Maybe anyone will help meLet f ≥ 0. Let μ ( { x : f ( x ) &gt; t } ) = 1 t 2 + 1 I'm trying to compute ∫ R f d μTo compute this, this is the approach I took. I know that the integral of a measurable function is ∫ f d μ = sup { ∫ g d μ : g simple , 0 ≤ g ≤ f } Further, I know that I can represent f as a non-decreasing sequence of simple functions that converges to f, as such: f n = n ⋅ 1 B n + ∑ k = 1 n ⋅ 2 n ( k − 1 ) 2 n 1 A n , k where B n = { x : f ( x ) &gt; n }and A n , k = { x : ( k − 1 ) 2 − n &lt; f ( x ) ≤ k 2 − n }Now, since this sequence is increasing and converges to f, the sup of the set earlier would just be lim n → ∞ ∫ f n But, I'm unsure how exactly to calculate this. I know that if g is a simple function such that g = ∑ k = 1 n c k Then ∫ g = ∑ k = 1 n c k μ ( A k )But, our sequence of f n has the n ⋅ 1 B n as an additional term outside the summation, and for the measure of A n , k , I'm unsure if I'm allowed to say that μ ( { x : ( k − 1 ) 2 − n &lt; f ( x ) ≤ k 2 − n } ) = = μ ( { x : f ( x ) &gt; ( k − 1 ) 2 − n } ) − μ ( { x : f ( x ) &gt; k ⋅ 2 − n } )

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Maybe anyone will help meLet   f  ≥  0. Let  μ      (    {    x    :    f    (    x    )    &amp;gt;    t    }    )    =      1                  t        2            +      1      I&#039;m trying to compute      ∫          R        f      d  μTo compute this, this is the approach I took. I know that the integral of a measurable function is  ∫  f      d  μ  =  sup      {    ∫    g        d    μ    :    g     simple    ,    0    ≤    g    ≤    f    }  Further, I know that I can represent   f as a non-decreasing sequence of simple functions that converges to   f, as such:      f    n    =  n  ⋅      1                  B        n              +      ∑          k      =      1              n      ⋅              2        n                        (      k      −      1      )              2      n            1                  A        n            ,      k      where      B    n    =  {  x  :  f  (  x  )  &amp;gt;  n  }and      A          n      ,      k        =  {  x  :  (  k  −  1  )      2          −      n        &amp;lt;  f  (  x  )  ≤  k        2          −      n        }Now, since this sequence is increasing and converges to f, the sup of the set earlier would just be      lim          n      →      ∞        ∫      f    n  But, I&#039;m unsure how exactly to calculate this. I know that if g is a simple function such that  g  =      ∑          k      =      1        n        c    k  Then  ∫  g  =      ∑          k      =      1        n        c    k    μ  (      A    k    )But, our sequence of       f    n   has the   n  ⋅      1                  B        n             as an additional term outside the summation, and for the measure of       A          n      ,      k      , I&#039;m unsure if I&#039;m allowed to say that                    μ                  (          {          x          :          (          k          −          1          )                      2                          −              n                                &amp;lt;          f          (          x          )          ≤          k                                2                          −              n                                }          )                                    =                                          =        μ        (        {        x        :        f        (        x        )        &amp;gt;        (        k        −        1        )                  2                      −            n                          }        )        −        μ        (        {        x        :        f        (        x        )        &amp;gt;        k        ⋅                  2                      −            n                          }        )

gudstrufy47j · Accepted Answer

On any measure space   (  X  ,      F    ,  μ  ), if   f  :  X  →  [  0  ,  ∞  ] is measurable, then      ∫          X        f    d  μ  =      ∫          0              ∞        μ  (  {  x  ∈  X  :  f  (  x  )  &amp;gt;  t  }  )    d  t  .When   μ is   σ-finite, this is an easy consequence of Tonelli&#039;s theorem. If   μ is not   σ-finite, then to prove it you can first prove it when   f is a simple function and then use monotone convergence theorem.

Maybe anyone will help me Let f ≥ 0 . Let μ

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