Full-measure sets containing non-empty null sets Let $(\Omega,\mathcal{F},\mu)$ a probability space and suppose $S\in\mathcal{F}$ is a full-measure set that is not a union of $\mathcal{F}$-atoms. In other words, we know that $\mu(S)=1$ and that there does not exist a collection $\{A_{i}\}_{i\in I}$ such that 
$$
S=\bigcup_{i\in I}A_{i}
$$
where each $A_{i}$ is an (algebraic) atom in $\mathcal{F}$ (in the sense that $A_{i}\in\mathcal{F}$ and there is no set $B\in\mathcal{F}$ with $\emptyset \neq B\varsubsetneq A_{i} $).
The question is: does $S$ necessarily contain some non-empty null subset (i.e. an $\mathcal{F}$-measurable set $X\subseteq S$ with $X\neq\emptyset$ and $\mu(X)=0$)?
 A: For clarity, let us separate the definition of $\mathcal{F}$-atom.

Definition: Let $(\Omega,\mathcal{F})$ a mesurable space. We say $A$ is a $\mathcal{F}$-atom  if there is not any set $B\in\mathcal{F}$ such that $\emptyset \neq B\varsubsetneq A$.

To answer the question, let us prove a result which is slightly more general than the question.

Let $(\Omega,\mathcal{F},\mu)$ be a measure space and suppose $S\in\mathcal{F}$ has positive finite measure (it means $0<\mu(S)<+\infty$) and it does not contain any $\mathcal{F}$-atom. Then $S$ contains a non-empty null subset.

Proof: Since $\mu(S)>0$, we have that $S\neq \emptyset$. So, let $x_0\in S$. 
Let us define 
$$H=\{A \in \mathcal{F} \:|\: x_0\in A \varsubsetneq S\}$$
Since $S$ does not contain any $\mathcal{F}$-atom, then $S$ itself is not a  $\mathcal{F}$-atom. So there is a set $B\in\mathcal{F}$ with $\emptyset \neq B\varsubsetneq S$, so we have either $x_0\in B \varsubsetneq S$ or $x_0\in S-B \varsubsetneq S$, which means either $B\in H$ or $S-B\in H$. So $H\neq \emptyset$.
Let $$m=\inf\{\mu(A) \:|\: A\in H\}$$ Since $H\neq \emptyset$, we have $0\leqslant m <+\infty$. 
For each $n\in\mathbb{N}$, let $A_n\in H$ such that $\mu(A_n)<m+\frac{1}{n+1}$. It is easy to see that $x_0\in \bigcap_{n\in\mathbb{N}}A_n \varsubsetneq S$. So, we have $\bigcap_{n\in\mathbb{N}}A_n \in H$. Thus, we have 
$$m \leqslant \mu\left(\bigcap_{n\in\mathbb{N}}A_n\right) \leqslant  \mu(A_n)<m+\frac{1}{n+1}$$ for all $n\in\mathbb{N}$. So we have, $\mu\left(\bigcap_{n\in\mathbb{N}}A_n\right)=m$. 
Since $S$ does not contain any $\mathcal{F}$-atom, $\bigcap_{n\in\mathbb{N}}A_n$ is not a $\mathcal{F}$-atom. So there is a set $B\in\mathcal{F}$ with $\emptyset \neq B\varsubsetneq \bigcap_{n\in\mathbb{N}}A_n$. Let us define 
$$D=B \:\:\:\textrm{ if  } x_0\in B$$ and  $$D=\left(\bigcap_{n\in\mathbb{N}}A_n \right) - B \:\:\: \textrm{ if  } x_0\notin B$$
It is easy to see that $$ x_0 \in D \varsubsetneq \bigcap_{n\in\mathbb{N}}A_n \varsubsetneq  S $$
So $D \in H$ and we have $m\leqslant \mu(D) \leqslant \mu\left(\bigcap_{n\in\mathbb{N}}A_n\right)=m$. So $\mu(D)=m$. 
So, we have that $D \varsubsetneq \bigcap_{n\in\mathbb{N}}A_n \varsubsetneq  S$ and $\mu(D) = \mu\left(\bigcap_{n\in\mathbb{N}}A_n\right) = m <+\infty$. So, 


*

*$ \left ( \bigcap_{n\in\mathbb{N}}A_n \right ) -D\neq \emptyset$ 

*$\left ( \bigcap_{n\in\mathbb{N}}A_n \right ) - D \subseteq S$  

*$\mu\left ( \left ( \bigcap_{n\in\mathbb{N}}A_n \right ) - D \right)=
    \mu\left (  \bigcap_{n\in\mathbb{N}}A_n  \right)  -
    \mu(D)=m-m=0$


So $ \left ( \bigcap_{n\in\mathbb{N}}A_n \right ) -D$ is a non-empty null subset of $S$. 
