A Lindeberg array with no identically distributed row

Each $0<p_{n,k}<1\mathrm{/}2$ defines a two-point probability. Index the pairs by $n(n-1)\mathrm{/}2+k$ and use [F1]–[F3] to construct all coordinates independently. By [F4], the centered entry has mean zero and variance $p_{n,k}(1-p_{n,k})$. Its values are $-p_{n,k}$ and $1-p_{n,k}$, both of absolute value less than one. Distinct p have distinct negative support points with positive mass, hence distinct centered laws. Normalizing every entry in a fixed row by the same positive s_n also preserves this distinction.

Since $1-p_{n,k}>1\mathrm{/}2$, $s_n^2\ge \frac{1}{2}{\sum }_{k=1}^{n}k\mathrm{/}(2n+2)=n\mathrm{/}8$. Here ${\sum }_{k=1}^{n}k=n(n+1)\mathrm{/}2$, obtained by pairing k with n+1-k and adding the n equal pair sums. Thus s_n is positive and tends to infinity. For any fixed epsilon>0, eventually $\epsilon s_n>1$, so every event $\mathrm{\mid }{X}_{n,k}\mathrm{\mid }>\epsilon s_n$ is empty. The Lindeberg sum is then exactly zero. All second moments are finite, so [F5] gives the claimed limit. AC is used only through the stated product construction and CLT suppliers; no cross-row independence is needed by the theorem.