Let $X$ be a set, $\mathcal{C} \subseteq X$ and $\mathcal{A}$ a $\sigma$-algebra on $X$. Is it true or false that $\sigma(\mathcal{C}) \cap \mathcal{A} = \sigma(\mathcal{C} \cap \mathcal{A})$?

Of course, $\supseteq$ is true.

I found a trivial counter-example for the other direction: Choose $\mathcal{C} = \{ A, B \}$ and $\mathcal{A} = \sigma(A \cap B)$, where $A \cap B \neq \emptyset$. Then $\mathcal{C} \cap \mathcal{A} = \emptyset$, $\sigma( \mathcal{C} \cap \mathcal{A} ) = \left\{ \emptyset , X \right\}$ but $\sigma(\mathcal{C}) \cap \mathcal{A} = \mathcal{A}$.

The problem with the counter-example is that the generator $\mathcal{C} \cap \mathcal{A}$ is an empty set. If we forbid an empty set as a generator, is the statement then true? Or more generally, under what typical circumstances does the equality of the $\sigma$-algebras hold?

I need to check, whether the following statement is true: Let $\mathcal{E} \subseteq \mathcal{C}$ be two generators. Then $\mathcal{C} \cap \mathcal{A} \subseteq \mathcal{E} \Rightarrow \sigma(\mathcal{C}) \cap \mathcal{A} \subseteq \sigma(\mathcal{E})$. I tried to prove with a monotone class argument, but had then problems to show the union property of the Dynkin class $\left\{ E \subseteq X \ | \ E \in \mathcal{A} \Leftrightarrow E \in \sigma(\mathcal{E}) \right\}$. The first statement above is just necessary in order to prove the second one.

  • $\begingroup$ Surely $A \cap B \in \sigma(A,B)$? (I presume by $\sigma(A,B)$ you mean the smallest sigma algebra containing $A$ and $B$). $\endgroup$ – copper.hat Sep 26 '14 at 6:22
  • $\begingroup$ The generated sigma algebra contains all countable and therefore finite intersections of the generating sets. $\endgroup$ – yadaddy Sep 26 '14 at 7:05
  • $\begingroup$ In your example, $\mathcal C\subseteq\mathcal A$. $\endgroup$ – Did Sep 26 '14 at 7:32
  • $\begingroup$ Oh, I interchanged the generators. It must be $\mathcal{C} = \{A, B\}$ and $\mathcal{A} = \sigma(\{A \cap B\})$. $\endgroup$ – yadaddy Sep 26 '14 at 8:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.