# Thread: A Problem about definition of mutual singularity for complex measures

1. ## 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!

2. 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 $\mu$.