Assume that ( X , E , μ ) is σ −−finite and write X = ∪ n A n where ( A n ) n is an increasing sequence of elements in A n ∈ E . Then for every f ∈ L ∞ ( μ ) I have to show that lim n → ∞ lim p → ∞ | | 1 A n f | | p = | | f | | ∞ A well-known result is that lim p → ∞ | | f | | p = | | f | | ∞ which I have to use but I am not sure other than that where to begin.I know that if p ∈ [ 1 , ∞ ) then | | f | | p = ( ∫ X | f | p d μ ) 1 / p but I am not sure how to proceed.

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Assume that   (  X  ,      E    ,  μ  ) is   σ  −−finite and write   X  =      ∪    n        A    n   where   (      A    n        )    n   is an increasing sequence of elements in       A    n    ∈      E  . Then for every   f  ∈            L        ∞    (  μ  ) I have to show that      lim          n      →      ∞            lim          p      →      ∞            |        |              1                      A        n              f      |              |        p    =      |        |    f      |              |        ∞  A well-known result is that       lim          p      →      ∞            |        |    f      |              |        p    =      |        |    f      |              |        ∞   which I have to use but I am not sure other than that where to begin.I know that if   p  ∈  [  1  ,  ∞  ) then       |        |    f      |              |        p    =            (              ∫        X                    |            f                        |                p            d      μ      )              1              /            p       but I am not sure how to proceed.

reflam2kfnr · Accepted Answer

Note that for the well-known result you reference, you do require more assumptions (like   μ  (      A    n    )  &amp;lt;  ∞ or that   f is in       L    r    (      A    n    ) for some   r).However, assuming this result,      lim          p      →      ∞        ‖            1                      A        n              f      ‖    p    =  ‖            1                      A        n              f      ‖    ∞    ,the question reduces to showing      lim          n        ‖  f            1                      A        n                  ‖    ∞    =  ‖  f      ‖    ∞    .This follows from the fact that   ‖  f            1                      A        n                  ‖    ∞   is increasing, and       sup          n        ‖  f            1                      A        n                  ‖    ∞    =  ‖  f      ‖    ∞  . To see the second of these, fix   ϵ  &amp;gt;  0 and find       B    ϵ    ⊂  X with       |    f      |    &amp;gt;  ‖  f      ‖    ∞    −  ϵ on       B    ϵ   and   μ  (      B    ϵ    )  &amp;gt;  0. By continuity of measure, there is an   n large enough so that   μ  (      A    n    ∩      B    ϵ    )  &amp;gt;  0, and  ‖  f            1                      A        n                  ‖    ∞    ≥  ‖  f            1                      A        n            ∩              B        ϵ                  ‖    ∞    &amp;gt;  ‖  f      ‖    ∞    −  ϵ  .Since   ϵ  &amp;gt;  0 was arbitrary and   ‖  f            1                      A        n                  ‖    ∞    ≤  ‖  f      ‖    ∞  ,       sup          n        ‖  f            1                      A        n                  ‖    ∞    =  ‖  f      ‖    ∞  .

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