Show that for any sequence $x_k$ in $[0,1]$ the series: $$S(x)=\sum_{k=1}^\infty\frac{1}{k^3|x-x_k|^2}$$ converges a.s. w.r.t Lebesgue measure on $[0,1].$
Iv'e tried to take intervals of size $k^{-\alpha}$ around $x_k$ and use Borel Canteli but no luck so far. To get convergence I need $\alpha < 1$ and to use Borel Canteli I need $\alpha > 1$.
Define $f_k(x) = \dfrac{1}{k^3|x-x_k|^2}.$
For $c>0,$ let $r_k = ck^{-3/2}, k=1,2,\dots$ Let $I_k$ be the interval $(x_k-r_k,x_k+r_k).$ Set $U=U_c= \cup_k I_k.$ Then
$$\tag 1 \int_{[0,1]\setminus U} (\sum_k f_k) = \sum_k \int_{[0,1]\setminus U}f_k \le \sum_k \int_{[0,1]\setminus I_k}f_k.$$
We can be free and easy in estimating $\int_{[0,1]\setminus I_k}f_k:$
$$\int_{[0,1]\setminus I_k}f_k \le 2k^{-3}\int_{r_k}^1 \frac{dt}{t^2}\,dt$$ $$ = 2k^{-3}(1/r_k -1) < 2k^{-3}/r_k= 2k^{-3/2}/c.$$
Since $\sum_k 2k^{-3/2}/c<\infty,$ we see the left side of $(1)$ is finite. This implies $\sum_k f_k <\infty$ a.e. on $[0,1]\setminus U.$ Now recall $U$ depends on $c.$ We can let $c\to 0^+$ to see $\sum_k f_k <\infty$ a.e. on $[0,1]$ as desired.