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Mutually Singular measures

  1. Dec 9, 2008 #1
    Suppose that [tex]\{g_n\}_{n=1}^\infty[/tex] is a sequence of positive continuous functions on [tex]I=[0,1][/tex], [tex]\mu[/tex] is a positive Borel measure on [tex]I[/tex], [tex]m[/tex] is the standard Lebesgue measure, and

    (i) [tex]\lim_{n\rightarrow\infty} g_n =0 \qquad a.e. [m][/tex];
    (ii) [tex]\int_{I}{g_n}\;dm = 1 \qquad[/tex] for all [tex]n\in\mathbb{N}[/tex];
    (iii) [tex]\lim_{n\rightarrow\infty}\int_{I} fg_n\;dm = \int_{I} f\;d\mu \qquad[/tex] for all [tex]f \in C(I).[/tex]

    Does it follow that the measures [tex]m[/tex] and [tex]\mu[/tex] are mutually singular?

    I kind of guess that it is a no and have been trying hard to come up with a counterexample. Some of my thoughts:
    1 . If [tex]\mu[/tex] is singular relative to [tex]m[/tex], then it has to be concentrated on a set which does not contain any open interval.
    2. A typical sequence that satifies both (i) and (ii) above is the familiar "triangular sequence", i.e. the sequence of functions whose graphs are triangles with increasing heights, each having area 1. But I don't know how to deal with (iii).

    Please help me. Thank you in advance.
    Last edited: Dec 9, 2008
  2. jcsd
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