14.381 F13 Recitation 1 notes: Modes of Convergence
نویسنده
چکیده
Proof. (⇒) Let Ω0 = {ω : limn Xn(ω) = X(ω)}. Suppose P (Ω0) = 1. Let > 0 be given. →∞ Let Am = ∩k=m{|Xk − X| ≤ }. Then Am ⊂ Am+1 ∀m and limm P (Am) = P (∪m=1Am) by →∞ continuity of probability measure. For each ω0, there exists m(ω0) such that |Xk(ω0)−X(ω0)| ≤ for all k ≥ m(ω0). Therefore, ∀ω0 ∈ Ω0, ω0 ∈ Am for some m and we can conclude that Ω0 ⊂ ∪m=1Am and 1 = P (Ω0) ≤ P (∪m=1Am) = limm P (A →∞ m) = 1. (⇐) Let Am( 1 ) be a set defined above with given = 1 . Suppose that limm P (A →∞ m( 1 )) = 1 n n n for all n. By continuity, we have P (A( 1 )) = 1 n where A( 1 ) = n ∪ ∞ m 1 =1Am( ) n . Let A = ∩ ∞ n=1A( 1 ) n . Then by the continuity, P (A) = 1 because A( 1 ) n ’s are monotone decreasing sequence of sets. Therefore, ∀ω0 ∈ A and ∀ > 0, there exists M such that |Xm(ω0) − X(ω0)| ≤ for all m ≥ M . We conclude that P ({ω : limm Xm(ω) = X(ω) →∞ }) = 1.
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