Absolutely continuity and Lipschitz-like continuity

Question

I'm trying to figure out if the following is true or false:

Let $\mu$ and $\nu$ be finite measures on Borel subsets of $[0,1]$,and $\nu \ll \mu$ ($\nu$ is absolutely cts with respect to $\mu$). Does there exist always a $C \in \mathbb{R}$ such that $\forall A \subseteq [0,1]$ that is Borel, we have $\nu(A)$ $\lt$ $C \mu (A)$.

The Radon-Nykodym theorem seems to almost suggest that is true, if I knew it was bounded, then I could say $\nu (A) \le \int_{A} g \ d\mu \le \int_{A} C \ d\mu \le C\mu (A)$, where $C$ is the bound. However, I don't know if it is bounded and in fact, I could take an unbounded function that is still integrable on $[0,1]$, such as $\frac{1}{\sqrt x}$.

I can't even be sure if it is true or false.

Thank you!

Answer 1

Your example is fine. Let $\mu$ be Lebesgue measure on $[0,1]$ and $\nu (E)=\int_E \frac 1 {\sqrt x} dx$. Then $\mu <<\mu$ and there is no $C$ such that $\nu(E) \leq C\mu (E)$. This can be seen by taking $E=(0,r)$ with $r$ sufficiently small.

Answer 2

Let $g(x)=\displaystyle\sum_{n=1}^{\infty}n\chi_{((n+1)^{-2},n^{-2}]}$ and $\nu(A)=\displaystyle\int_{A}g(x)dx$, then $\nu([0,1])<\infty$ but if it were $\nu(A)\leq C|A|$, then $n=\nu(((n+1)^{-2},n^{-2}])\leq C(n^{-2}-(n+1)^{-2})$, taking $n\rightarrow\infty$ will obtain a contradiction.