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Math Help - A Problem about definition of mutual singularity for complex measures

  1. #1
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    A Problem about definition of mutual singularity for complex measures

    First a few definitions.
    A measurable set is called null for signed or complex measure \nu if each of its measurable subset has zero \nu-value. This is equivalent to the condition that the value of total variation |\nu| of this measurable set is 0.
    For two measure \lambda_1 and [/tex]\lambda_2[/tex], we say [tex]\lambda_1\perp\lambda_2[tex] if there are two measurable sets E and F such that E\cap F=\emptyset, E\cup F=X and \lambda_1(E)=\lambda_2(F)=0.
    For two signed measure \lambda_1 and \lambda_2, we say \lambda_1\perp\lambda_2 if there are two measurable sets E and F such that E\cap F=\emptyset, E\cup F=X and E is null for \lambda_1 and F is null for \lambda_2. The definition of mutual sigularity for measures agrees with that for signed measures.

    Now I have found two kinds of definitions of mutual singularity for complex measures:
    For two complex measures \nu and \mu, they are said to be mutually singular if
    Definition one: \nu_r\perp\mu_r,\nu_r\perp\mu_i,\nu_i\perp\mu_r,\n  u_i\perp\mu_i where \nu_r and \nu_i are real and imaginary parts of \nu, respectively, \mu_r and \mu_i are real and imaginary parts of \mu, respectively. Note that \nu_r, \nu_i, \mu_r, \mu_i are all (finite) signed measures.
    Definition two: |\nu|\perp|\mu| where |\bullet| denotes total variation. Note that |\nu| and |\mu| are both (finite) measures. This definition can be proved to be equivalent to: there are two measurable sets E and F such that E\cap F=\emptyset, E\cup F=X and E is null for \nu and F is null for \mu.

    I'm trying to prove that these two definitions are equivalent. I have proved that Definitin two can imply Definition one. But I met with difficulties when proving the converse. The follwing is my attempt: For \nu_r\perp\mu_r, suppose E_1 and F_1 make the mutual regularity, i.e., E\cap F=\emptyset, E\cup F=X and E is null for \nu_r and F is null for \mu_r. For \nu_r\perp\mu_i, suppose E_2 and F_2 make the mutual regularity. Likewise, suppose E_3 and F_3 for \nu_i\perp\mu_r, E_4 and F_4 for \nu_i\perp\mu_i. I wish to find E and F such that E is null for \nu and F is null for \mu to meet the condition in Definition two. I construct E=E_1\cap E_2\cap E_3\cap E_4, F=E^c(the complement of E) =E_1^c\cup E_2^c\cup E_3^c\cup E_4^c. Since E is a subset of E_1, E is null for \nu_r. Since E is a subset of E_3, E is null for \nu_i, so E is null for \nu. But for F, subadditivity shows that |\mu|(F)\leq|\mu|(E_1^c)+|\mu|(E_2^c)+|\mu|(E_3^c)  +|\mu|(E_4^c), if I can obtain |\mu|(E_1^c) (and other three terms) is equal to 0, we can arrive at the desired consequence. But E_1^c is only null for \mu_r so I have only |\mu_r|(E_1^c)=0. I hope I can prove |\mu|\leq|\mu_r|, but Unfortunately, like the amplitude of real part is always smaller than the amplitude of the complex number, the realtion between |\mu_r| and |\mu| is |\mu_r|\leq|\mu|. So the above construction does not work. I can not find other methods to construct the desired E and F and I was stuck here. It's not a homework or exercise and I have thought several days for it. I think I am unable to work it out by myself so I post this question here hoping someone could help me. Any clue or hint is welcomed, thanks!

    If you need relevant information such as a specific definition, please tell me and I'll post it (after 15 hours because I have to go to bed now). Thanks!
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  2. #2
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    OK, solved.
    Construct E=(E_1\cup E_2)\cap(E_3\cup E_4) and F=E^c=(F_1\cap F_2)\cup(F_3\cap F_4). We can check that E is null for \nu and F is null for \mu.
    Last edited by zzzhhh; March 23rd 2011 at 10:28 AM.
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