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I am struggling for quite some time, because of a problem involving Radon-Nikodým derivatives. I will try to describe the main features and perhaps somebody has an idea how to solve it.

I consider two sequences of measures on some compact set of $\mathbb{R}^2$: $\Xi_n$ and $\Lambda_n$ which are absolutely continuous in the sense $\Xi_n \ll \Lambda_n \ll \mathrm{Leb}$. I know that $\Lambda_n$ converges to some measure $\Lambda$ which is not absolutely continuous wrt. to Lebesgue measure anymore. (I suspect that it has a non-zero absolutely continuous part, but I am not quite sure how to prove this.) Furthermore, $\Xi_n$ also converges to some measure $\Xi$ satisfying $\Xi \ll \Lambda$.

The weird thing is that I need to integrate $\dfrac{d\Xi}{d\Lambda}$ against Lebsegue measure. In the prelimit case, there is no problem, because $\Xi_n$ and $\Lambda_n$ both have densities $\xi_n > 0$ and $\lambda_n > 0$ with respect to $\mathrm{Leb}$, so that simply $$ \dfrac{d\Xi_n}{d\Lambda_n} = \dfrac{\xi_n}{\lambda_n} \quad \mathrm{Leb}-\text{a.e.}\quad\text{(because $\lambda_n > 0$)} $$ Finally, $0\leq \dfrac{\xi_n}{\lambda_n}\leq 1$ so that the measure given by $d\kappa_n(x) := \dfrac{\xi_n(x)}{\lambda_n(x)}\cdot d\mathrm{Leb}(x)$ so that $(\kappa_n)$ is tight. For the moment, suppose that there is only one limiting point, i.e. $\kappa_n \to \kappa$, where $\kappa$ is some finite measure. I am wondering if one could conclude that $$ d\kappa(x) = \dfrac{d\Xi}{d\Lambda}(x)\cdot d\mathrm{Leb}. $$ In particular that would imply that $\Xi$ and $\Lambda$ have a non trivial absolutely continuous part. (Note that we may assume that $\mathrm{Leb}\ll\Lambda$ so that the Radon-Nikodým derivative is defined $\mathrm{Leb}$-a.e.)

I am not deep into measure theory, so I am really struggling with this problem. Especially because I find it strange to integrate the Radon-Nikodým derivative of $\Xi$ wrt. to $\Lambda$ against Lebesgue measure. I would be glad for any ideas or references that might go into this direction.

To give a larger context: I am working with a sequence $\Lambda_n(t)$ of measure-valued stochastic process that converges to a white noise process $\Lambda(t)$. And now I am looking at a related process $\Xi_n(t)$ and want to understand its limiting behaviour. So I would also be very glad about any ideas or references wrt. absolutely continuous parts of random fields and random measures wrt. to Lebesgue measure.

Edit: The convergence takes place at least in the distributional sense. I think I can also get it for Lipschitz continuous functions so that it is weak convergence.

Edit2: Initially, the integral $$ \int \dfrac{\xi_n}{\lambda_n}(x) \phi_n(x) dx $$ for some sequence of continuous (or smooth if you like) functions converging pointwise to some continuous (or smooth) limit phi appears in the problem. Since I could not figure out what will happen, I translated it towards the Radon-Nikodym derivative, hoping that since I require only weak convergence, the measure theoretic setting would be more helpful.

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  • $\begingroup$ I'm curious: what do you need to integrate $d\Xi/d\Lambda$ for? Either I missed it or you left it out on purpose. Note that $d\Xi/d\Lambda$ is only defined a.s. up to $\Lambda$-nullsets, so you may obtain different results when integrating against Lebesgue measure depending on which version you choose. $\endgroup$
    – S.Surace
    Jul 9, 2020 at 13:01
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    $\begingroup$ In what sense does $\Lambda_n$ "converge" to $\Lambda$? $\endgroup$
    – Nik Weaver
    Jul 9, 2020 at 13:01
  • $\begingroup$ @S.Surace The integral with respect to Lebesgue measure appears in the infinitesimal generator of the process I am considering. Since $\Lambda$ can be decomposed into an absolutely continuous part and a singular part wrt. Lebesgue measure, I was thinking that a) for the absolutely continuous part, there is no problem and b) the singular part only changes things on a Lebesgue-nullset, so that different versions of this derivative shouldn't yield different results... $\endgroup$
    – Mushu Nrek
    Jul 9, 2020 at 13:55
  • $\begingroup$ @NikWeaver I edited my post. It is convergence in the distributional sense or weak convergence of measures. $\endgroup$
    – Mushu Nrek
    Jul 9, 2020 at 13:56
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    $\begingroup$ You still have the problem that $d\Xi/d\Lambda$ is defined only up to $\Lambda$-nullsets, giving potentially very different results when integrated against Lebesgue measure (some versions are not even integrable). For example, take $\Xi_n$ and $\Lambda_n$ each to be a weighted sum of a Gaussian centered at $0$ with variance $1/n$, and a uniform distribution on $[1,2]$. Then $d\Xi/d\Lambda$ may take arbitrary values on $(-\infty,0)\cup(0,1)\cup(2,\infty)$, which is a $\Lambda$ but not a Lebesgue nullset. Maybe try to multiply by indicator of $\text{supp}(\Lambda)$. $\endgroup$
    – S.Surace
    Jul 21, 2020 at 8:21

1 Answer 1

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There is no reason for the limit measure $\kappa$ to be related in any way to the limit measures $\Xi$ and $\Lambda$ (and, in particular, to their Radon-Nikodym derivative).

More precisely, if your sequences $\Xi_n,\Lambda_n$ on a compact $X\subset\mathbb R^2$ are such that $\Lambda=\lim\Lambda_n$ is singular with respect to $\text{Leb}$, then for any prescribed measure $\kappa$ on $X$ there are sequences $$ \Xi'_n\ll\Lambda'_n\ll \text{Leb} $$ with $$ \|\Xi_n-\Xi'_n\|, \|\Lambda_n-\Lambda'_n\|\to 0 $$ (so that, in particular, $\Xi'_n\to\Xi, \Lambda'_n\to\Lambda$), and such that the measures $$ \kappa'_n = \frac{d\Xi'_n}{d\Lambda'_n}\,\text{Leb} $$ weakly converge to $\kappa$. In fact, the presence of an ambient Euclidean space is completely irrelevant here, and instead of the Lebesgue measure one can talk about any reference measure on $X$.

The idea of the construction is very simple (I skip the details). Since $\lim\Lambda_n$ is singular with respect to the measure $\text{Leb}$, there are subsets $X_n\subset X$ with $\text{Leb}(X_n)\to\text{Leb}(X)$, whereas $\Xi(X_n),\Lambda_n(X_n)\to 0$. Fix a sequence $\epsilon_n\to 0$. Then the restrictions of $\Xi'_n$ and $\Lambda'_n$ to $X\setminus X_n$ are the multiples of $\Xi_n|_{X\setminus X_n}$ and $\Lambda_n|_{X\setminus X_n}$, respectively, chosen in such a way that $\Xi'_n(X\setminus X_n)=\Lambda'_n(X\setminus X_n)=1-\epsilon_n$, whereas on $X_n$ the measures $\Xi'_n$ and $\Lambda'_n$ can be defined in such a way that the Lebesgue measure multiplied by their ratio converges to $\kappa$.

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  • $\begingroup$ Thank you very much! I see that in general, there is no way of getting a positive result. But in my case, I know (strongly suspect) that Lebesgue measure is absolutely continuous wrt $\Lambda$. This means that $\Lambda$ and Lebesgue are not completely singular, and in particular, I don't see how in that case, we can find the sets $X_n$ you mention. It would be great if you could elaborate on that. $\endgroup$
    – Mushu Nrek
    Jul 24, 2020 at 7:54

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