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I recently learned about algebras, $\sigma$-algebras, and monotone classes. The algebras and $\sigma$-algebras I feel are quite clear, but I think I may be a bit confused about certain aspects of monotone classes. The following is the definition of monotone classes given:

A non-empty family $\Phi \subset \mathcal{P}(X)$ which is closed under countable increasing unions and countable decreasing intersections is called a monotone class in X, i.e. $\Phi$ is a monotone class if

  • $E_1\subset E_2\subset ...$ where each $E_j\in\Phi$ implies that $\bigcup_{j=1}^{\infty} E_j \in \Phi$;

  • $D_1\supset D_2\supset ...$ where each $D_j\in\Phi$ implies that $\bigcap_{j=1}^{\infty} D_j \in \Phi$.

Now, suppose that $X=\left\{1,2,3\right\}$, and let $\Phi_1=\left\{\left\{1\right\},\left\{1,2\right\}\right\}$, $\Phi_2=\left\{\emptyset, X\right\}$. Then $\Phi = \Phi_1 \cap \Phi_2=\emptyset$, so $\Phi$ is empty, and therefore, not a monotone class.

But this confuses me a bit, as I don't see clearly from this definition why either of $\Phi_1$ and $\Phi_2$ shouldn't be a monotone class.

My guess is that $\Phi_1$ is not a monotone class and $\Phi_2$ is a monotone class, since $X\in\Phi_2$, and I've read that $X$ is always an element of any monotone class in $X$. But I don't see why that is. What am I not seeing in the definition?

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Both $\Phi_1$ and $\Phi_2$ are monotone classes.

Pick an infinite sequence $A_n$ of elements of $\Phi_1$. Then for every $n$ either $A_n=\{1\}$ or $A_n = \{1,2\}$, so the infinite intersection and union is going to be either $\{1,2\}$ or $\{1\}$, both of which belong to $Phi_1$.

A similar argument can be made regarding $\Phi_2$.

Thus we see that it is not necessary to include $X$ in order to be a monotone class.

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  • $\begingroup$ Ok, but then this should not be possible to prove, if I'm correct: "If $\left\{\Phi_i\right\}_{i\in I}$ is a collection of monotone classes, then $\Phi=\bigcap_{i\in I} \Phi_i$ is a monotone class as well." This bothers me, since I have an exercise asking me to prove that statement, and the example I gave in my question is a direct counterexample. Is the definition of monotone classes given to me wrong? $\endgroup$ – Scounged Sep 18 '16 at 12:31
  • $\begingroup$ @Scounged You are right as long as you require non-emptiness in the definition. $\endgroup$ – Jsevillamol Sep 18 '16 at 15:55

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