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By H.; Beckert, H. Schumann

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Left-continuity: limx↑x0 F (x) = F (x0 ). Proof Since for x1 ≤ x2 one has {ξ < x1 } ⊆ {ξ < x2 }, F1 immediately follows from property 3 of probability (see Sect. 2). To prove F2, consider two number sequences {xn } and {yn } such that {xn } is decreasing and xn → −∞, while {yn } is increasing and yn → ∞. Put An = {ξ < xn } and Bn = {ξ < yn }. Since xn tends monotonically to −∞, the sequence of sets An decreases monotonically to An = ∅. By the continuity axiom (see Sect. 1), P(An ) → 0 as n → ∞ or, which is the same, limn→∞ F (xn ) = 0.

E. the collection of all the sets which belong simultaneously to all the σ algebras) is again a σ -algebra. It is the smallest σ -algebra containing all intervals and is called the Borel σ -algebra. Roughly speaking, the Borel σ -algebra could be thought of as the collection of sets obtained from intervals by taking countably many unions, intersections and complements. This is a rather rich class of sets which is certainly sufficient for any practical purposes. The elements of the Borel σ -algebra are called Borel sets.

Permutations are equally likely). What is the probability that at least one element retains its position? There are n! different permutations. Let Ak denote the event that the k-th item retains its position. This event is composed of (n − 1)! outcomes, so its probability equals (n − 1)! n! The event Ak Al means that the k-th and l-th items retain their positions; hence P(Ak ) = P(Ak Al ) = (n − 2)! , n! , P(A1 · · · Ak ) = (n − (n − 1))! 1! = . n! n! Now nk=1 Ak is precisely the event that at least one item retains its position.

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