I have a puzzling problem and hope some of you can give me a hint how to solve it.
$F \geq 0$ is a distribution function of a LS-measure $\nu_F$. The following things are to do:
(i) $G:=F^2$ is a distribution function of a LS-measure $\nu_G$ too
(ii) It holds $\nu_g \ll \nu_F$
(iii) Find the Density $\frac{d\nu_G}{d\nu_F}$
Well, using the properties of a LS-measure and using that $\phi: t\mapsto t^2$ is a continous, monotoniously growing function I managed to show, that $G$ is indeed a contribution function of another LS-measure.
Furthermore I already proofed (ii) via using the properties of a distribution function and the given connex between $F$ and $G$.
My problems start with (iii). I know that due to Radon-Nikodym (requirements fulfilled) a Density $g$ exists, such that $\mu_G(A)=\int_A g d\mu_F$ with $g\geq0$ and $g = \frac{d\nu_G}{d\nu_F}$. Now I don't have an idea how to find an expression for $g$.
I guess, that $g(x):=G'(x)=2F(x)F'(x)$ might work, but I can't proof that this is true and I even don't know, wether $F'(x)$ does exist.
Does anybody have a hint, how I can proceed?
Thanks a lot!
I would suggest the following:
$g:=\frac{G'(x)}{F'(x)} = \frac{2F(x)F'(x)}{F'(x)} = 2F(x)$ if $F$ is differentiable at $x$
and otherwise
$g = \frac{G(x)-G^{-}(x)}{F(x)-F^{-}(x)} = \frac{F^2(x)-(F^{-}(x))^2}{F(x)-F^{-}(x)} = F(x) + F^{-}(x)$
I hope that helps!