Let $m$ be Lebesgue measure on $[0,1]$ and define $\mu(E)=m(E)+im(E\cap[0,\frac12]).$ Let $m$ be Lebesgue measure on $[0,1]$ and define $\mu(E)=m(E)+im(E\cap[0,\frac12]).$  Compute the following: $\frac{d\mu}{dm}$, $ \frac{d|\mu|}{dm}$, $\frac{d\mu}{d|\mu|}$.
Man this stuff is really bugging me. I've been turning Royden upside down and rightside up trying to make sense of what exactly is going on here, but alas, I'm really struggilng. If somebody could help me with this one you'd be a lifesaver!!
 A: It is easy to see that, for any Lebesgue measurable set $E \subseteq [0,1]$, 
$$ \mu(E)=m(E)+im(E\cap[0,\frac12]) = \int_E 1 + i \chi_{[0,\frac12]} dm $$
where $\chi_{[0,\frac12]}$  is the indicator function of $[0,\frac12]$. By the uniqueness of Radon–Nikodym derivative, we have 
$$ \frac{d\mu}{dm} = 1 + i \chi_{[0,\frac12]}  \phantom{aaa} \textrm{ a.e.}$$ 
Since $|1 + i \chi_{[0,\frac12]}| = \sqrt{2} \chi_{[0,\frac12]}+  \chi_{(\frac12,1]}$, we have 
$$ \frac{d|\mu|}{dm} = \sqrt{2} \chi_{[0,\frac12]}+  \chi_{(\frac12,1]} \phantom{aaa} \textrm{ a.e.}$$
Since we also have that $m \ll |\mu|$, we can conclude that 
$$ \frac{dm}{d|\mu|} = \frac{1}{\sqrt{2}} \chi_{[0,\frac12]}+  \chi_{(\frac12,1]} \phantom{aaa} \textrm{ a.e.}$$
Finally, since we have that $\mu \ll  m \ll |\mu|$, we have: 
$$ \frac{d\mu}{d|\mu|} = \frac{d\mu}{dm} \frac{dm}{d|\mu|} =\left (1 + i \chi_{[0,\frac12]}\right ) \left (\frac{1}{\sqrt{2}} \chi_{[0,\frac12]}+  \chi_{(\frac12,1]} \right)= \frac{1+i}{\sqrt{2}} \chi_{[0,\frac12]}+  \chi_{(\frac12,1]} \phantom{aaa} \textrm{ a.e.}$$
