and suggestions, which he truly believes improved the presentation of
this paper.

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