Decreasing sequence on a measure space I have the following problem:
Let $\mu$ be a measure on the measure space $(X,\mathbf{A})$. Let $\{A_{n}\}_{n=1}^{\infty}$ be a decreasing sequence in $\mathbf{A}$ so that $A_{n} \in \mathbf{A}$ for all $n\geq1$ and $A_{1} \supseteq A_{2} \supseteq A_{3}...,$
Let $A = \cap_{n=1}^{\infty}A_{n}$
(a) Assume that $\mu(A_{n})\leq \frac{1}{n}$ for all $n\geq 1$. Is it true that $\mu(A)=0$. (Provide a proof or counter example.)
I don't know how to approach such a problem and I have more of them so I would be really thankful for a general way to solve these types of problems.
Anyone with some knowledge of this?
More generally we have a theorem that states the following:
Let $\mathcal{A}$ be an algebra and let $\mu:\mathcal{A} \to [0,\infty]$ be a set function for which $\mu(\emptyset)=0$. If we have a decreasing sequence of sets $A_n\subseteq A_{n-1}$ in $\mathcal{A}$ such that $\mu(A_n)&lt;\infty$ for some $n\in \mathbb{N}$, then:
$$\mu\left(\bigcap_{n=1}^{\infty} A_n\right)=\lim_{n\to \infty}\mu(A_n)$$
We say that the measure(or set function more generally) is continuous from above in the set $\bigcap_{n=1}^{\infty} A_n$. If you are not familiar with algebras or more general measure theory then just know that every $\sigma$-algebra is an algebra and since it is given that $\mu$ is a measure it will satisfy the conditions of this theorem.
Hint
$\mu(A) \leq \mu(A_n)$ for every $n\in \mathbb{N}.$