Example of a set and monotone class where monotone class is not a $\sigma$-algebra? What is an example of a set $X$ and a monotone class $\mathcal{M}$ consisting of subsets of $X$ such that $\emptyset \in \mathcal{M}$, $X \in \mathcal{M}$, but $\mathcal{M}$ is not a $\sigma$-algebra?
 A: Recall that a σ-algebra for $X$ is a collection of subsets $\Sigma$ of $X$ such that:


*

*The empty set and the whole set $X$ belong to $\Sigma$.

*$\Sigma$ is closed under all countable unions.

*$\Sigma$ is closed under all countable intersections.

*$\Sigma$ is closed under complementation.


Recall that a monotone class for $X$ is a collection of subsets $\mathcal{M}$ of $X$ such that:


*

*The whole set $X$ belongs to $\mathcal{M}$.

*$\mathcal{M}$ is closed under unions of monotonically increasing collections $A_1 \subseteq A_2 \subseteq \ldots $ of sets in $\mathcal{M}$.

*$\mathcal{M}$ is closed under intersections of monotonically decreasing collections $B_1 \supseteq B_2 \supseteq \ldots $ of sets in $\mathcal{M}$.


So suppose $\mathcal{M}$ is a monotone class for the set $X$, and suppose furthermore that $\varnothing \in \mathcal{M}$. Then $\mathcal{M}$ satisfies the first property of being a σ-algebra, but may fail in the other three. Let's find examples for each.
Failing the countable union property
Put $X = \{1,2,3\}$ and $\mathcal{M} = \{\varnothing, \{1\},\,\{2\},\{3\}, X\}$.
We can easily show that $\mathcal{M}$ is a monotone class. However, $\mathcal{M}$ fails the countable union property, since $\{1\} \in \mathcal{M}$ and $\{2\}\in \mathcal{M}$ but $\{1\}\cup \{2\} = \{1,2\} \notin \mathcal{M}$.
The trick here was to generate an example that is closed under unions of sets in a monotone collection, but not closed under unions of sets like $\{1\}$ and $\{2\}$ which are not related by the $\subseteq$ relation.
Failing the countable intersection property
Put $X = \{1,2,3\}$ and $\mathcal{M} = \{\varnothing, \{1,2\},\,\{2,3\}, X\}$.
We can easily show that $\mathcal{M}$ is a monotone class. However, $\mathcal{M}$ fails the countable intersection property, since $\{1,2\} \in \mathcal{M}$ and $\{2,3\}\in \mathcal{M}$ but $\{1,2\}\cap \{2,3\} = \{2\} \notin \mathcal{M}$.
I generated this example using essentially the same trick as before.
Failing the complementation property
In fact, both of our examples above fail to have the complementation property, as you can show. Another example, mentioned in the comments, is:
$X = \{1,2\}$, $\mathcal{M} = \{\varnothing, \{1\}, X\}$.
Here, $\{1\} \in \mathcal{M}$ but its complement $\{2\} \notin \mathcal{M}$.
Hope this helps!

Side note: the "complementation" example above has all the properties of a σ-algebra except for complementation. You might try to construct analogous examples of, e.g. a monotone class that satisfies all the properties of a σ-algebra except countable unions— however, this is impossible. For example, if a collection $\mathcal{M}$ is closed under complementation and countable unions, then it is necessarily closed under countable intersections:

For any sequence $\{A_i\}_{i=1}^\infty$  in $\mathcal{M}$,
$$\bigcap_{i=1}^\infty A_i = \left(\bigcup_{i=1}^\infty A_i^\mathsf{C}\right)^\mathsf{C}.$$
All of the $A_i^\mathsf{C}$ are in $\mathcal{M}$ since $\mathcal{M}$ is closed under complementation; hence, so is their union $A=\cup_i A_i^\mathsf{C}$, since $\mathcal{M}$ is closed under countable unions. Finally, then so is $A^{\mathsf{C}} = \cap_{i=1}^\infty A_i$ since $\mathcal{M}$ is closed under complementation.
