Suppose in a two player zero-sum game, player I chooses a number $m$ from $\mathbb{Z}$, and player II chooses $n$ from $\mathbb{Z}$.
- If $m-n =1$, then player I recieves a payoff of $1$, while player II recieves a payoff of $-1$.
- If $m-n =-1$, then player I recieves a payoff of $-1$, while player II recieves a payoff of $1$.
- Otherewise, both player I and player II recieve a payoff of $0$.
What are the equilibrium strategies for player I and player II?
It seems to me, one obvious way is that player I and player II choose the same mixed strategy that assigns each number in $\mathbb{Z}$ the equal probability. The problem is that there isn't a countably additive probability measure to fulfill that task. Is finitely additive probability measure useful in this scenario?
The problem with allowing finitely additive measures as mixed strategies is that it is not clear how one evaluates payoffs. In particular, Fubuini's theorem on interchanging integrals can fail.
Here is a simple example: Ann and Bob pick natural numbers. If they pick the same number, there is a draw, otherwise the person who picked the highest number wins. Suppose both Ann and Bob choose a finitely additive mixed strategy that puts zero probability on a finite set (such measures can be constructed from a free ultrafilter). No matter which number Bob chooses, with probability $1$, Ann chooses a higher number. So integrating over all numbers Bob can choose, we see that Ann win with probability $1$. But by reversing the roles of Ann and Bob, we can argue in exactly the same way that Bob wins with probability $1$. This makes no sense.
Some games are just not meant to have a Nash equilibrium. Take the one-player game in which Ann can pick a natural number $n$ and gets a payoff of $n$. This game is a simple maximization problem and I don't think anyone believes it has a solution.