It's a theorem that
Let $(X, M, m)$ be a measure space and let $\{A_k\}$ be a sequence of measurable sets. If $A_1 \supset A_2 \supset\dots$ and $m(A_N)<\infty$ for some $N \in \mathbb{N}$ then $m(\bigcap_{k=1}^{\infty} A_k) = \lim_{k \to \infty} m(A_k)$.
I was wondering if measurablity of $\{A_k\}$ is a must or is there an example of non-measurable $\{A_k\}$ such that $m^{*}(\cap_{k=1}^{\infty} A_k) \ne \lim_{k \to \infty} m^{*}(A_k)$ even if for each $k$, $m^{*}(A_k) < \infty$?
The answer is YES. Here is a simple example.
Let $(\Bbb N, \Sigma, m)$ be a measure space where $\Sigma =\{\emptyset, \Bbb N\}$ and $m$ is a measure defined as $m(\emptyset)=0$ and $m(\Bbb N)=1$.
It is easy to see that, $m^*(\emptyset)=0$ and, for all $E \subseteq \Bbb N$, if $E\neq \emptyset$, then $m^*(E)=1$.
For each $k \in \Bbb N$, let $A_k=\{k, k+1, k+2, \cdots \} \subseteq \Bbb N$.
We have $\bigcap_k A_k = \emptyset$. So, we have $$m^*\left (\bigcap_k A_k \right) = m^*(\emptyset)= 0 $$
and, for each $k \in \Bbb N$, $A_k \neq \emptyset$. So, $m^*(A_k)=1$. So, we have $$m^*\left (\bigcap_{k=1}^{\infty} A_k \right) =0 \ne 1= \lim_{k \to \infty} m^{*}(A_k)$$