Let { f k } k = 1 ∞ be a sequence of S -measurable real-valued functions on a measure space ( M , S ). Then the functions lim sup k → ∞ f k ( x ) and lim inf k → ∞ f k ( x )are also S -measurable.I already have proven this statement and my question is about something else. In the script it says that the reader may use the hint:For a sequence { a k } k = 1 ∞ of real numbers the following is true: lim sup k → ∞ a k = lim k → ∞ lim m → ∞ max { a k , a k + 1 , … , a m } Now, I have not used this in my proof, but I can assure you, that my proof is airtight nonetheless. Anyway, can someone point out why this holds? I have not seen this equality before and did not know that one can present lim sup , lim inf like that. I know that for a set of limit points H we can say that lim sup H = max H and vice versa. I do not even see how this property can be useful for our proof.

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Let             {              f                  k                    }              k      =      1              ∞       be a sequence of       S  -measurable real-valued functions on a measure space   (  M  ,      S    ). Then the functions      lim sup          k      →      ∞            f          k        (  x  )     and         lim inf          k      →      ∞            f          k        (  x  )are also       S  -measurable.I already have proven this statement and my question is about something else. In the script it says that the reader may use the hint:For a sequence             {              a                  k                    }              k      =      1              ∞       of real numbers the following is true:      lim sup          k      →      ∞            a          k        =      lim          k      →      ∞            lim          m      →      ∞        max      {          a              k              ,          a              k        +        1              ,    …    ,          a              m              }  Now, I have not used this in my proof, but I can assure you, that my proof is airtight nonetheless. Anyway, can someone point out why this holds? I have not seen this equality before and did not know that one can present   lim sup  ,  lim inf like that. I know that for a set of limit points   H we can say that   lim sup  H  =  max  H and vice versa. I do not even see how this property can be useful for our proof.

charringpq49u · Accepted Answer

By definition      lim sup          k      →      ∞            a    k    =      lim          k      →      ∞        (      sup          m      ≥      k            a    m    )  ,so to prove the claim, it suffices to prove for any   k  ∈      N   that      sup          m      ≥      k            a    m    =      lim          m      →      ∞        max  (      a    k    ,      a          k      +      1        ,  …  ,      a          k      +      m        )  .The reason that this representation is useful, is that   m  ↦  max  (      a    k    ,  …  ,      a          k      +      m        ) is increasing, and   k  ↦      sup          m      ≥      k            a    m   is decreasing. So to prove that       lim sup          k      →      ∞            f    k   is measurable, it suffices to show that the limit of monotone sequences of functions are measurable.

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