Let ( Ω , F , μ ) be a measure space, with μ a finite measure. Let X : Ω → R be a measurable function. Suppose lim n → ∞ ∫ | X | &gt; n | X | d P = 0In proving the integrability of X, the following step is made. Take N ∈ N , such that ∫ | X | &gt; N | X | d P ≤ 1.Now: ∫ R | X | d P = ∫ | X | &gt; N | X | d P + ∫ | X | ≤ N | X | d P ≤ 1 + N ⋅ μ ( | X | ≤ N )Where μ ( | X | ≤ N ) := μ ( { ω ∈ Ω : | X ( ω ) | ≤ N } ). Why is it that the second integral can be bounded by N ⋅ μ ( | X | ≤ N )?My first thought was by somehow using the Markov inequality, but this doesn't seem to work.

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Let   (  Ω  ,      F    ,  μ  ) be a measure space, with   μ a finite measure. Let   X  :  Ω  →      R   be a measurable function. Suppose      lim          n      →      ∞            ∫                  |            X              |            &amp;gt;      n            |    X      |       d      P    =  0In proving the integrability of   X, the following step is made. Take   N  ∈      N  , such that       ∫                  |            X              |            &amp;gt;      N            |    X      |       d      P    ≤  1.Now:      ∫          R            |    X      |       d      P    =      ∫                  |            X              |            &amp;gt;      N            |    X      |       d      P    +      ∫                  |            X              |            ≤      N            |    X      |       d      P    ≤  1  +  N  ⋅  μ  (      |    X      |    ≤  N  )Where   μ  (      |    X      |    ≤  N  )  :=  μ  (  {  ω  ∈  Ω  :      |    X  (  ω  )      |    ≤  N  }  ). Why is it that the second integral can be bounded by   N  ⋅  μ  (      |    X      |    ≤  N  )?My first thought was by somehow using the Markov inequality, but this doesn&#039;t seem to work.

Schetterai · Accepted Answer

By definition, for the domain of integration in the second integral we have that       |    X      |    ≤  N everywhere. So clearly,  ∫      X                  |            X              |            ≤      N        d      P    ≤      ∫                  |            X              |            ≤      N        N  d      P    =  N      ∫                  |            X              |            ≤      N        d      P    =  N  μ  {      |    X      |    ≤  N  }  .

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