Let $\Omega$ be a non empty set, and $\mathcal{A}$ be an algebra of sets of $\Omega$. Then, the Monotone Class Theorem asserts that the generated $\sigma$-algebra $\mathcal{F}(\mathcal{A})$ coincides with the generated monotone class $\mathcal{M}(\mathcal{A})$.
I was trying to understand limit cases of this theorem. Lets say that the original class $\mathcal{A}$ does not need to be an algebra. Then, it is easy to cook up an example where $\mathcal{A}$ itself is a monotone class that is not a sigma algebra. For instance, $\Omega = \{0,1\}$ and $\mathcal{A} = \{\emptyset, \Omega, \{1\}\}$.
Then, I tried to understand if the Monotone Class Theorem is still true in the case where $\mathcal{A} = \mathcal{S}$ is a semi-algebra of sets. It is easy to see that $\mathcal{F}(\mathcal{S}) = \mathcal{F}(\mathcal{A}(\mathcal{S}))$. But then, the first problem appears.
- Is it true that $\mathcal{M}(\mathcal{S}) = \mathcal{M}(\mathcal{A}(\mathcal{S}))$? Of course, one of the implications is immediate, it only remains to check that $\mathcal{M}(\mathcal{S}) \supset \mathcal{M}(\mathcal{A}(\mathcal{S}))$.
To show this, I tried to argue that $\mathcal{M}(\mathcal{S})$ is a monotone class containing the algebra generated $\mathcal{A}(\mathcal{S})$. However, if $B\in\mathcal{A}(\mathcal{S})$, then $B$ is written as the disjoint union:
\begin{equation} B = \bigcup_{j=1}^n A_j
\end{equation}
for $A_j \in \mathcal{S}$, $A_j\cap A_k = \emptyset$ if $j\neq k$.
- However, having the example of the sigma algebra of intervals in $\mathbb{R}$ in mind, it seems unreasonable that $B$ could be written as the increaasing or decreasing union of intervals in a way that we could conclude that $B\in\mathcal{M}(\mathcal{S})$. This lead me to believe that this could be a counter example to the Monotone Class Theorem for semi algebras. I believe that this could be a counter example: the monotone class generated by the algebra of intervals is the Borel sigma algebra, however, I believe that the monotone class generated by the semi algebra of intervals is strictly smaller then this, and it would be simply the class of intervals and singletons, but I was not able to prove this.
Any help is appreciated. This is not homework.
Remark: I posted this as a comment but I'll add here. If $\Omega = \{0,1,2\}$, consider $\mathcal{M} = \{\emptyset, \Omega, \{0\}, \{1\}, \{2\}\}$. Then $\mathcal{M}$ is a monotone semi algebra. In particular, its generated monotone class is itself, but it is strictly contained in the generated $\sigma$-algebra.
Best Answer
Let $\mathcal{S}= \{ I \subseteq \mathbb{R} : I \textrm{ is an interval}\} $ (we consider the empty set, the singletons and $\mathbb{R}$ as intervals). It is easy to prove that $\mathcal{S}$ is a monotone semi-algebra. On the other hand, the $\sigma$-algebra generated by $\mathcal{S}$ is the Borel $\sigma$-algebra in $\mathbb{R}$.
In other other words, let $\mathcal{B}$ be the Borel $\sigma$-algebra in $\mathbb{R}$. Then we have: $$\mathcal{M}(\mathcal{S}) = \mathcal{S} \subsetneq \mathcal{B} = \mathcal{F}(\mathcal{S}) = \mathcal{M}(\mathcal{A}(\mathcal{S}))$$
(So the answer to the question "Is it true that $\mathcal{M}(\mathcal{S}) = \mathcal{M}(\mathcal{A}(\mathcal{S}))$?" is "No").
Remark: To prove that $\mathcal{S}$ is a monotone semi-algebra, first prove that $\mathcal{S}$ is a semi-algebra and then, it is enough to remark that any increasing union of intervals and any decreasing intersection of intervals are also intervals (as we consider the empty set, the singletons and $\mathbb{R}$ as intervals).