The Lindeberg condition implies Feller negligibility

Fix $\epsilon >0$. For each entry split its square on $\mathrm{\mid }{X}_{n,k}\mathrm{\mid }\le \epsilon$ and its complement. The first integral is at most ${\epsilon }^2\mathbb{P}(\mathrm{\mid }{X}_{n,k}\mathrm{\mid }\le \epsilon )\le {\epsilon }^2$. The second is one nonnegative term of $L_n(\epsilon )$. Centering identifies variance with second moment, hence $0\le {\max }_k\mathrm{Var}(X_{n,k})\le {\epsilon }^2+L_n(\epsilon )$. The maximum exists because the row is finite and nonempty.

For any $\eta >0$, take $0<\epsilon <\sqrt{\eta \mathrm{/}2}$. Lindeberg gives an index after which $L_n(\epsilon )<\eta \mathrm{/}2$. The preceding bound is then less than eta, proving convergence to zero. Values at the cutoff stay in the small part, zero entries satisfy the bound, and a single variance-one summand in every row would contradict this conclusion and hence could not satisfy Lindeberg. The proof uses only given finite rows and explicit inequalities, not independence or choice.