I am looking for the following statement. Let $X$ be a topological space and let $\mu$, $\nu$ be Radon-Borel measures in the same measure class i.e. the the zero-sets are identical. Denote then by $f$ the Radon derivative.

We can assume a lot of regularity on $X$ i.e. it is compact, locally compact... Also we can assume some regularity on $\mu$, $\nu$ i.e. atom-freeness the support is X...

Is then the following statement true.

For each $x$ in $X$ outside a measure zero-set and for each sequence of nested open sets $U_i$ whose intersection is only $x$ it follows that $f(x)=\liminf \mu(U_i)/\nu(U_i) $

Are there any further assumption for $U_i$?

Thanks

the restrictionof $f$ to some set $E$ of almost full measure is continuous or, if you prefer, $f$ equals to some other continuous function $g$ outside a set of small measure. This is very different from the continuity of $f$ itself. The theorem on Lebesgue points also requires very particular shapes of $U_i$ to employ covering lemmas. @Klaus The desired property is currently stated so sloppily that it doesn't hold even at points of continuity: take $X=\mathbb R$, $x=0$, and $U_i=(-1/i,1/i)\cup (i,+\infty)$. – fedja Jan 11 '12 at 15:22