This question stems from [The kinetic limit of a system of coagulating Brownian particles] (https://arxiv.org/abs/math/0408395), specifically the last step in the proof of Lemma 4.2.

The setting is as follows. We fix some finite time $T<\infty$, some constant $K\in\mathbb R$, a positive mollifier $(\eta^\delta)_{\delta>0}$, $\psi(x)=(x-K)_+ $, and a probability measure $\mathbb P$ on the space of non-negative finite measures on $[0,T]\times \mathbb R^d$. We then show that \begin{align} \int \mathbb P (d\mu) \int_{\mathbb R^d} \psi \left( \int_{[0,T]\times \mathbb R^d}\eta^\delta (x-y)\gamma(t)\mu(dt,dy)\right) dx =0 \end{align} for any continuous function $\gamma : [0,T]\to\mathbb R_+$ of compact support with $\int _0^T \gamma(t)dt=1$.

The paper deduces from this ("as $\gamma$ is arbitrary) that $\mathbb P(d\mu)$-almost surely, $\mu$ is absolutely continuous with respect to Lebesgue on $[0,T]\times\mathbb R^d$ with its density bounded by our $K$. I don't see how they draw this conclusion. I'm familiar with the Radon-Nikodym theorem but don't know if it's helpful here.

Has anyone dealt with something similar or have an idea here?

Thanks in advance for any help.

Edit for additional information: The probability measure $\mathbb P$ is the limit as $n\to\infty$ of probability measures $\mathbb P_n$ induced by random measures \begin{align} \xi_n(dt,dx)=\frac 1 n dt \sum_{j\leq n(t)}\delta_{x_j^n(t)}(dx), \end{align} where $x_j^n$ are random variables, moving through $\mathbb R^d$ as independent Brownian motions. The population size $n(t)$ grows with $n$.

The reason I'm adding this information is to point out that each of these $\xi_n$ is just Lebesgue on $[0,T]$. Maybe this tells us something about its limit $\mathbb P$'s distribution.


1 Answer 1


Maybe not the most rigorous answer, you can do a chain of arguments:

  • $\psi$ is a nonnegative function for all $x$ and $\mu$. It integrates to $0$ with respect to Lebesgue$\otimes \mathbb{P}$, so for Lebesgue-almost every $x$ and $\mathbb{P}$-almost every $\mu$ we have $\psi=0$.
  • Next, observe that for every $\mu$, $\psi(\ldots)$ is continuous in $x$. Because it is $0$ for almost every $x$, it is always $0$ (This holds $\mathbb{P}$-almost surely).
  • By knowing what exactly $\psi$ is, $\psi$ being $0$ as random variable means that for every $x$ and $\mathbb{P}$ almost every $\mu$ we have that $$\int_{[0,T]\times \mathbb{R}^d}\eta^\delta(x-y)\gamma(t)\mu(dt,dy) \leq K.$$
  • Because $\eta$ and $\gamma$ are arbitrary (and continuous!), this implies that $\mu$ is absolutely continuous with respect to the Lebesgue measure (restricted to a ball around $x$ but $x$ is arbitrary too, so its true everywhere). Then Radon-Nikodym implies that $\mu$ has a density and the above inequality gives you the bound on this density. (This argument works for $\mathbb{P}$-almost every $\mu$.
  • Extra detail about absolut continuity: Assume $\mu$ is not absolutely continuous. Let $A$ be a Lebesgue-nullset contained in $B_\varepsilon(x)$ ($\varepsilon$ should depend on the choice of mollifier but doesn't really matter) and assume that with some probability $\mu(A)>0$. Then we can can choose $\gamma$ such that $\eta \gamma >0$ on $A$, hence $\int_A \eta \gamma d\mu >0$ with non-zero probability. Multiply $\gamma$ by a constant to scale it up such that $\int_A \eta \gamma d\mu > K$ with positive probability. By continuity of $\eta$ this works uniformly for all $x$ in a little ball around $x_0$ (this basically turns a null-set into something that is not null with respect to Leb). But once we have that, $\psi$ is not almost--surely equal to $0$ anymore which is a contradiction.

I hope this helps!

  • $\begingroup$ Thanks for helping! Could you elaborate on the last step, why $\mu$ is absolutely continuous wrt lebesgue (on a ball around x)? Can I show this via the definition, that $\mu$ is zero when lebesgue is zero? $\endgroup$ Commented Apr 20 at 21:46
  • 1
    $\begingroup$ Basically yes, added a paragraph. Basically you can exploit the continuity of $\eta$. $\endgroup$
    – David
    Commented Apr 22 at 9:49

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