Countable sum of atomic measures is atomic?

Question

Let $(X,\Sigma)$ be a measurable space and $(\mu_n)$ a sequence of atomic measures defined on this space. Recall that a measure $\mu$ is atomic if for any measurable $A$ of measure $\mu(A)>0$ there is some measurable $E \subset A$ that is an atom, which means that $\mu(E)>0$, and for any measurable $F\in X$, either $\mu(E\cap F)$ or $\mu(E-F)=0$. Consider the measure $\mu=\sum_{n\in \mathbb{N}} \mu_n$. Is it true that $\mu$ must be atomic? This question is raised (but not answered) here.

EDIT: The answer below by @George reflects the question (here) when it was previously required that the measure not take on infinite value. This is not the correct definition, and does not reflect the original question of Roy A. Johnson.

Answer 1

The answer is no. Indeed,for $ k \in N$ let $\mu_k$ be a Dirac measure concentrated at $x=0$(we consider measures on the real axis $R$). Then $\sum_{k \in N}\mu_k$ is not atomic measure in the above-mentioned sense

Answer 2

It seems that the answer is positive: If $(\mu_n)_{n \in \mathbb{N}}$ is a sequence of atomic measures, then the sum $\sum_{n \in \mathbb{N}} \mu_n$ is atomic. This has been shown by P. Capek in his paper The atoms of a countable sum of set functions (Mathematica Slovaca 1989, No. 1, p.81-89; link).