On $[0,1]$ let m be lebesgue measure and $\mu$ a positive borel measure with $\mu \perp m$. Show that there exists a measure $\nu$ such that
$\|\mu-\nu\|< \epsilon$, and $m(\text{support}(\nu)) = 0$.
Attempt/Explication:
This question derives from p.63 of Schlag and Muscalu's Harmonic Analysis, Volume 1. I've reduced the issue to the positive measure case. Using inner regularity, we can pick K compact so that $\mu([0,1] \setminus K) < \epsilon$ and define $\nu$ via $\nu(A) = \mu(A \cap K)$ for all borel sets A; this is done in the text.
I'm struggling to show that $\nu$, so chosen, has lebesgue-null support.
Let $\mu (E)=0$ and $m(E^{c})=0$. There exits $K$ compact, $K \subset E^{c}$ such that $\mu (E^{c}\setminus K)<\epsilon$ Define $\nu (A)=\mu (A \cap K)$. Then support of $\nu$ is contained in $K$, hence in $E^{c}$ which has Lebesgue measure $0$.