Consider a complete, finite measure space ( X , M , μ ). Let { f n } n = 1 ∞ ⊂ L 1 ( μ ) be a sequence such that sup n ∫ | f n | d μ &lt; ∞. Furthermore, assume that for every ε &gt; 0 there exists δ &gt; 0 such that ∫ E | f n | d μ &lt; ε whenever μ ( E ) &lt; δ. I want to show the followings: ( a ) sup n ∫ { | f n | ≥ M } | f n | d μ → 0 as M → ∞ . ( b ) sup n ∫ { | f n | ≤ δ } | f n | d μ → 0 as δ → 0.I have the following proof for (a):Let E = { x ∈ X : | f n ( x ) | ≥ M }. By Chebyshev's inequality, μ ( E ) ≤ 1 M ∫ X | f n | d μ ∀ n ∈ N .Since μ ( E ) → 0 as M → ∞, ∀ δ &gt; 0, μ ( E ) &lt; δ and therefore, ∫ E | f n | d μ &lt; ε .Since this is true for all ε &gt; 0 and n ∈ N , ∫ E | f n | d μ = ∫ { | f n | ≥ M } | f n | d μ → 0 as M → ∞ .Taking the supremum on the left-hand side, we get sup n ∫ { | f n | ≥ M } | f n | d μ → 0 as M → ∞ .I wonder if my proof is coorect, and for (b), I wonder how to show that the measure of the set { | f n | ≤ δ } is small so that a similar argument can be stated as in (a)?

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Consider a complete, finite measure space   (  X  ,      M    ,  μ  ). Let   {      f    n        }          n      =      1              ∞        ⊂      L    1    (  μ  ) be a sequence such that       sup    n    ∫      |        f    n        |      d  μ  &amp;lt;  ∞. Furthermore, assume that for every   ε  &amp;gt;  0 there exists   δ  &amp;gt;  0 such that       ∫          E            |        f    n        |      d  μ  &amp;lt;  ε whenever   μ  (  E  )  &amp;lt;  δ. I want to show the followings:  (  a  )         sup    n        ∫          {              |                    f        n                    |            ≥      M      }            |        f    n        |      d  μ  →  0   as   M  →  ∞  .  (  b  )         sup    n        ∫          {              |                    f        n                    |            ≤      δ      }            |        f    n        |      d  μ  →  0   as   δ  →  0.I have the following proof for (a):Let   E  =  {  x  ∈  X  :      |        f    n    (  x  )      |    ≥  M  }. By Chebyshev&#039;s inequality,  μ  (  E  )  ≤      1    M        ∫          X            |        f          n            |      d  μ     ∀  n  ∈      N    .Since   μ  (  E  )  →  0 as   M  →  ∞,   ∀  δ  &amp;gt;  0,   μ  (  E  )  &amp;lt;  δ and therefore,      ∫          E            |        f          n            |      d  μ  &amp;lt;  ε  .Since this is true for all   ε  &amp;gt;  0 and   n  ∈      N  ,      ∫          E            |        f          n            |      d  μ  =      ∫          {              |                    f        n                    |            ≥      M      }            |        f    n        |      d  μ  →  0   as   M  →  ∞  .Taking the supremum on the left-hand side, we get      sup          n            ∫          {              |                    f        n                    |            ≥      M      }            |        f    n        |      d  μ  →  0   as   M  →  ∞  .I wonder if my proof is coorect, and for (b), I wonder how to show that the measure of the set   {      |        f    n        |    ≤  δ  } is small so that a similar argument can be stated as in (a)?

rederijkd6h1c · Accepted Answer

Your answer to a) looks good. b) is in fact simpler. Recall that the space has finite measure, so      ∫          {              |                    f        n                    |            ≤      δ      }            |        f    n        |    d  μ  ≤  δ      ∫          {              |                    f        n                    |            ≤      δ      }        d  μ  ≤  δ  μ  (  X  )so      sup          n            ∫          {              |                    f        n                    |            ≤      δ      }            |        f    n        |    d  μ  ≤  δ  μ  (  X  )  →  0as   δ  →  0.

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