Let $\mathscr{R}$ be a $\sigma$-ring and $\mu$ be a (positive) measure on $\mathscr{R}$. Suppose that $\{A_n\}$ is in $\mathscr{R}$ with $A_1 \supseteq A_2 \supseteq A_3 \supseteq\cdots$ and $A = \bigcap_{n=1}^{\infty} A_n$. Given this information, I need to provide a counter-example for the following equality when $\mu(A_1)=\infty$:
$$ \mu(A)=\lim_{n\to\infty}\mu(A_n). $$
The counter-example I came up with is: $A_n=[n, \infty)$. It's clear that $\mu(A_1)=\infty$. Also, $\mu(A) = 0$ since $\bigcap_{n=1}^{\infty}\ A_n=\varnothing$. To see the latter, assume that $x\in A$ then $x\in A_n$ and thus $x\ge n$, for any $n\in\mathbb{N}$ but this is absurd since there always exists $n\in\mathbb{N}$ such that $1 \lt n\frac{1}{x}$. Moreover, $\mu(A_n)=\mu((n, \infty))=\infty$ for any $n$ thus $\lim_{n\to\infty}\mu(A_n)=\infty$. This leads to the desired counter-example for the equality.
Is anything wrong with this the reasoning above, especially the part where I concluded that $\lim_{n\to\infty}\mu(A_n)=\infty$ from $\mu(A_n)=\mu((n, \infty))=\infty$ for any $n$?