Let f : E → [ 0 , + ∞ ] be measurable. Show that ∫ E f = ∑ i = 1 n c i m ( { x ∈ E ∣ f ( x ) = c i } )when f ( E ) is countable.I suppose I have to use simple functions here to show this? The rhs of the equality seems to be the integral of some simple function over E and E i = { x ∈ E ∣ f ( x ) = c i }. Since f is non-negative I have that ∫ E f = sup { ∫ E s ∣ 0 ≤ s ≤ f } where s is a simple function. So in essence I have to show that sup { ∫ E s ∣ 0 ≤ s ≤ f } = ∑ i = 1 n c i m ( { x ∈ E ∣ f ( x ) = c i } ) ?

Question

Let   f  :  E  →  [  0  ,  +  ∞  ] be measurable. Show that      ∫    E    f  =      ∑          i      =      1        n        c    i    m  (  {  x  ∈  E  ∣  f  (  x  )  =      c    i    }  )when   f  (  E  ) is countable.I suppose I have to use simple functions here to show this? The rhs of the equality seems to be the integral of some simple function over   E and       E    i    =  {  x  ∈  E  ∣  f  (  x  )  =      c    i    }. Since   f is non-negative I have that      ∫    E    f  =  sup      {          ∫      E        s    ∣    0    ≤    s    ≤    f    }  where s is a simple function. So in essence I have to show that  sup      {          ∫      E        s    ∣    0    ≤    s    ≤    f    }    =      ∑          i      =      1        n        c    i    m  (  {  x  ∈  E  ∣  f  (  x  )  =      c    i    }  )  ?

rass1k6s · Accepted Answer

Since   f  (  E  ) is countable we have  f  =      ∑          i      =      1        ∞        c    i        χ                  E        i              ,where   {      c    i    } is an enumeration of   f  (  E  ) and       E    i    =      f          −      1        (      c    i    ). Then   f  =  lim      f    n  , where   {      f    n    } is the increasing sequence (since       c    i    ≥  0)      f    n    =      ∑          i      =      1        n        c    i        χ                  E        i              .By monotone convergence theorem,      ∫    E    f  =      lim    n        ∫    E        f    n    =      lim    n        ∑          i      =      1        n        c    i    m  (      E    i    )  =      ∑          i      =      1        ∞        c    i    m  (      E    i    )  .

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