Does there exist a decreasing sequence of sets ($A_{n+1}\subseteq A_n$) ; $\{A_n\}$ of real numbers such that
$m^*(A_n)<\infty , \forall n \in \mathbb N$ and $m^*(\cap_{n\in \mathbb N} A_n) < \lim_{n\to \infty} m^*(A_n)$ ? Here $m^*$ is the Lebesgue outer measure on real line
Let $V$ be the Vitali set which is formed by chooseing a representative within $[0, 1]$ for each coset of $\Bbb{R}/\Bbb{Q}$. Also, let us enumerate $\Bbb{Q} \cap [-1, 1] = \{q_1, q_2, \cdots \}$ and define $\{V_n\}$ by
$$ V_n = q_n + V = \{ v + q_n : v \in V\}. $$
Here are some observations:
Since the Lebesgue outer measure is translation-invariant, $\{V_n\}$ has the common outer measure $m^*(V_n) = m^*(V)$.
For each $x \in [0, 1]$ there is a unique $v \in V$ such that $x - v \in \Bbb{Q}$. Since $x - v \in [-1, 1]$, this tells that $x - v = q_n$ for a unique $n$ and hence $x \in V_n$.
From the previous observation, we have
$$ [0, 1] \subseteq \cup_{n=1}^{\infty} V_n \subseteq [-1, 2]. $$
If $V$ were measurable, we would have had $1 \leq \infty \cdot m(V) \leq 3$, which is impossible. Therefore $V$ is non-measurable. Also notice that this implies $m^*(V) > 0$, for otherwise $V$ is a null-set and hence measurable.
Finally, define $\{ A_n \}$ by
$$ A_n = \cup_{k=n}^{\infty} V_k. $$
Then $\{ A_n \}$ decreases to $\varnothing$, and $V_n \subset A_n$ implies that $m^*(A_n) \geq m^*(V_n) = m^*(V)$. Therefore
$$ m^*(\varnothing) = 0 < m^*(V) \leq \lim_{n\to\infty} m^*(A_n). $$
An alternative approach:
Let $B$ be a basis of $\mathbb{R}$ over $\mathbb{Q}$ such that $1\in B$. Define $G=\mathrm{span}_{\mathbb{Q}}(B \backslash \{1\})$. Then $\{G+q | q\in\mathbb{Q}\}$ forms countably many cosets of $G$ partitioning $\mathbb{R}$. Let $$ A_n=[0,1]\cap\left( \cup \{ G+q | q\in\mathbb{Q}, \ q>n \}\right). $$ Then $$ \cap_{n=1}^{\infty} A_n = \emptyset \ \mathrm{and} \ \ m^*(A_n)=1 \ \ \mathrm{for} \ \mathrm{all} \ n\in\mathbb{N}.$$ The last statement follows from the answer to your other question: $G$ be a non-measurable subgroup of $(\mathbb R,+)$ ; $I$ be a bounded interval , then $m^*(G \cap I)=m^*(I)$?
In fact, $m^*([0,1]\cap (G+q)) = m^*([0,1]\cap G)= 1$ for all $q\in\mathbb{Q}$.