Let $\nu $ be the Borel measure define by $\nu(A)=\sum_{n=1}^{\infty}\frac{ \lambda((n,3n/2) \cap A)}{3^n}$. Show that $\nu \ll \lambda$ with $\lambda$ the Lebesgue measure of $\mathbb{R}$ and compute $\frac{d\nu}{d \lambda}$.
First if $A \in \mathcal{B}$ and $\lambda(A)=0$ then is $\lambda((n,3n/2) \cap A)=0, \forall n \in \mathbb{N}$ (because $(n,3n/2) \cap A \subset A, \forall n \in \mathbb{N}$ and the lebesgue measure is complete) and these implies that $\nu (A)=0$
And for compute $\frac{ d \nu}{d \lambda}$ i define $g= \sum_{n=1}^{\infty} \frac{ \chi_{(n,3n/2)}}{3^n}$ then given $A \in \mathcal{B}$
$$ \int_{A} g \ d \lambda = \int_{\mathbb{R}} g \chi_{A} \ d\lambda= \int_{\mathbb{R}} \sum_{n=1}^{\infty} \frac{\chi_{((n,3n/2)\cap A)}}{3^n} \ d\lambda= \sum_{n=1}^{\infty}\frac{ \lambda((n,3n/2) \cap A)}{3^n}=\nu (A)$$
Is these correct? any hint or help i will be very grateful