In Rudin's Real and Complex Analysis, section 7.9, the definition of nicely shrinking sets is given as follows:
Let $x\in \Bbb R^k$. A sequence $\{E_n\}$ of Borel sets in $\Bbb R^k$ is said to shrink to $x$ nicely if there is a number $\alpha>0$ with the following property: There is a sequence of balls $B(x,r_n)$, with $\lim r_n=0$, such that $E_n\subset B(x,r_n)$ and $$ m(E_n) \geq \alpha \cdot m(B(x,r_n))$$ for $n=1,2,\cdots$. (Here $m$ is the Lebesgue measure)
Then Rudin says that
(1) A nested sequence of $k$-cells whose longest edge is at most 1,000 times as long as its shortest edge and whose diameter tends to $0$ shrinks nicely.
(2) But a nested sequence of rectangles (in $\Bbb R^2$) whose edges have lengths $1/n$ and $1/n^2$ does not shrink nicely.
But I can't see why. I think the biggest difference is there is a upper limit of the ratio between the edges in (1), while not in (2), but how can I prove these using the definitions?
Based on @Toffomat's comment: If $B(x,r)$ is any ball containing $E_n$ of case 2, then $$ 2r = \text{diam} (B(x,r)) \geq \text{diam} (E_n) \geq 1/n \, . $$ Thus, for the intended inequality to hold we must have $$ m(E_n) =n^{-3} \geq \alpha \, m(B(x,r)) \geq \alpha \, \pi \, (2n^{-1})^2 \, , $$ which cannot hold for all $n$ if $\alpha > 0.$
In case 1, a ball of diameter $\sqrt{1+(k-1)1000^2}$ times the shortest edge does contain the square, and its volume is comparable with a fixed ratio to that of the square. (That number for the diameter comes from Pythagorian theorem applied to any two points within our square.)