For Ω ⊂ R n compact and some regular Borel measure μ : B ( Ω ) → R , is the mapping f ↦ ∫ Ω f d μa linear and continuous functional on L 1 ( Ω , B ( Ω ) , λ )?Can we show that the integral expression is bounded, or can we find a counterexample to it?Edit: To avoid misunderstandings I would like to emphasize again that we are dealing here with two different measures λ and μ. Primarily I am looking for a counterexample of an L 1 -function f (with respect to λ) and a measure μ such that the integral of f w.r.t. μ is no longer finite. I have already played around with various L 1 functions and measures, but have never come to a contradiction.

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For   Ω  ⊂            R        n   compact and some regular Borel measure   μ  :      B    (  Ω  )  →      R  , is the mapping  f  ↦      ∫    Ω    f  d  μa linear and continuous functional on       L    1    (  Ω  ,      B    (  Ω  )  ,  λ  )?Can we show that the integral expression is bounded, or can we find a counterexample to it?Edit: To avoid misunderstandings I would like to emphasize again that we are dealing here with two different measures   λ and   μ. Primarily I am looking for a counterexample of an       L    1  -function   f (with respect to   λ) and a measure   μ such that the integral of   f w.r.t.   μ is no longer finite. I have already played around with various       L    1   functions and measures, but have never come to a contradiction.

svirajueh · Accepted Answer

Consider any point       x    0    ∈  Ω and let       δ                  x        0             be the Dirac measure, i.e.       δ                  x        0              (  A  )  =  1  (      x    0    ∈  A  ) for any   A  ∈      B    (  Ω  ). This is a Borel regular measure. Define the function   f  (  x  )  =  1  (  x  =      x    0    ), so   f  (  x  )  =  1 for   x  =      x    0   and   f  (  x  )  =  0 otherwise. Then  ∫  f  d      δ                  x        0              =  1but  ∫      |    f      |    d  λ  =  0.

Kiana Dodson · Accepted Answer

You don&#039;t need to decide whether the integral of   f is finite or not. If       ∫    Ω    f         d    μ is not finite, then   f is not       L    1  ! Now that we know all these integrals are finite, the linearity of the integral shows this map is linear, and hence is a member of the dual space of       L    1    (  Ω  ,      B    (  Ω  )  ,  μ  ).

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