Is there a decreasing sequence of sets in $\mathbb{R}^{n}$ with these outer-measure properties?

Question

If $(E_{k})$ is a decreasing sequence of measurable (in the Lebesgue sense) sets in $\mathbb{R}^{n}$ such that the measure $m(E_{k}) < + \infty$ for some $k$, then a suitable telescoping decomposition immediately allows us to write $$ \lim_{k \to \infty}m(E_{k}) = m\big( \bigcap_{k}E_{K} \big). $$

Now, if measurability is not required, is there a decreasing sequence $(E_{k})$ of sets in $\mathbb{R}^{n}$ such that the outer measure $m^{*}(E_{k}) < + \infty$ for all $k$ and $$ \lim_{k \to \infty}m^{*}(E_{k}) > m^{*}\big( \bigcap_{k}E_{k} \big)? $$ The inequality is of prime interest; the opposite inequality is welcome also. Just to see how the equality could break.

Clearly, the result by which this post starts implies that such a sequence cannot be such that each of its elements is measurable. I considered some examples but got at most two conditions satisfied. For example, if $V$ is a Vitali set and if $E_{k} :=B(0, 1/k) \cap V$ for all $k$, then each $E_{k}$ is nonmeasurable and $(E_{k})$ is decreasing with $m^{*}(E_{k}) < + \infty$ for all $k$. But this $(E_{k})$ gives an equality...

Answer 1

Instead of considering one Vitali set, consider all of them: $\{V_q: q ∈ ℚ\}$ and put $E_n := ⋃\{V_{q_k}: k ≥ n\}$ where $ℚ =: \{q_n: n ∈ ℕ\}$.

Answer 2

There exists a family $F=\{F_j:j\in\Bbb N\}$ of subsets of $[0,1]$ with $F_j\cap F_{j'}=\emptyset$ whenever $j\ne j'$ and with the $1$-dimensional Lebesgue outer measure $m_1^*(F_j)=1$ for every $j.$

Let $G_k=\cup_{j\ge k}F_j.$

If $n=1$ let $E_k=G_k.$ If $n>1$ let $E_k=G_k\times [0,1]^{n-1}.$

Then the $n$-dimensional Lebesgue outer measure $m_n^*(E_k)=1$ for every $k$ but $\cap_{k\in\Bbb N}E_k=\emptyset.$