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?


  • 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 $\mathcal{S}$
  • 1
    $\begingroup$ 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? $\endgroup$
    – supinf
    Commented Sep 11, 2019 at 15:39

1 Answer 1


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 $\mathcal{S}\subset \mathcal{D}_1$.

  • $\begingroup$ Apologies for not clarifying notation, but yes, you're right, $d(\mathcal{S})$ is the minimal $d$-system containing $\mathcal{S}$. Thank you! $\endgroup$
    – harisf
    Commented Sep 17, 2019 at 7:52

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .