# 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 $\mu(X)=0$)?

## 1 Answer

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,

1. $\left ( \bigcap_{n\in\mathbb{N}}A_n \right ) -D\neq \emptyset$
2. $\left ( \bigcap_{n\in\mathbb{N}}A_n \right ) - D \subseteq S$
3. $\mu\left ( \left ( \bigcap_{n\in\mathbb{N}}A_n \right ) - D \right)= \mu\left ( \bigcap_{n\in\mathbb{N}}A_n \right) - \mu(D)=m-m=0$

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