Show that $\mathcal A_1$ $\cap$ $\mathcal A_2$ is also a $\sigma$-algebra

Question

I am currently studying for a measure theory final and have come across a past short exam question that reads "Let $\mathcal A_1$ and $\mathcal A_2$ be a $\sigma$-algebra of subsets of a set X. Show that $\mathcal A_1$ $\cap$ $\mathcal A_2$ is also a $\sigma$-algebra." Conceptually I cannot quite grasp this. Since $\mathcal A_1$ and $\mathcal A_2$ both have the set X as an element by the definition of a $\sigma$-algebra, how can we know that the intersection of $\mathcal A_1$ and $\mathcal A_2$ also has X as an element? A proof of this would be really great or even just a push in the right direction. Thanks in advance.

Answer 1

As mentioned by @Reveillark, you shall need the definition of intersection of sets.

Based on it, we can proceed as follows.

Let $\Omega$ be a nonempty set and $\mathcal{A}_{1}$ and $\mathcal{A}_{2}$ be $\sigma$-algebras on $\Omega$.

Then clearly $\Omega\in\mathcal{A}_{1}\cap\mathcal{A}_{2}$, because $\Omega\in\mathcal{A}_{i}$ for $i = 1,2$.

Now let us assume that $A\in\mathcal{A}_{1}\cap\mathcal{A}_{2}$. Thus $A\in\mathcal{A}_{1}$ and $A\in\mathcal{A}_{2}$. But $\mathcal{A}_{i}$ are $\sigma$-algebras.

Then $A^{c}\in\mathcal{A}_{1}$ and $A^{c}\in\mathcal{A}_{2}$. Consequently, $A^{c}\in\mathcal{A}_{1}\cap\mathcal{A}_{2}$.

At last but not least, let us suppose that $A_{n}\in\mathcal{A}_{1}\cap\mathcal{A}_{2}$ for $n\in\mathbb{N}$. Since $\mathcal{A}_{i}$ are $\sigma$-algebras, one has that \begin{align*} \left(\bigcup_{n\in\mathbb{N}}A_{n}\in\mathcal{A}_{1}\right)\wedge\left(\bigcup_{n\in\mathbb{N}}A_{n}\in\mathcal{A}_{2}\right) \Rightarrow \bigcup_{n\in\mathbb{N}}A_{n}\in\mathcal{A}_{1}\cap\mathcal{A}_{2} \end{align*}

and we are done.

Hopefully this helps.