In this exercise, we construct an example of a sequence of random variables Zn such that
but

That is, Zn converges in probability to 0, but Zn does not converge to 0 with probability 1. Indeed, Zn converges to 0 with probability 0.
Let X be a random variable having the uniform distribution on the interval [0, 1].We will construct a sequence of functions hn(x) for n = 1, 2, . . . and define Zn = hn(X). Each function hn will take only two values, 0 and 1. The set of x where hn(x) = 1 is determined by dividing the interval [0, 1] into k nonoverlappling subintervals of length 1/k for k = 1, 2, . . . , arranging these intervals in sequence, and letting hn(x) = 1 on the nth interval in the sequence for n = 1, 2, . . . . For each k, there are k nonoverlapping subintervals, so the number of subintervals with lengths 1, 1/2, 1/3, . . . , 1/k is
1+ 2 + 3 + . . . + k = k(k + 1)/2.
The remainder of the construction is based on this formula. The first interval in the sequence has length 1, the next two have length 1/2, the next three have length 1/3, etc.
a. For each n = 1, 2, . . ., prove that there is a unique positive integer kn such that
(kn ˆ’ 1)kn/2 b. For each n = 1, 2, . . . , let jn = nˆ’(kn ˆ’1)kn/2. Show that jn takes the values 1, . . . , kn as n runs through 1+ (kn ˆ’ 1)kn/2, . . . , kn(kn + 1)/2.
c. Define

Show that, for every x ˆˆ [0, 1), hn(x) = 1 for one and only one n among 1+ (kn ˆ’ 1)kn/2, . . . , kn(kn + 1)/2.
d. Show that Zn = hn(X) takes the value 1 infinitely often with probability 1.
e. Show that Pr(Zn = 0) = 1ˆ’ 1/kn and limn†’ˆž kn = ˆž.
f. Show that

Pm, (62.18) lim Z00 (6.2.18) hn(x) = 0 ifnot.