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'asymmetric' equality

  1. Oct 24, 2012 #1
    Hi all,

    Apologies if this is stupid question, but I have the following situation. Given two measures [itex]u(x)[/itex] and [itex]v(x)[/itex] if [itex]u(x)[/itex] is absolutely continuous to [itex]v(x)[/itex] ( [itex]u<<v[/itex]) I have a result such that

    [itex]\int_A f(x)dv(x) [/itex] always takes the value [itex]\int_B g(x)du(x)[/itex]

    But strictly [itex]\int_B g(x)du(x)\neq\int_A f(x)dv(x)[/itex]

    If that makes sense... Basically I want to express this idea of 'taking the value of without being equal' but don't know how to express it mathematically (in words and symbols).

    In case it still isn't clear, performing the first integral will necessarily return the value of the second, but performing the second will never give the value of the first. It appears to be a non symmetric equality.. Or is this nonsense?!

    I then want to contrast this to when the measures are equivalent ([itex]u(x)\sim v(x)[/itex]) when the two integrals are equal y'see.

    Many thanks!

    R
     
  2. jcsd
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