# Prove that algebra is sigma-algebra

Let's assume that $$\mathcal{A}$$ is an algebra over $$\mathit{X}$$, and $$\mathcal{A}$$ is closed under countable intersections of descending set sequences; if $$A_1 \supseteq A_2 \supseteq A_3 \supseteq ...$$ and $$A_i \in \mathcal{A}, i \in \mathbb{N}$$, then $$\bigcap\limits^{\infty}_{i=1}A_i \in \mathcal{A}$$. Prove that $$\mathcal{A}$$ is $$\sigma$$-algebra on $$X$$.

My attempt: Let $$\forall n \in \mathbb{N} \; B_n \in \mathcal{A}$$. Then define $$A_n = B_n \setminus \bigcup\limits_{i=1}^{n-1} B_i$$. Then $$A_n$$ is descending family of sets, $$A_n \in \mathcal{A}$$ as this is algebra. We know that $$\bigcap\limits^{\infty}_{i=1}A_i \in \mathcal{A} \leftrightarrow \bigcap\limits^{\infty}_{i=1}(B_n \setminus \bigcup\limits_{i=1}^{n-1} B_i) \in \mathcal{A}$$. Now, I was thinking about using De Morgan's laws, etc., but I can't obtain the result and I don't know if I haven't messed up in the beginning.

You want the unions of countable increasing chains of sets to be in the algebra, and you have decreasing intersections, which in a way, the complement. So given such an increasing chain, think of the chain of complements, show why it is decreasing, etc, consider the obtained intersection, and then look at the complement of the intersection.

In rigorous terms: if $${A_i}$$ is the increasing chain, consider $$A_i^c$$ which is a decreasing chain, in the algebra, and then $$\bigcap A_i^c$$ is in the algebra. Finish by noting that $$(\bigcap A_i^c)^c=\bigcup A_i$$

• Turn out I was overthinking this :) Thank you! – Nerwena Jan 23 at 18:17

You need to show $$\mathcal{A}$$ is closed under countable unions. We can first show it is closed under countable intersections, and if $$A_n, n \in \omega$$ are in $$\mathcal{A}$$ we can write $$\bigcap_n A_n = \bigcap_n B_n$$ where $$B_n = \bigcap_{i=0}^n A_i$$ forms a decreasing sequence of subsets of $$\mathcal{A}$$, and so $$\bigcap_n A_n$$ is in $$\mathcal{A}$$.

Now to get unions we apply de Morgan:

$$\bigcup_n A_n = \left( \bigcap_n A_n^\complement \right)^\complement$$

and we already know the algebra is closed under countable intersections...