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5in Convergence to the Independent Increment Processes linwang 53 Here and in the rest of the proof, we always assume ε is chosen so that ε is not a jump point of τ (·). Define the restriction map Mp ([0, ∞) × [−∞, ∞] \ {0}) → Mp ([0, ∞) × {x : |x| > ε}) by m → m|[0,∞)×{x:|x|>ε}. It is almost surely continuous with respect to the distribution of P RM (λ × ν). We claim that the summation functional ψ (ε) : Mp ([0, ∞) × {x : |x| > ε}) → D([0, T ], R) defined by ∞ ψ (ε) ( δ(τk ,Jk ) )(t) → k=1 Jk , τk ≤t is continuous respect to the distribution of P RM (λ × ν) almost surely.

Ii)⇒ (iii). Let C+ K ((0, ∞]) be the space of continuous functions which have compact supports. For f ∈ C+ K ((0, ∞]), the support of f is contained in (δ, ∞] for some δ > 0. From (ii), nP[ ξ > x] → x−α = να ((x, ∞]) for any x > 0. 5in Convergence to the Independent Increment Processes linwang 47 On (δ, ∞], define Pn (·) = P[ bξn ∈ ·] P[ bξn > δ] and P (·) = να (·) , να ((δ, ∞]) which are probability measures on (δ, ∞]. Then for y ∈ (δ, ∞], Pn ((y, ∞]) → P ((y, ∞]). In R, convergence of distribution functions is equivalent to weak convergence, so for bounded continuous function f on (δ, ∞], we have Pn (f ) → P (f ) which means that nEf ( ξ ) → να (f ).

So sup |λn (s) − s| ≤ 3γ. s∈[0,T ] Thus ψ (ε) (mn ) → ψ (ε) (m) in D[0, 1] according to the J1 topology. Hence, ψ (ε) is continuous almost surely respect to the distribution of N. Then, we have ∞ n 1[|Xk |>an ε] k Xk (n , an k=1 ⇒ ) 1[||jk ||>ε] (tk ,jk ) k=1 in Mp ([0, ∞) × {x : |x| > ε}), and [n·] k=1 Xk 1[|Xk |>an ε] ⇒ an jk 1[|jk |>ε] tk ≤(·) in D([0, T ]). Similarly, [n·] k=1 Xk Xk 1 ⇒ an [ε<| an |≤1] jk 1[ε<|jk |≤1] . tk ≤(·) Then, taking expectations, we have [n·] E(X1 1[ε<| X1 |≤1] ) ⇒ (·) an an xν(dx) {x:ε<|x|≤1} in D([0, T ]).