Absolute continuity of probability measure

Question

Let $([0,1), \mathcal{B}([0,1))$ be a probability space and define the filtration $\mathcal{F}_n=\sigma([0,\frac{1}{2^n}),[\frac{1}{2^n}\frac{2}{2^n}),\ldots,[\frac{2^n-1}{2^n},1))$. Let $\mu_n$ and $\nu_n$ be the restrictions to $\mathcal{F}_n$ of $\mu=$ any probability measure on $(\Omega, \mathcal{F})$ and $\nu=$ Lebesgue measure respectively. Why is it true for all probability measures $\mu$ that $\mu_n \ll \nu_n$?

Answer 1

The sigma algebra consists of just unions of the $2^{n}$ intervals. It is obvious that the only set $E$ in this sigma algebra with $\nu_n(E)=0$ is the emprty set. Of course, $\mu_n(\emptyset)=0$.