# How do we deduce that a measure is absolutely continuous with respect to Lebesgue?

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.

• $$\psi$$ 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.
• 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? Commented Apr 20 at 21:46
• Basically yes, added a paragraph. Basically you can exploit the continuity of $\eta$. Commented Apr 22 at 9:49