Player $i$ chooses an effort level, $e_i \in [0, 10]$. Let player $i$ have the following payoff function: $90 -e_i$ if $e_i > e_j$ and $80 -e_i$ if $e_i \leq e_j$. What is the Nash Equilibrium (NE) of this game? (I actually got confused at mixed-strategy part.)
For the pure-strategy NE - Let player $i$ choose an arbitrary effort level, say $8$, then he/she gets $72$ if the other player chooses the same effort level or less. Then the player $i$ is better off by choosing an higher effort level, say $9$, but indeed here we have the same situation. Hence, the player is better off by choosing the highest effort, $10$, which allows him to get $80$ at most. However, choosing $0$ effort is superior to this because he can get $80$ and there is no risk for getting $70$. This procedure continues on and we conclude that there is no pure-strategy NE.
Now we need to consider mixed-strategy NE. Here is the part where I got confused. Player $i$ is mixing over infinitely many strategies. So, we should consider a distribution where $\int F(.) =1.$ Thus, what should be the probability distribution that allows us to be indifferent among strategies?