Let $(X,\Sigma,\mu)$ be a finite measure space. Call $\mu$ nonatomic if for any $A\in\Sigma$ with $\mu(A)>0$ there exists $B\in\Sigma$ with $B\subset A$ and $0<\mu(B)<\mu(A)$.
Suppose that $\mathcal S$ be the algebra that generates $\Sigma$. I am asking to what is the relation between nonatomicity of $\mu$ and $\mu|_\mathcal S$.
On
I'm no measure theory wizard, so check my argument well for errors!
Consider the collection of sets $\mathcal A\subset\Sigma$ defined by: $A\in\mathcal A$ if and only if the following two conditions both hold
If we assume that $\mu|_\mathcal S$ is 'nonatomic', then $\mathcal A$ is a Dynkin class containing $\mathcal S$ (check this! I found the second definition found on Wikipedia to be simplest to work with). Since $\mathcal S$ is clearly stable under finite intersections, it follows from Dynkin's Lemma that $\Sigma\subset \mathcal A$.
**Fact **. If $\mu$ is non-atomic then so is $\mu|_{S}$.
Proof. Assume the contrary and let $\mu|_{S}$ is atomic. Then there will be a set(atom) $A \in S$ with $\mu(A)>0$ such that for each element of the $S$ with $B \subseteq A$ we have $\mu|_{S}(B)=0$ or $\mu|_{S}(B)=\mu|_{S}(A)$. Since $\mu$ is non-atomic and extension of $\mu|_{S}$ by the condition $\mu(A)=\mu|_{S}(A)$ we claim that there is $B_0 \in \sum$ such that $0<\mu(B_0)<\mu(A)$. By the Charatheodory measure extension theorem from the algebra to the minimal algebra we have $ \mu(B_0)= \inf \{\sum_{k=1}^{\infty} \mu|_{S}(A_k) : A_k \in S \& k \in N \& $ $ B_0 \subseteq \cup_{k =1}^{\infty}A_k \in S\}. $ For $\epsilon=((\mu(A)-\mu(B_0))/2$ there is a sequence $\{ A_k \in S:k \in N\}$ such that $B_0 \subseteq \cup_{k =1}^{\infty}A_k \in S$ and $\sum_{k=1}^{\infty}\mu(A_k)<\mu(B_0)+\epsilon$. Then we have:
(i) $\cup_{k=1}^{\infty}A_k\cap A \in S$;
(ii) $\mu|_{S}(\cup_{k =1}^{\infty}A_k\cap A)=\mu(\cup_{k =1}^{\infty}A_k\cap A) \le \sum_{k=1}^{\infty}\mu(A_k)<\mu(B_0)+\epsilon < \mu(A)=\mu|_{S}(A)$;
(iii) $\mu|_{S}(\cup_{k =1}^{\infty}A_k\cap A)=\mu(\cup_{k =1}^{\infty}A_k\cap A)\ge \mu(B_0)>0$;
So The conditions (i)-(iii) contradicts to the our assumption that $A$ is atom for $\mu|_{S}$.
**Example( The converse of the Fact above is not valid). We have to show there is $S$ and $\mu$ such that $\mu|_{S}$ is non-atomic but $\mu$ is atomic. Indeed, let consider $E=Q\cap[0,1]$ where $Q$ is a set of all rational numbers; $\sum=\{A:A\subseteq E\}$. Let $(q_k)_{k \in N}$ be a numeration of the $E$; Let $m_k$ be a sequence of positive numbers with $\sum_{k=1}^{\infty}m_k=1$. For $A \in \sum$ we set $\mu(A)=\sum_{q_k \in A}m_k$. Let $S$ be a class of subsets of $E$ which can be presented in the form $\sum_{k=1}^n[a_k,b_k[\cap Q , (n\in N)$, where $\{[a_k,b_k[:1 \le k \le n\}$ is a sequence of pairwise disjoint intervals closed from the left and open from the right. Then it is not hard to show that
(j) $S$ is an algebra of subsets of $E$ which generates $\sum$;
(jj) $\mu$ is atomic measure on $\sum$;
(jjj) $\mu|_{S}$ is non-atomic measure on $S$.