why integral converges

Question

Let $v$ be a probability measure on $\mathbb{R}$ such that $v$ has no atom. Let $\left(x^{i, N}\right)_{1 \leqslant i \leqslant N}$ be the sequence of real numbers defined by: $$ \begin{aligned} x^{1, N} & =\inf \left\{x \mid v((-\infty, x]) \geqslant \frac{1}{N+1}\right\} \\ x^{(i+1), N} & =\inf \left\{x \geqslant x^{i, N} \mid v\left(\left(x^{i, N}, x\right]\right) \geqslant \frac{1}{N+1}\right\} ,\quad 1 \leqslant i \leqslant (N-1) \end{aligned} $$ We have, $$ -\infty<x^{1, N}<x^{2, N}<\cdots<x^{N, N}<+\infty, $$ How does following follows $$ \varliminf_{N \rightarrow \infty} \left\{ \int_{x^{1, N} \leqslant x<y \leqslant x^{N, N}} \log (y-x) \, dv(x) \, dv(y) \right\} \geqslant \frac{1}{2}\iint \log|y-x| \, dv(x) \, dv(y) $$