Let $(\Omega, \Sigma, \mu)$ be any measure space and let $\mathcal{C} \subset \Sigma$ be the class of sets $A \in \Sigma$ such that either $\mu(A) = 0$ or $\mu(A^c) = 0$. Then $\mathcal{C}$ is a $\sigma$-algebra:
- Since $\mu$ is a measure we have $\mu(\emptyset) = 0$, so $\emptyset \in \mathcal{C}$.
- By definition $A \in \mathcal{C}$ if and only if $A^c \in \mathcal{C}$.
- If $A_1,A_2, \ldots \in \mathcal{C}$ then either $\mu(A_n) = 0$ for all $n$, so $\mu(\bigcup A_n) \leq \sum \mu(A_n) = 0$ and thus $\bigcup A_n \in \mathcal{C}$ or there is some $k$ such that $\mu(A_{k}^c) = 0$. But then $\Omega \smallsetminus \bigcup A_n = \bigcap A_{n}^c \subset A_{k}^c$, so $\bigcup A_n$ is conull and hence in $\mathcal{C}$. In both cases the union is in $\mathcal{C}$ and we are done.
Now let $\mathcal{N} = \{N \subset \Omega\,:\,\text{there is }E \in \Sigma \text{ such that }N \subset E\text{ and }\mu(E)=0\}$ be the class of negligible sets. Then $\mathcal{N}$ is a $\sigma$-ideal, that is to say
- $\emptyset \in \mathcal{N}$,
- If $N \in \mathcal{N}$ and $B \subset N$ then $B \in \mathcal{N}$,
- If $N_1,N_2,\ldots \in \mathcal{N}$ then $\bigcup N_n \in \mathcal{N}$.
Example: If $\Omega = \mathbb{R}$, $\Sigma = $ Borel sets and $\mu =$ Lebesgue measure then $\mathcal{N}$ is the class of Lebesgue null sets.
The structure of a $\sigma$-ideal plays a rôle in the theory of measure algebras and on a basic level in the context of completion of a measure space.