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Radon measures

  1. Mar 25, 2007 #1


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    Find an example of a sequence (mn) of complex Radon measures on R that tends to 0 vaguely such that for some bounded measurable function g with compact support, [itex]\int g\, dm_n \not\to 0[/itex].

    Definitions and facts

    . complex measures are finite
    . Radon measures are only defined on Borel sets
    . if m is a Radon measure and E is Borel, m(E) is the infimum of the m(U), where U are open sets containing E
    . if m is Radon and U is open, m(U) is the supremum of the m(K), where K are compact sets contained in U
    . if m is a complex measure, there exists a positive measure m' and an m'-measurable function f such that [itex]m(E) = \int _E f\, dm'[/itex]. We express this relation by dm = fdm'.
    . if m, m', and f are as above, then the total variation of m, denoted |m| is defined by the relation d|m| = |f|dm' (this is well-defined, i.e. if m'' is another positive measure, g an m''-measurable function, such that dm = gdm'', then |f|dm' = |g|dm'')
    . the norm of a complex measure m, ||m|| is defined to be [itex]\int d|m|[/itex]
    . (mn) tends to 0 vaguely iff for every continuous function f from R to C that vanishes at infinity, [itex]\int f\, dm_n \to 0[/itex]
    . if (mn) is a sequence of complex radon measures on R and Fn(x) = mn({y in R : y < x}), then if ||mn|| are uniformly bounded and the Fn converge pointwise to 0, then (mn) tends to 0 vaguely
    . (there are lots theorems in this section, but none that immediately strike me as relevant)


    It seems to me that if we can find a sequence of measures and a corresponding bounded measurable function with compact support, we can do so with the restriction that this measurable function g have compact support contained in [0,1]. I've tried a few things but they haven't worked.
    Last edited: Mar 25, 2007
  2. jcsd
  3. Mar 26, 2007 #2


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    How about thinking of delta function type measures? Take m_n to be delta(0)-delta(1/n). Then it goes to zero for any continuous function, but if we take g(x)=1 at x=0, 0 otherwise then the integral of g is 1 for any m. Does that work?
  4. Mar 26, 2007 #3


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    Yes, very nice! Thanks.
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