The Minkowski content is defined as
$\displaystyle M_{\beta}(A)=\lim_{\delta \rightarrow 0} \frac{\mu(A_{\delta})}{(2\delta)^{1-\beta}}$
where $0 < \beta < 1$, $A \subset \mathbb{R}$, and $A_{\delta}$ is the $\delta$ - neighbourhood around $A$.
Evaluate $M_{1/(1+\alpha)}(F_{\alpha})$ where
$F_{\alpha}=\{0,1,\frac{1}{2^{\alpha}},\frac{1}{3^{\alpha}},\frac{1}{4^{\alpha}},\dots\}$.
I am struggling to see how this can be answered.
If $(\delta_k)$ is a sequence of positive numbers tending to zero, then $$ M_{1/(1+\alpha)}(F_\alpha)=\lim_{k\to\infty}(2\delta_k)^{-\alpha/(1+\alpha)}\mu(F_{\alpha,\delta_k}). $$ Choose $2\delta_k=k^{-\alpha}-(k+1)^{-\alpha}$. Then $$ F_{\alpha,\delta_k} = (-\delta_k,(k+1)^{-\alpha}+\delta_k)\cup\bigcup_{i=1}^k(i^{-\alpha}-\delta_k,i^{-\alpha}+\delta_k). $$ Note that these open intervals are disjoint; we have chosen $\delta_k$ so that $(k+1)^{-\alpha}+\delta_k=k^{-\alpha}-\delta_k$. Now it is easy to evaluate the Lebesgue measure of $F_{\alpha,\delta_k}$: $$ \mu(F_{\alpha,\delta_k}) = [(k+1)^{-\alpha}+2\delta_k]+k\times2\delta_k = (k+1)^{-\alpha}+(k+1)(k^{-\alpha}-(k+1)^{-\alpha}). $$ Therefore $$ M_{1/(1+\alpha)}(F_\alpha) = \lim_{k\to\infty}[k^{-\alpha}-(k+1)^{-\alpha}]^{-\alpha/(1+\alpha)}[(k+1)^{-\alpha}+(k+1)(k^{-\alpha}-(k+1)^{-\alpha})]. $$ The problem has thus reduced to evaluating a limit of an explicit sequence. Note that \begin{eqnarray} k^{-\alpha}-(k+1)^{-\alpha} &=& k^{-\alpha}(1-(1+1/k)^{-\alpha}) \\&=& k^{-\alpha}(\alpha k^{-1}+O(k^{-2})) \end{eqnarray} and \begin{eqnarray} (k+1)^{-\alpha} &=& k^{-\alpha}(1+1/k)^{-\alpha} \\&=& k^{-\alpha}(1-\alpha k^{-1}+O(k^{-2})). \end{eqnarray} These give \begin{eqnarray} (k+1)^{-\alpha}+(k+1)(k^{-\alpha}-(k+1)^{-\alpha}) &=& k^{-\alpha}(1-\alpha k^{-1}+O(k^{-2})) \\&& +(k+1)k^{-\alpha}(\alpha k^{-1}+O(k^{-2})) \\&=& k^{-\alpha}(1+\alpha+O(k^{-1})). \end{eqnarray} Also \begin{eqnarray} [k^{-\alpha}-(k+1)^{-\alpha}]^{-\alpha/(1+\alpha)} &=& [k^{-\alpha}(\alpha k^{-1}+O(k^{-2}))]^{-\alpha/(1+\alpha)} \\&=& \alpha^{-\alpha/(1+\alpha)}k^\alpha(1+O(k^{-1})). \end{eqnarray} Combining these estimates we get \begin{eqnarray} M_{1/(1+\alpha)}(F_\alpha) &=& \lim_{k\to\infty}[\alpha^{-\alpha/(1+\alpha)}k^\alpha(1+O(k^{-1}))][k^{-\alpha}(1+\alpha+O(k^{-1}))] \\&=& \lim_{k\to\infty}\alpha^{-\alpha/(1+\alpha)}(1+\alpha)(1+O(k^{-1})) \\&=& \alpha^{-\alpha/(1+\alpha)}(1+\alpha) . \end{eqnarray} It is of course possible that I have made a mistake in the calculations somewhere. If you find a mistake, let me know.