MathOverflow Asked on November 3, 2021
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 $dfrac{dXi}{dLambda}$ 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
$$
dfrac{dXi_n}{dLambda_n} = dfrac{xi_n}{lambda_n} quad mathrm{Leb}-text{a.e.}quadtext{(because $lambda_n > 0$)}
$$
Finally, $0leq dfrac{xi_n}{lambda_n}leq 1$ so that the measure given by $dkappa_n(x) := dfrac{xi_n(x)}{lambda_n(x)}cdot dmathrm{Leb}(x)$ 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
$$
dkappa(x) = dfrac{dXi}{dLambda}(x)cdot dmathrm{Leb}.
$$
In particular that would imply that $Xi$ and $Lambda$ have a non trivial absolutely continuous part. (Note that we may assume that $mathrm{Leb}llLambda$ 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.
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 $Xsubsetmathbb R^2$ are such that $Lambda=limLambda_n$ is singular with respect to $text{Leb}$, then for any prescribed measure $kappa$ on $X$ there are sequences $$ Xi'_nllLambda'_nll text{Leb} $$ with $$ |Xi_n-Xi'_n|, |Lambda_n-Lambda'_n|to 0 $$ (so that, in particular, $Xi'_ntoXi, Lambda'_ntoLambda$), and such that the measures $$ kappa'_n = frac{dXi'_n}{dLambda'_n},text{Leb} $$ 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 $limLambda_n$ is singular with respect to the measure $text{Leb}$, there are subsets $X_nsubset X$ with $text{Leb}(X_n)totext{Leb}(X)$, whereas $Xi(X_n),Lambda_n(X_n)to 0$. Fix a sequence $epsilon_nto 0$. Then the restrictions of $Xi'_n$ and $Lambda'_n$ to $Xsetminus X_n$ are the multiples of $Xi_n|_{Xsetminus X_n}$ and $Lambda_n|_{Xsetminus X_n}$, respectively, chosen in such a way that $Xi'_n(Xsetminus X_n)=Lambda'_n(Xsetminus X_n)=1-epsilon_n$, 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 $kappa$.
Answered by R W on November 3, 2021
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