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In the context of Corollary A.6, let { ( z$_{m}$, r$_{m}$ ) }$_{m}$$_{≥}$$_{1}$ be a sequence with values in ℓ $^{∞}$( F ) × [0 , ∞ ) and such that sup$_{f}$$_{∈F}$ | z$_{m}$ ( f ) − z ( f ) | + | r$_{m}$ | → 0. Lemma A.5 implies that h ( z$_{m}$, r$_{m}$ ) → h ( z, 0) ≡ 0. Since the sequence was arbitrarily chosen, h is indeed continuous at ( z, 0) and the proof of Corollary A.6 is complete. 

Lemma A.7. If Π is Donsker, then F is also Donsker. 

Proof. Similar to (A.1), the key is the following remark: for every policy π : X → {− 1 , 1 } , 

<img src='content_image/327680.jpg'>

For future use, we first note that (A.1) and (A.2) imply, for any π$_{1}$, π$_{2}$ ∈ Π, 

hence 

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<img src='content_image/327684.jpg'>

Introduce φ : R 5 → R given by 2 φ ( u ) = u$_{1}$ | u$_{2}$ + u$_{3}$ | + (1 + u$_{3}$ ) u$_{4}$ + (1 + u$_{3}$ ) u$_{5}$ and f$_{1}$, f$_{2}$, f$_{4}$, f$_{5}$ be the function given by f$_{1}$ ( o ) ≡ ( y − Q ( a, x )) /P ( A = a | X = x ), f$_{2}$ ( o ) ≡ x , f$_{4}$ ( o ) ≡ Q (1 , x ) and f$_{5}$ ( o ) ≡ Q ( − 1 , x ). Let F$_{1}$ , F$_{2}$ , F$_{4}$ , F$_{5}$ be the singletons { f$_{1}$ } , { f$_{2}$ } , { f$_{4}$ } and { f$_{5}$ } , each of them a Donsker class. Let ˜ F ≡ F$_{1}$ × F$_{2}$ × Π × F$_{4}$ × F$_{5}$ and note that φ ◦ f = ˜ f$_{π}$ if f ∈ ˜ F writes as f = ( f$_{1}$, f$_{2}$, π, f$_{4}$, f$_{5}$ ). In light of (A.10), observe now that, for every f$_{1}$ = ( f$_{1}$, f$_{2}$, π$_{1}$, f$_{4}$, f$_{5}$ ) , f$_{2}$ = ( f$_{1}$, f$_{2}$, π$_{2}$, f$_{4}$, f$_{5}$ ) ∈ ˜ F , it holds that 

<img src='content_image/327686.jpg'>

(the bound on Y implies that ‖ γ ‖$_{∞}$ is finite). By [13, Theorem 2.10.6], whose conditions are obviously met, φ ◦ ˜ F = { ˜ f$_{π}$ : π ∈ Π } is Donsker. Because Λ ≡ {V ( π ) : π ∈ Π } (viewed 

