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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

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  • $\begingroup$ By Luzin's theorem, $f$ is continuous except for a set of arbitrarily small measure; if $f$ is continuous at $x$ then your limit equals $f(x)$. This has to be made more accurate though. $\endgroup$
    – Yulia Kuznetsova
    Commented Jan 11, 2012 at 13:32
  • $\begingroup$ Try to search for material on Lebesgue points. If $\nu$ is the Lebesgue measure in $\mathbb R^n$, then almost every point is a Lebesgue point and so has your property. $\endgroup$
    – Yulia Kuznetsova
    Commented Jan 11, 2012 at 14:03
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    $\begingroup$ @Yulia: By Luzin's theorem, the restriction of $f$ 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 $U_i=(-1/i,1/i)\cup (i,+\infty)$. $\endgroup$
    – fedja
    Commented Jan 11, 2012 at 15:22
  • $\begingroup$ @yulia: many thank's, embarassingly I did not know the notion of Lebesgue points. As X is also metric and one measure is the Hausdorff measure there might be some theorem that almost every point is lebesgue. @fedja: Many thank's for correcting me, my question is missleading. Actually, I am looking for a criterion for the shape of $U_i$. $\endgroup$
    – Klaus
    Commented Jan 11, 2012 at 15:44
  • $\begingroup$ @fedja: thank you, I forgot much of it since using. @Klaus: I remember reading a good review close to this topic: A. Bruckner, “Differentiation of integrals,” Amer. Math. Monthly 78 (9, part II) (1971). $\endgroup$
    – Yulia Kuznetsova
    Commented Jan 11, 2012 at 16:11

1 Answer 1

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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 $f\log^+(f)\in L^1$

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  • $\begingroup$ To clarify: what makes the counterexample above possible is that the sides of the rectangles collapse at different rates. Indeed, if there exist constants $a,b$ such that $0<a\leq \frac{b_n-a_n}{d_n-c_n}\leq b<\infty$ for all $n$ then for $\mu$ a.e. $x$ there exists $ \lim_n \frac{\mu(A_n)}{\nu(A_n)}$ and equals $f(x)$ (a more general statement is shown in theorem 7.10 in the book `Real and complex analysis' by Rudin for the case where $\nu$ is the Lebesgue measure). $\endgroup$
    – pietro siorpaes
    Commented Sep 20, 2018 at 10:16

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