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4 1. BASICS ON LARGE DEVIATIONS Similarly, one can show that lim sup n→∞ 1 bn log P{Yn ≤ α } ≤ −l. Summarizing our argument, we have lim sup n→∞ 1 bn log P{Yn ∈ F } ≤ −l. Letting l → inf λ∈F Λ∗(λ) on the right hand side leads to (1.1.4). The proof of lower bound (1.1.5) is harder. To do this we claim that for any closed (bounded or unbounded) interval I ⊂ R and θ ∈ Do Λ , lim sup n→∞ 1 bn log E exp θbnYn 1{Y n ∈I} ≤ sup λ∈I θλ − Λ∗(λ) . (1.1.6) Due to similarity, we only deal with the case θ ≥ 0. We first consider the case when I = [a, b] is bounded. For 0, let a = x0 x1 · · · xm = b be a partition of I such that xj − xj−1 . Thus, E exp θbnYn 1{Y n ∈I} ≤ m j=1 exp θbnxj P{Yn ∈ [xj−1,xj]}. Consequently, by (1.1.4) we get lim sup n→∞ 1 bn log E exp θbnYn 1{Y n ∈I} ≤ max 1≤j≤m θxj − inf xj−1≤λ≤xj Λ∗(λ) ≤ θ + sup λ∈I θλ − Λ∗(λ) . Letting → 0+ leads to (1.1.6). We now prove (1.1.6) in the case when I is unbounded. By duality, Λ(θ) = sup λ∈R λθ − Λ∗(λ) θ ∈ R. Consequently, (1.1.6) is equivalent to Assumption 1.1.1 when I = (−∞, ∞). The remaining case is when I is a half line. For similarity we let I be unbounded from above. Given a large M 0, write IM = I ∩ (−∞,M]. We have E exp θbnYn 1{Y n ∈I} ≤ E exp θbnYn 1{Y n ∈IM } + E exp θbnYn 1{Y n ≥M} . By what has been established, lim sup n→∞ 1 bn log E exp θbnYn 1{Y n ∈I} ≤ max lim sup n→∞ 1 bn log E exp θbnYn 1{Y n ≥M} , sup λ∈I θλ − Λ∗(λ) .