Feller negligibility cannot be removed from the converse

On the standard-normal probability space supplied by [F7], let Z be its coordinate and define $X_{n,1}=Z$, $X_{n,k}=0$ for $2\le k\le n$. By [F1], every entry is centered and the total variance is one. Each row is independent: if an event for a zero coordinate excludes zero, both the intersection probability and the product are zero; otherwise all those events are the full space and the identity reduces to the event for Z. Thus the row sum is exactly Z for every n, its law is N(0,1), and the maximum summand variance equals one. Independence between rows is not required.

For each fixed epsilon>0, [F2] gives the Lindeberg quantity $\mathbb{E}[Z^2{\mathbf{1}}_{\{\mathrm{\mid }Z\mathrm{\mid }>\epsilon \}}]$, independent of n. On $[\epsilon +1,\epsilon +2]$, the integrand against the normal density is at least $(\epsilon +1{)}^2{e}^{-(\epsilon +2{)}^2\mathrm{/}2}\mathrm{/}\sqrt{2\pi }>0$, by [F3] and [F5]. Integrating this bound over the length-one interval with [F4]–[F6] proves strict positivity. Hence the Lindeberg limit is not zero for any epsilon>0, despite exact normality of all row sums. The example works for n=1 as well. AC is inherited solely through construction of the normal law and its Lebesgue density; one Z is reused and no independent sequence across rows is constructed.