Let f : Ω → B be Bochner-measurable, i.e. the point-wise limit of a sequence of simple (i.e. countably-valued measurable) functions ( s n ). I know that if ∫ Ω | | f ( ω ) | | d μ ( ω ) &lt; ∞ ( ∗ )then one can show that f is Bochner-integrable, i.e. there is even a Bochner-integrable sequence of simple functions ( s ~ n ) such that lim n → ∞ ∫ Ω | | f ( ω ) − s ~ n ( ω ) | | d μ ( ω ) = 0 ( ∗ ∗ )I am unsure, however, how to derive this sequence when the measure space Ω is not finite. Apparently one is supposed to make use of the fact that due to ( ∗ ), the set A := { f ≠ 0 } = ⋃ n = 1 ∞ { | | f | | &gt; 1 n } ⏟ =: A n is σ-finite, as each A n has finite measure. But how to proceed? Setting s ~ n = 1 A s n would not be enough to make the functions Bochner-integrable, as the entire set A might still have infinite measure. Setting s ~ n = 1 A n s n would do it and maintain pointwise convergence, but then I don't know how to show the convergence in ( ∗ ∗ ) anymore...any tips are much appreciated.

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Let   f  :  Ω  →  B be Bochner-measurable, i.e. the point-wise limit of a sequence of simple (i.e. countably-valued measurable) functions   (      s    n    ). I know that if      ∫          Ω            |        |    f  (  ω  )      |        |      d  μ  (  ω  )  &amp;lt;  ∞    (  ∗  )then one can show that f is Bochner-integrable, i.e. there is even a Bochner-integrable sequence of simple functions   (                    s        ~              n    ) such that      lim          n      →      ∞              ∫          Ω            |        |    f  (  ω  )  −                    s        ~              n    (  ω  )      |        |      d  μ  (  ω  )    =    0    (  ∗  ∗  )I am unsure, however, how to derive this sequence when the measure space   Ω is not finite. Apparently one is supposed to make use of the fact that due to   (  ∗  ), the set  A  :=  {  f  ≠  0  }  =      ⋃          n      =      1              ∞                                      {                      |                                |                    f                      |                                |                    &amp;gt;                      1            n                    }                ⏟                    =:              A        n            is   σ-finite, as each       A    n   has finite measure. But how to proceed? Setting                     s        ~              n    =      1          A            s    n   would not be enough to make the functions Bochner-integrable, as the entire set   A might still have infinite measure. Setting                     s        ~              n    =      1                  A        n                  s    n   would do it and maintain pointwise convergence, but then I don&#039;t know how to show the convergence in   (  ∗  ∗  ) anymore...any tips are much appreciated.

Schetterai · Accepted Answer

Think of       A    n   as a measure space with the restriction of the   σ  −−algebra on   Ω and the restriction of the measure   μ. Since you already know the result for finite measure space you can find a simple function       t    n   on this space such that       ∫                  A        n              ‖  f      χ                  A        n              −      t    n    ‖  d  μ  &amp;lt;      1    n  . Let       s    n    =      t    n   on An and 0 outside. Then       s    n   is a simple function on   Ω and   ∫  ‖  f  −      s    n    ‖  d  μ  ≤      ∫                  A        n              ‖  f      χ                  A        n              −      t    n    ‖  +      ∫          Ω      ∖              A        n              ‖  f  ‖  d  μ. Now       ∫          Ω      ∖              A        n              ‖  f  ‖  d  μ  →  0 because       lim    n        ∫                  A        n              ‖  f  ‖  d  μ  =  ∫  ‖  f      χ          {      f      ≠      0      }        ‖  d  μ  (  ≡  ∫  ‖  f  ‖  d  μ  ) by Monotone ConvegenceTheorem.

Let f : <mi mathvariant="normal">Ω → B be Bochne

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