Let I 1 , I 2 . . . be an arrangement of the intervals...

Garrett Black

Garrett Black

Answered

2022-07-01

Let I 1 , I 2 . . . be an arrangement of the intervals [ i 1 2 n , i 2 n ) in a sequence. If X n = n on I n and 0 elsewhere then s u p n E X n < but P ( sup n X n < ) = 0. My basic space is [0,1] with Lebesgue measure.
A := { sup n X n < } = { ω Ω : M > 0 , n N , X n ( ω ) M }
hence the complement is
A ¯ = { ω Ω : M N , n N , X n ( ω ) > M } = M N n N { X n > M } = M N n M { X n > M }
How do i go on from here?

Answer & Explanation

nuvolor8

nuvolor8

Expert

2022-07-02Added 32 answers

The example provided there is as follows:
X n ( ω ) = { m ω [ k 2 m , k + 1 2 m ] 0 otherwise
where m = log ( n ) and k = n 2 m .
To prove sup n X n = almost surely, consider a single point ω [ 0 , 1 ]. You can easily verify that X n ( ω ) will go to infinity. Therefore, the set of ω such that sup n X n = is [0,1]. This simple example really does not need more advanced technique like Borel–Cantelli lemma.

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