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 $\mu(X)=0$)?


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 $S$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.