such that | F ( a ) − F ( a $^{′}$) | ≤ 1 for every a, a ′ ∈ A such that ρ ( a, a $^{′}$) = 1 . Then for every r > 0 µ ({ a : F ( a ) ≥ ∫ F d µ + r }) ≤ e − r $^{2}$/ 2 $^{n}$. 3.2. Remark. (1) If the Lipschitz constant for F is δ (instead of 1 as in the above formu- lation) then by considering the function F/δ the bound on the probability becomes e − r $^{2}$/ 2 nδ $^{2}$. (2) By applying the same bound to − F one gets µ ({ a : ∣ − ∫ F d µ ∣ ≥ r }) ≤ 2 e − r $^{2}$/ 2 $^{n}$. µ ( E c $_{i,h}$) ≤ 2 exp( − ϵ $^{2}$/ 8( n − 1) δ $^{2}$) < 1 /nm ∣$^{f$_{i}$}$$^{(}$$^{h, a$_{−}$$_{i}$}$$^{)}$ From Proposition 3.1, the Lipschitz property of f$_{i}$ ( h, · ) and the choice of δ it follows that E$_{i,h}$ = A$_{i}$ × { a$_{−}$$_{i}$ ∈ A$_{−}$$_{i}$ : ∣ − ∫ f$_{i}$ ( h, τ ) µ$_{−}$$_{i}$ (d τ ) ∣ ≤ ϵ/ 2 } . ∣$^{F}$$^{(}$$^{a}$$^{)}$ Consider a game in L ( n, m, δ ) with δ = ϵ/ √ 8 n log(2 mn ). Let ( µ$_{1}$, . . . , µ$_{n}$ ) be a mixed strategy Nash equilibrium of the game. Thus, each µ$_{i}$ is a probability distribution over A$_{i}$ and (1) support( µ$_{i}$ ) ⊆ arg max a$_{i}$ ∈ A$_{i}$ ∫ f$_{i}$ ( a$_{i}$, τ ) µ$_{−}$$_{i}$ (d τ ) , where µ$_{−}$$_{i}$ = ∏ j = $_{i}$µ$_{j}$ . For every player i For every player i and every strategy h ∈ A$_{i}$ let E$_{i,h}$ ⊆ A be the set of all strategy profiles a such that, if player i plays h against a$_{−}$$_{i}$ her payoff is roughly the same as her expected payoff when she plays h and the opponents play their Nash equilibrium strategy: for every player i and every h ∈ A$_{i}$ . Since there are at most mn such events E$_{i,h}$ , it follows that µ ( ∩ E$_{i,h}$ ) > 0. Let a ∗ be a strategy profile such that a ∗ ∈ support( µ ) and a ∗ ∈ ∩ E$_{i,h}$ . We claim that a ∗ is an ϵ –equilibrium. Indeed, for every player i and every deviation d ∈ A$_{i}$ 7