# Integrate Radon-Nikodým derivatives against Lebesgue measure

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.

• 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. Jul 9, 2020 at 13:01
• In what sense does $\Lambda_n$ "converge" to $\Lambda$? Jul 9, 2020 at 13:01
• @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... Jul 9, 2020 at 13:55
• @NikWeaver I edited my post. It is convergence in the distributional sense or weak convergence of measures. Jul 9, 2020 at 13:56
• 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 $\text{supp}(\Lambda)$. Jul 21, 2020 at 8:21

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 $$\kappa$$.
• 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. Jul 24, 2020 at 7:54