Show that if $\sum_{n=1}^\infty\mu(A_n) <\infty$ we have that $\mu(B) = 0$.

Question

Exercise: Let $(S,\mathcal{A},\mu)$ be a measurable space and let $A_1,A_2,\ldots\in \mathcal{A}$. Define $B\subseteq S$ as $B = \bigcap\limits_{k = 1}^\infty\bigcup\limits_{n = k}^\infty A_n$. Show that $B\in\mathcal{A}$ and show that if $\sum_{n = 1}^\infty\mu(A_n)<\infty$ we have that $\mu(B) = 0$.

What I've tried: I first tried to show that $B\in\mathcal{A}$. I know that $\bigcup_{n =k}^\infty A_n\in\mathcal{A}$ for any $k\in\mathbb{N}$. Unfortunately, I cannot conclude that the infinite intersection of sets that are in $\mathcal{A}$ is in $\mathcal{A}$ as well. Though I thus far haven't been able to prove that $B\in\mathcal{A}$, it makes a lot of sense. $B = (A_1\cup A_2\cup\ldots)\cap (A_2\cup A_3\cup\ldots)\cap \ldots$ so I know that $B\subseteq \bigcup_{n =1}^\infty A_n$. I feel this is the direction I need to be looking, but I'm kind of stuck at this point.

To show that if $\sum_{n = 1}^\infty \mu(A_n) <\infty$ we have that $\mu(B) = 0$ I tried to use the fact that $B_n = \bigcup_{j = n}^\infty A_j$ is a decreasing sequence. Since $B_n$ is a decreasing sequence and $B = \bigcap_{n = 1}^\infty B_n$, we have that if $\mu(B_1) <\infty$, then $\mu(B_n)\downarrow \mu(B)$. Since we have $\mu(B_1) = \sum_{n = 1}^\infty \mu(A_n) <\infty$, $\mu(B_n)\to 0$ implies $\mu(B) = 0$, which is what we want to show. However, I'm not sure how to show that $\mu(B_n)\to 0$. It again makes a lot of sense, but I'm not quite sure how I should prove it.

Question: How do I solve this exercise?

Thanks!

Answer 1

\begin{align*} \mu(B_{n})&\leq\sum_{j=n}^{\infty}\mu(A_{j})=\sum_{j=1}^{\infty}\mu(A_{j})-\sum_{j=1}^{n-1}\mu(A_{j})\rightarrow 0. \end{align*}

Answer 2

If $D_n$ are measurable for $n=1,2,\dots$ then also $D_n^{\complement}$ is measurable for $n=1,2,\dots$, since $\sigma$-algebras are closed under complementation.

Then for $D=\bigcap_{n=1}^{\infty}D_n$ we find that $D^{\complement}=\bigcup_{n=1}^{\infty} D_n^{\complement}$ is measurable since $\sigma$-algebras are closed under the formation of countable unions.

Then $D=\left(D^{\complement}\right)^{\complement}$ is measurable, again because $\sigma$-algebras are closed under complementation.

Proved is now that $\sigma$-algebras are closed under the formation of countable intersections.

Further see the answer of user284331.