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t→∞ t (i) Suppose π is generated by Φ. For every p ∈ Δn, the tangent vector v = (v1, . . . , vn) of Δn given by
for all open sets G such that G ∩ supp(ν0) 6= ∅. These inequalities and Lemma 4.11 imply the LDP.
Since Pt(Y ) → P(Y ) > 0 as A is a Pcontinuity set, the measures Pet are well defined for t sufficiently large.
We claim that Pet converges weakly to Pe. This implies the statement because f is bounded continuous on Y and Z 1 Z 1 Z Z To prove the claim, it suffices by the Portmanteau theorem to show that Pet(A) → 1
Pe(A) for all A ⊂ Y with Pe(∂Y A) = P(Y ) P (∂Y A ∩ Y ) = 0. Note that ∂Y A ⊂ Y , so P (∂Y A) = 0. By Lemma A.1, we have ∂S A ⊂ ∂Y A ∪ ∂S Y , and so P (∂S A) = 0 [Bil09] P. Billingsley, Convergence of probability measures, vol. 493, John Wiley & Sons, 2009.
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