Let $\Omega\subset\mathbb{R}^N$ be a open, unbounded and connected set ($N\ge 2$).
Let $m$ and $\mathcal{H}^{N-1}$ denote respectively, Lebesgue and $(N-1)$-Hausdorff measures. Suppose that $$\lim_{r\to\infty}\frac{m(\Omega\cap B(0,r))}{m(B(0,r))}=0,$$
where $B(0,r)=\{x\in \mathbb{R}^N:\ |x|<r\}$. Is it true that for all sequence $\delta_n\to 0$, $\delta_n>0$, $\delta_n>\delta_{n+1}$, there is a sequence $r_n\to \infty$ such that $$\lim_{n\to\infty}\mathcal{H}^{N-1}(\Omega\cap\partial B(0,r_n))<\delta_n$$
No, that need not be the case. Consider
$$\Omega = \left\{ x\in \mathbb{R}^N : x_1 > 0, \sum_{i=2}^N x_i^2 < 1 \right\}.$$
Then you have
$$\frac{m(\Omega\cap B(0,r))}{m(B(0,r))} \sim \frac{\tau_{N-1}r}{\tau_Nr^N} \xrightarrow{r\to\infty} 0,\tag{1}$$
and
$$\mathcal{H}^{N-1}\left(\Omega \cap \partial B(0,r)\right) \to \tau_{N-1} > 0,\tag{2}$$
where $\tau_k$ is the $k$-dimensional measure of the $k$-dimensional unit ball. (If you normalise $\mathcal{H}^{N-1}$ differently than $\mathcal{H}^{N-1}([0,1]^{N-1}\times\{0\}) = 1$, you get a constant factor in $2$.)