# Prove that the intersection of two $\sigma$-algebra is a $\sigma$-algebra

My proof goes:

Let {$$A_1 , A_2$$} be a family of two $$\sigma$$ -algebras and let $$A \in A_1 \cap A_2$$ Which means $$A \in A_1$$ and $$A \in A_2$$ and implies $$A^c \in A_1$$ and $$A^c \in A_2$$ so $$A^c \in A_1 \cap A_2$$

Now let $$A_j \in A_1 \cap A_2$$ for $$j\in J$$ Then $$A_j \in A_1 , A_j \in A_2$$ $$\forall j$$ Therefore $$\cup A_j \in A_1$$ and $$\cup A_j \in A_2$$ Hence $$\cup A_j \in A_1 \cap A_2$$ and intersection of two $$\sigma$$-algebras is $$\sigma$$-algebra

Would this be correct? Is there another way of proving this?

• It seems correct and most probably the only way to prove. Commented Apr 14, 2021 at 4:36
• Notice that you can use the same argument to prove that the intersection of an arbitrary number of $\sigma$-algebras is still a $\sigma$-algebra Commented Apr 14, 2021 at 5:26
• @gonzalo benavides: another way would be to use the identity $\sigma\left(\bigcap_{i\in I}\mathcal A_i\right) = \bigcap_{i\in I}\sigma(\mathcal A_i)$ Commented Apr 14, 2021 at 9:04
• @lmaosome But that identity is actually a consequence of the countable version of what is asked here Commented Apr 15, 2021 at 15:23

Don't forget to mention that $$|J| \leq |\mathbb N|$$, otherwise this seems to be a correct proof.