Let $K$ be a compact subset of $\mathbb{R}^d$, denote the Lebesgue measure of K by $|K|$. For any $r>0$ and let $\mathcal{U}_r=\lbrace B(x,s)\rbrace_{x\in K,s<r}$. We can always find a countable family $\lbrace B(x_n,s_n)\rbrace_{n=1}^\infty\subset\mathcal{U}_r $, such that $|\bigcup\limits_{n=1}^\infty B(x_n,s_n)|< |K|+r$. By compactness of $K$, there exists $N_{r}\in\mathbb{N}$ such that $K\subset\bigcup\limits_{n=1}^{N_{r,}}B(x_n,s_n)$. Denote $U_r=\bigcup\limits_{n=1}^{N_{r}}B(x_n,s_n)$, clearly $|U_r\setminus K|<r$.
Now let $\tilde{U}_r=\bigcup\limits_{n=1}^{N_{r}}B(x_n,2s_n)$.
Is it true that $\lim\limits_{r\to 0}|\tilde{U}_r\setminus U_r|=0$?