As a special phenomenon of high dimension, we know that for arbitary $\epsilon>0$, as $n \rightarrow ∞$. we have:
Where $m$ denotes the Jordan measure and $B_n$ denotes the $n$-dimensional unit ball.
We say that the mass of the ball $B_n$ concentrates in an $O(n^α)$ boundary layer if 
Now I am asked to find a infimum for $\alpha$ to make the mass concentrates in an $O(n^α)$ boundary layer.
That is, find the $\beta$ defined as:
$\beta:=inf\lbrace α < 0 :$ Mass of $B_n$ concentrates in an $O(n^α)$ boundary layer.$\rbrace$.
And once we find the $\beta$, may I ask what can we say about the concentration of mass if (1)$α < β$, (2)$ \alpha=\beta$, (3) $\alpha>\beta$?
I have not exposed to some convoluted high-dimensional geometry context and I find myself lost in this question so far. I searched but I cannot find any material which can lead me to a start point. May I please ask for an answer (together with some explaination) if possible? Or any help or reference would be appreciate. Thanks in advance!
The question boils down to evaluating the limit and see for which values of $\alpha$ does the limit evaluates to 1.
First we have to know how the formula for $m(B_n(1-n^\alpha))$ relates to $m(B_n)$. Since the n-balls of radius $1$ and $(1-n^\alpha)$ and are similar, their volumes are proportional to the $n^{th}$ power of their radii. In particular, $$\frac{m(B_n(1-n^\alpha))}{m(B_n)} = \left(\frac{(1-n^\alpha)}{1}\right)^n $$ (Think about how areas are proportional to the square of the ratio of radii; volumes in 3D are proportional to the cube of the ratio, etc.)
Then $$\frac{m(B_n \backslash B_n(1-n^\alpha))}{m(B_n)} = \frac{m(B_n)-m(B_n(1-n^\alpha)}{m(B_n)} =1-(1-n^\alpha)^n$$
We want to find out for which $\alpha$ this limit tends to 1 as $n \to \infty$, in other words, the questions is for which $\alpha$, $$\lim \limits_{n\to\infty} (1-n^\alpha)^n = 0?$$
Now it's time for some algebra. $$\lim \limits_{n\to\infty} (1-n^\alpha)^n = \lim \limits_{n\to\infty} \left(1-\frac{1}{n^{-\alpha}}\right)^n$$ $$=\lim \limits_{n\to\infty} \left(1-\frac{1}{n^{-\alpha}}\right)^{n^{-\alpha}\cdot n^{\alpha+1}} $$ $$=\lim \limits_{n\to\infty} e^{-n^{\alpha+1}}$$
The rest is left for you to figure out! For what $\alpha$ does the above expression evaluates to 0?
And once you figure out the range of $\alpha$, the infimum $\beta$ is obvious. :)
As for the conditions (1)$\alpha < \beta$, (2)$\alpha = \beta$, and (3) $\alpha > \beta$ go, think about what the limit evaluates to for all of the above cases. What implications does that have for the original ratio that we were interested in, i.e. $\frac{m(B_n \backslash B_n(1-n^\alpha))}{m(B_n)}$? Is that still 1? Or other values? What does that mean geometrically?
Hope that helps! Good luck!