**Unformatted text preview: ** X (that is, Xn increases to X ). 18–1 Proof. Let X ≥ 0 be given and deﬁne the sequence (Xn ) by
k
+1
, if 2k ≤ X (ω ) < k2n and 0 ≤ k ≤ n2n − 1,
n
2n
Xn ( ω ) =
n, if X (ω ) ≥ n.
Then it follows that Xn ≤ Xn+1 for every n = 1, 2, 3, . . . and Xn → X which completes the
proof.
Proposition 18.2. If X ≥ 0 and (Xn ) is a sequence of simple random variables with Xn ↑ X ,
then E(Xn ) ↑ E(X ). That is, if Xn ≤ Xn+1 and
lim Xn = X, n→∞ then E(Xn+1 ) ≤ E(Xn ) and lim E(Xn ) = E(X ). n→∞ Proof. Suppose that X ≥ 0 is a random variable and let (Xn ) be a sequence of simple
random variables with Xn ≥ 0 and Xn ↑ X . Observe that since the Xn are increasing, we
have E(Xn ) ≤ E(Xn+1 ). Therefore, E(Xn ) increases to some limit a ∈ [0, ∞]; that is,
E(Xn ) ↑ a
for some 0 ≤ a ≤ ∞. (If E(Xn ) is an unbounded sequence, then a = ∞. However, if E(Xn )
is a bounded sequence, then a < ∞ follows from the fact that increasing, bounded sequences
have unique limits....

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