How to prove that a logarithmic divergence implies a measure has no atoms? Let $\mu$ be a probability measure on $\mathbb{R}^2$, and suppose that
$$
-\int_{\mathbb{R}^2} \int_{\mathbb{R}^2} \log |z-w|\,du(z)\,du(w)<\infty
$$
is finite, where $|z-w|$ is the Euclidean distance between the points $z,w$ in the plane.
How to prove that $\mu$ has no atoms?
Edit:
It is relatively easy to see that there are no "atomic points". I wonder whether there are no atomic sets, more generally.

I guess that the specific choice of $\log$ here is not crucial, and the conclusion should holf for other functions with similar growth properties.
(I found this claim in a paper, and I had trouble verifying it).
 A: Let $\lambda$ be the Lebesgue measure. By Lebesgue-Radon-Nykodim, we can write $d\mu$ as $fd\lambda +d\mu_{sc}+d\mu_{sp}$, where $\mu_{sc}$ is continuous, i.e. $\mu_{sc}(\{x\})=0$ for every point. To see this, it suffices to define, given the usual decomposition $d\mu=fd\lambda+d\nu$ with $\nu\perp \lambda$, $\mu_{sp}:=\sum_E \nu(\{x\})\delta_x$ where $E:=\{x:\nu(\{x\})>0\}$ and it's clearly at most countable.
Now the same proof as the one in @GiuseppeNegro's answer implies that $\mu_{sp}=0$. It remains to prove that $\mu$ is atomless (i.e. diffuse). We prove a stronger statement, namely that

A finite measure $\mu$, on a complete separable space $X$ is atomless
iff it does not have atomic points

Let $A$ be a atom of $\mu$. Let $Q$ be a countable dense set in $X$, $\mathcal{B}=\{\bar B(x,r):r\in \mathbb{Q}, x\in Q\}$ and let $\mathcal{B}_\rho$ be the same family, only with the restriction $r\le\rho$. Let $A^1_1,\dots, A^1_n,\dots$ be the intersersections of $A$ with the elements of $\mathcal{B}_1$. Since $A$ is an atom, either $\mu(A^1_i)=0$ or $\mu(A_i^1)=\mu(A)$. Since it can't be that it's always the first case, let $j$ be an index such that $\mu(A_j^1)=\mu(A)$. We write $A^1:=A^1_j$. We can iterate the construction on $A^1$, this time with $A^2_1,\dots, A^2_n,\dots$ the intersections of $A^1$ with $\mathcal{B}_\frac12$. Again, there must be a $A_2^j:\mu(A_2^j)=\mu(A_1)$. Iterating this, we get a sequence $\{A^1,\dots, A^n,\dots\}$ of elements such that $\mu(A)=\mu(A^n)$ for every $n$, $A^n\subset A^{n-1}$ and $\text{diam}(A^n)\to 0$. Completeness implies that there one and only one element in $\cap A^n$, $x$. Since $\mu$ is finite, $\mu(\{x\})=\lim \mu(A_n)=\mu(A)$, which implies that $\{x\}$ is an atom.
Another way of proving this result, albeit longer, is by proving that on a locally compact $T_2$ space such that every open set is $\sigma-$compact every borel locally finite measure is Radon (hence regular, since every Radon measure is regular on its $\sigma-$finite Borel sets). A proof of these results is found in Folland's "Real analysis".
A: Suppose that has an atom at the origin, so that
$$\mu(x)=f(x)dx + M \delta(x),$$
where $f\ge 0$ is in $L^1$ and $M>0$. This is not the most general case but I think it would be enough to understand what is going on. The integral reads
$$
-\iint \log\lvert z-w\rvert \, f(z)f(w)\, dzdw -  M\iint\log\lvert z-w\rvert\,\delta(z)\delta(w)\, dzdw.$$
The second summand is $+\infty$. Indeed, using Fubini and the defining property of the $\delta$ distribution,
$$
\int\delta(z)\left(\int\log\lvert z-w\rvert\delta(w)\,dw\right)\,dz = \int \delta(z) \log\lvert z\rvert \,dz = -\infty.$$
This assumes that $\mu$ is a positive measure, which is your case.
