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 &lt; ∞ but P ( sup n X n &lt; ∞ ) = 0. My basic space is [0,1] with Lebesgue measure. A := { sup n X n &lt; ∞ } = { ω ∈ Ω : ∃ M &gt; 0 , ∀ n ∈ N , X n ( ω ) ≤ M }hence the complement is A ¯ = { ω ∈ Ω : ∀ M ∈ N , ∃ n ∈ N , X n ( ω ) &gt; M } = ⋂ M ∈ N ⋃ n ∈ N { X n &gt; M } = ⋂ M ∈ N ⋃ n ≥ M { X n &gt; M } How do i go on from here?

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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    &amp;lt;  ∞ but   P  (      sup    n        X    n    &amp;lt;  ∞  )  =  0. My basic space is [0,1] with Lebesgue measure.  A  :=  {      sup    n        X    n    &amp;lt;  ∞  }  =  {  ω  ∈  Ω  :  ∃  M  &amp;gt;  0  ,  ∀  n  ∈      N    ,      X    n    (  ω  )  ≤  M  }hence the complement is                              A          ¯                                    =        {        ω        ∈        Ω        :        ∀        M        ∈                  N                ,        ∃        n        ∈                  N                ,                  X          n                (        ω        )        &amp;gt;        M        }                                          =                  ⋂                      M            ∈                          N                                                ⋃                      n            ∈                          N                                      {                  X          n                &amp;gt;        M        }                                          =                  ⋂                      M            ∈                          N                                                ⋃                      n            ≥            M                          {                  X          n                &amp;gt;        M        }            How do i go on from here?

nuvolor8 · Accepted Answer

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.

Let I 1 </msub> , I 2 </msub> . . . be an arrangeme

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