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Answered question


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 ) < ( )
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 | | > 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.

Answer & Explanation



Beginner2022-06-25Added 25 answers

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 < 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.

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