Suppose f n : X → [ 0 , ∞ ] is measurable for n = 1 , 2 , 3 , …, f 1 ≥ f 2 ≥ f 3 ≥ ⋯ ≥ 0, f n ( x ) → f ( x ) as n → ∞, for every x ∈ X, and f 1 ∈ L 1 ( μ ). Prove that then (*) lim n → ∞ ∫ X f n d μ = ∫ X f d μ and show that this conclusion does not follow if the condition " f 1 ∈ L 1 ( μ )" is omitted.Let E consist of the points x ∈ X at which f 1 ( x ) &lt; ∞. By the dominated convergence theorem, ∫ E f n d μ → ∫ E f d μ . Since f 1 ∈ L 1 ( μ ), μ ( E c ) = 0, and hence (*) follows.Let X = { 1 , 2 , 3 , … }, and let μ be the counting measure. For each n, define f n : X → [ 0 , ∞ ] by f n ( x ) = { ∞ ( x ≥ n ) 0 ( x &lt; n ) . Then lim f n = 0, and ∫ X f n d μ = ∞ for all n.Is this correct?

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Suppose       f    n    :  X  →  [  0  ,  ∞  ] is measurable for   n  =  1  ,  2  ,  3  ,  …,       f    1    ≥      f    2    ≥      f    3    ≥  ⋯  ≥  0,       f    n    (  x  )  →  f  (  x  ) as   n  →  ∞, for every   x  ∈  X, and       f    1    ∈      L    1    (  μ  ). Prove that then                    (*)                              lim                      n            →            ∞                                    ∫          X                          f          n                        d        μ        =                  ∫          X                f                d        μ            and show that this conclusion does not follow if the condition &quot;      f    1    ∈      L    1    (  μ  )&quot; is omitted.Let   E consist of the points   x  ∈  X at which       f    1    (  x  )  &amp;lt;  ∞. By the dominated convergence theorem,      ∫    E        f    n      d  μ  →      ∫    E    f    d  μ      .  Since       f    1    ∈      L    1    (  μ  ),   μ  (      E    c    )  =  0, and hence (*) follows.Let   X  =  {  1  ,  2  ,  3  ,  …  }, and let   μ be the counting measure. For each   n, define       f    n    :  X  →  [  0  ,  ∞  ] by      f    n    (  x  )  =      {                            ∞                          (          x          ≥          n          )                                      0                          (          x          &amp;lt;          n          )          .                        Then   lim      f    n    =  0, and       ∫    X        f    n      d  μ  =  ∞ for all   n.Is this correct?

Erzrivalef6 · Accepted Answer

Since   0  ≤  f  ≤      f    n    ≤      f    1    ∈      L    1    ,    ∀  n and       f    n    ↓  f then you can use DCT and the conclusion follows. However that&#039;s probably not what you&#039;re supposed to do. As suggested in the comments, the sequence given by       u    k    :=      f    1    −      f    k   is nonnegative and increasing s.t. you can use MCT for nonnegative increasing sequences      sup    n    ∫  (      f    1    −      f    n    )  d  μ  =  ∫  (      f    1    −  f  )  d  μSince       f    n    ,      f    1    ,  f  ∈      L    1   we get      sup    n    ∫  (  −      f    n    )  d  μ  =      sup    n    ∫  (  (      f    1    −      f    n    )  −      f    1    )  d  μ  =  ∫  (      f    1    −  f  )  d  μ  −  ∫      f    1    d  μ  =  ∫  (  −  f  )  d  μand the claim follows since       sup    n    (  −  ∫      f    n    )  =  −      inf    n    ∫      f    n  .

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