By Debnath & Mikusinski

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**Example text**

Set δ = inf{E b (σa0 ∧ 1) : ρ(b, a) ≥ ε}. Then δ > 0 by (A0 -3). 4 The Existence and Uniqueness Theorem 31 Observe that h(u) ∧ 1 n X (du) ≥ U h(u) ∧ 1 n X (du) Vε ≥ (h(u) − σε (u)) ∧ 1 n X (du) Vε = (h(u) − σε (u)) ∧ 1 V = S = S k(db)Pb (X 0 ∈ du) S k(db)E b [(σa0 − σε (X 0 )) ∧ 1, σa0 > σε (X 0 )] k(db)E b [E X (σε (X 0 )) (σa0 ∧ 1), σa0 > σε (X 0 )] ≥δ S k(db)Pb (σa0 > σε (X 0 )) k(db)Pb (X 0 ∈ Vε ) =δ S = δn X (Vε ) and that h(u) ∧ 1 n X (du) = U h(u) ∧ 1 U k(db)Pb (X 0 ∈ du) S k(db)E b (h(X 0 ) ∧ 1) = S = S k(db)E b (σa0 ∧ 1).

Since we fix ε for the moment, we omit ε in S i,ε , U i,ε etc. Let h(X s ). J (t, X) = s≤t s∈D X Similarly for J (t, X i ). X 1 is discrete. Let σ be the first element in D X 1 . Noticing that s ∈ D X , s < σ =⇒ s ∈ D X 2 , we have J (σ −, X) = J (σ −, X 2 ), X σ = X σ1 and Yt = Yt2 for t < mσ + J (σ −, X) = mσ + J (σ −, X 2 ). f (a) ≡ Rα g(a) ∞ =E e−αt g(Yt ) dt 0 mσ +J (σ −,X) =E e−αt g(Yt ) dt 0 h(X σ ) + E e−αmσ −α J (σ −,X) e−αt g(X σ (t)) dt 0 +E e ∞ −αmσ −α J (σ −,X)−αh(X σ ) e−αt g(Y (t, θσ X|d (0, ∞))) dt 0 mσ +J (σ −,X 2 ) =E 0 e−αt g(Yt2 ) dt + E e−αmσ −α J (σ −,X 2 h(X σ1 ) ) 0 + E e−αmσ −α J (σ −,X 2 e−αt g(X σ1 (t)) dt ∞ )−αh(X σ1 ) 0 = I1 + I2 + I3 .

S. for every ε > 0. s. , namely that n X (Vε ) < ∞. Set δ = inf{E b (σa0 ∧ 1) : ρ(b, a) ≥ ε}. Then δ > 0 by (A0 -3). 4 The Existence and Uniqueness Theorem 31 Observe that h(u) ∧ 1 n X (du) ≥ U h(u) ∧ 1 n X (du) Vε ≥ (h(u) − σε (u)) ∧ 1 n X (du) Vε = (h(u) − σε (u)) ∧ 1 V = S = S k(db)Pb (X 0 ∈ du) S k(db)E b [(σa0 − σε (X 0 )) ∧ 1, σa0 > σε (X 0 )] k(db)E b [E X (σε (X 0 )) (σa0 ∧ 1), σa0 > σε (X 0 )] ≥δ S k(db)Pb (σa0 > σε (X 0 )) k(db)Pb (X 0 ∈ Vε ) =δ S = δn X (Vε ) and that h(u) ∧ 1 n X (du) = U h(u) ∧ 1 U k(db)Pb (X 0 ∈ du) S k(db)E b (h(X 0 ) ∧ 1) = S = S k(db)E b (σa0 ∧ 1).