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This question is related to Radon-Nikodym derivatives as limits of ratios.

Let $F$, $G$ be sigma-finite measures (or at least probability measures) on $\mathbb{R}$ such that $F \ll G$.

The theorem quoted in the link tells that the Radon-Nikodym derivative checks $$\frac{\mathrm{d}F}{\mathrm{d}G} (x) = \lim_{h\to 0^+} \frac{F(x-h, x+h)}{G(x-h, x+h)}$$ for $G$-almost every $x$.

Do we have a similar equality with one-sided balls? In other words, is the following equality $$\frac{\mathrm{d}F}{\mathrm{d}G} (x) = \lim_{h\to 0^+} \frac{F [x, x+h)}{G [x, x+h)}$$ true for $G$-almost every $x$?

Thank you very much.

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  • $\begingroup$ Actually, I am unable to find a answer that use nicely shrinking sets when G is not the Lebesgue measure. Is this notion defined somewhere for more general measures? $\endgroup$
    – PhilippeC
    Commented Feb 17, 2016 at 13:37
  • $\begingroup$ That's a good point. For example, $(x,x+h)$ also shrinks nicely, but won't work for pure point measures. $\endgroup$
    – Christian Remling
    Commented Feb 17, 2016 at 18:22
  • $\begingroup$ OK, the Lebesgue differentiation theorem seems to work arbitrary Radon measure G and nicely shrinking sets, provided that "nicely shrinking" is defined with respect to the measure G, not the Lebesgue measure. Unfortunately, finding a reference to this fact is hard. The only one I have found is in this online supplementary material of the book A First Course in Sobolev Spaces. Does anybody know another reference? $\endgroup$
    – PhilippeC
    Commented Feb 18, 2016 at 0:10
  • $\begingroup$ So in my case, one still has to prove that for G-almost all x, there exists a constant C > 0 independent of h > 0 such that G [x,x+h) > C G [x-h,x+h]. If x is a jump point of G, it is OK. For (Lebesgue-)almost all continuity points, G'(x) exists and is the limit of G [x,x+h) /h and G [x-h,x+h] /(2h). But does the ratio G [x,x+h) / G [x-h,x+h] converges to 1/2? It would be OK if G'(x) were not 0 on all that points. But unfortunately that is not sure as G might be singular... $\endgroup$
    – PhilippeC
    Commented Feb 18, 2016 at 0:45

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