Approximation of the radon-derivative 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     
 A: Here some references that show that (surprisingly!) the result you stated is wrong  if $\liminf$ is replaced by $\limsup$ and $X=\mathbb{R}^2$.
More precisely, if $\nu$ is the Lebesgue measure on the unit square $[0,1]^2$, there exists an integrable function $0\leq f\in L^1(\nu)$ such that the measure $\mu$ given by $\mu(A):=\int_A f d\nu$ satisfies the following property:
for $\nu$ a.e. $x$ there exist open rectangles $A_n=(a_n,b_n)\times (c_n, d_n)$ such that $x\in A_n$, $b_n-a_n\to 0$, $d_n-c_n\to 0$ and $\limsup \frac{\mu(A_n)}{\nu(A_n)}=\infty$. Notice that by replacing $A_n$ with a subsequence you can assume without loss of generality that the $A_n$ are nested.
This example is taken from  Theorem 7.2.10 of the book Stopping Times and Directed Processes' by Edgar and Sucheston. You can also find it (although it is much less readable) in 1.5.1. Proposition in the bookDerivation and Martingales' by Hayes and Pauc. In these books you will also find a lot of info on additional conditions under which the conclusion you stated holds (with $\lim$, not just $\liminf$); for example, if you use open rectangles and $\nu,\mu$ are as above, then the result holds if $f$ satisfies a stronger integrability property, namely $f\log^+(f)\in L^1$
