# Monotone class theorem - proof

Reading a proof of the monotone class theorem, I'm stuck at a particular step.

First, here's the theorem as stated in Karr, Alan F., Probability. (1993)

Let $$\mathcal{S}$$ be a $$\pi$$-system. Then $$\sigma(\mathcal{S}) = d(\mathcal{S}).$$

The part I'm stuck at in the proof relates to showing that the generated $$d$$-system, $$d(\mathcal{S})$$, is also a $$\pi$$-system. A step in that process concerns the set $$\mathcal{D}_1 = \{ B \in d(\mathcal{S}) : B \cap C \in d(\mathcal{S}) \; \text{for all} \; C \in \mathcal{S} \}.$$

The particular part I'm having trouble understanding is the following statement: $$\mathcal{S} \subseteq \mathcal{D}_1.$$

Why is this true?

Edit:

• a $$\pi$$-system is closed under finite intersections
• a $$d$$-system is closed under proper differences and countable increasing unions
• $$\sigma(\mathcal{S})$$ and $$d(\mathcal{S})$$ are the unique minimal $$\sigma$$-algebra and $$d$$-system, respectively, containing the family of subsets $$\mathcal{S}$$
• This question could be improved by explaining some of the mathematical notions. For example, what is a $d$-system, or how is $d(\mathcal S)$ defined? – supinf Sep 11 '19 at 15:39

I gather that $$d(\mathcal{S})$$ is the minimal $$d$$-system containing $$\mathcal{S}$$. Notice that if $$C\in\mathcal{S}$$ and $$B\in\mathcal{S}$$, then $$B\cap C\in\mathcal{S}$$ because $$\mathcal{S}$$ is a $$\pi$$-system. Since $$\mathcal{S}\subset d(\mathcal{S})$$, and $$B\in\mathcal{S}$$ implies $$B\cap C\in\mathcal{S}$$ whenever $$C\in\mathcal{S}$$, it follows that $$\mathcal{S}\subset \mathcal{D}_1$$.
• Apologies for not clarifying notation, but yes, you're right, $d(\mathcal{S})$ is the minimal $d$-system containing $\mathcal{S}$. Thank you! – harisf Sep 17 '19 at 7:52