Proof of Lemma 9. Set arbitrarily π$_{1}$, π$_{2}$ ∈ Π and m ∈ [ V ( π$_{2}$ ) ± c ]. We suppose without loss of generality that V ( π$_{1}$ ) ≤ V ( π$_{2}$ ). We will use the fact that (15) implies that F$_{π}$$_{1}$ ( V ( π$_{1}$ )) = F$_{π}$$_{2}$ ( V ( π$_{2}$ )) = 1 / 2 several times in this proof. Firstly, note that where k$_{1}$ is a finite, positive constant that only depends on the lower bound on P ( A | X ) from the strong positivity assumption. Secondly, the continuous differentiability of F$_{π}$$_{2}$ and (15) imply the existence of ˜ m ∈ [ m, V ( π$_{2}$ )] such that Therefore, there exists a finite, positive constant k$_{2}$ such that The remainder of this proof is broken into two parts: we will show that (i) V ( π$_{2}$ ) −V ( π$_{1}$ ) ≤ c implies V ( π$_{2}$ ) −V ( π$_{1}$ ) ≤ k$_{1}$k$_{2}$ ‖ π$_{1}$ − π$_{2}$ ‖ , and (ii) ‖ π$_{1}$ − π$_{2}$ ‖ < c/k$_{1}$k$_{2}$ yields V ( π$_{2}$ ) −V ( π$_{1}$ ) ≤ c . By combining these two results, we will thus prove that V ( π$_{2}$ ) − V ( π$_{1}$ ) ≤ k$_{1}$k$_{2}$ ‖ π$_{1}$ − π$_{2}$ ‖ for all ‖ π$_{1}$ − π$_{2}$ ‖ sufficiently small, and this will complete the proof ( k$_{1}$k$_{2}$ does not depend on π$_{1}$ or π$_{2}$ ). Recall that F$_{π}$$_{1}$ ( V ( π$_{1}$ )) = F$_{π}$$_{2}$ ( V ( π$_{2}$ )) = 1 / 2. If V ( π$_{2}$ ) − V ( π$_{1}$ ) ≤ c , then combining (B.19) 33 | F$_{π}$$_{1}$ ( m ) − F$_{π}$$_{2}$ ( m ) | = ∣ E [ 1 { A = π$_{1}$ ( X ) } − 1 { A = π$_{2}$ ( X ) } P ( A | X ) 1 { Y ≤ m } ]∣ ∣ ≤ 1 2 E | π$_{1}$ ( X ) − π$_{2}$ ( X ) | P ( A | X ) 1 { Y ≤ m } ] ≤ k$_{1}$ ‖ π$_{1}$ − π$_{2}$ ‖ (B.19) | F$_{π}$$_{2}$ ( V ( π$_{2}$ )) − F$_{π}$$_{2}$ ( m ) | = |V ( π$_{2}$ ) − m | × ˙ F$_{π}$$_{2}$ ( ˜ m ) ≥ |V ( π$_{2}$ ) − m | × inf π ∈ Π inf m ∈ [ V ( π ) ± c ] ˙ F$_{π}$ ( m ) . |V ( π$_{2}$ ) − m | ≤ k$_{2}$ | F$_{π}$$_{2}$ ( V ( π$_{2}$ )) − F$_{π}$$_{2}$ ( m ) | . (B.20)