Radon-Nikodym derivative: Lebesgue measure

Question

If we have two functions $F(t)=1/2(1-\exp(-t))$ and $G(t)=(1/2)(1+t)\exp(-t)$, where $2F$ and $G$ are two cumulative distribution functions. We know that, if $F=∫_{0}^{t} k(x)dG(x)$, $k$ becomes the Radon-Nikodym derivative. What is the exact value of $k$ (Lebesgue measure)? Is there a way to calculate this?

Answer 1

You forgot to say that $F$ and $G$ are supported by $(0,\infty)$.

$\frac1 2(1-e^{-t})=\int_0^{t} k(x) \frac 1 2 (e^{-x}-(1+x)e^{-x}) dx$ so you get $k(.)$ by differentiating both sides w.r.t. $t$. This gives $k(x)=\frac 1 x, 0<x<\infty$.