Working through some measure theory, I came upon the Lebesgue-like decomposition for monotonic functions. In that context, I've cooked up a singular measure $\nu$ on $[0,1]$ about which I know only that it is regular Borel, continuous, and satisfies $\nu K = 0$ for some $K \subset [0,1]$ of full Lebesgue measure.
I would like to conclude that $K$ contains a subset $K'$ of full Lebesgue measure for which each element $x$ satisfies $$ \nu\bigl((x-1/n,x+1/n)\bigr) = \mathcal O(1/n)~~. $$ This would be sufficient to demonstrate that the function mapping $x \mapsto \nu\bigl([0,x]\bigr)$ is singular, in the sense of having an a.e. defined and vanishing derivative (of course, it is continuous as well by continuity of $\nu$).
I'd appreciate any comments on how to show the above bound, and whether that's a good strategy for approaching the problem of defining a singular function from a singular measure. My other thought was that perhaps there is a standard way to 'remove' density from the complement of $K$ (using continuity) sufficiently to form an open set containing $K$ with zero $\nu$-measure, from which the result follows (and the singular set looks much more like the complement of a Cantor set).