Lindeberg-Feller central limit theorem: sufficiency

Put $Y_{n,k}=X_{n,k}\mathrm{/}{s}_n$ and $v_{n,k}=\mathbb{E}{Y}_{n,k}^2$. By [F1], the normalized row has total variance one, remains centered and independent, and its tail sum $L_n(\epsilon )$ tends to zero. Fix real t and write $w_{n,k}={\phi }_{Y_{n,k}}(t)-1$. The centering and scalar bound in [F2] give $\mathrm{\mid }{w}_{n,k}\mathrm{\mid }=\mathrm{\mid }\mathbb{E}(e^{it{Y}_{n,k}}-1-it{Y}_{n,k})\mathrm{\mid }\le t^2{v}_{n,k}$, using [F8]–[F9]. Consequently ${\sum }_k\mathrm{\mid }{w}_{n,k}\mathrm{\mid }\le t^2$, ${\max }_k\mathrm{\mid }{w}_{n,k}\mathrm{\mid }\le t^2{\max }_k{v}_{n,k}\to 0$, and ${\sum }_k\mathrm{\mid }{w}_{n,k}{\mathrm{\mid }}^2\le t^4{\max }_k{v}_{n,k}\to 0$ by [F3].

With r as in [F2], linearity gives ${\sum }_k{w}_{n,k}=-t^2\mathrm{/}2+{\sum }_k\mathbb{E}r(t{Y}_{n,k})$. On $\mathrm{\mid }{Y}_{n,k}\mathrm{\mid }\le \epsilon$ the cubic bound gives $\mathrm{\mid }r(t{Y}_{n,k})\mathrm{\mid }\le \mathrm{\mid }t{\mathrm{\mid }}^3\epsilon Y_{n,k}^2\mathrm{/}3$; on the complement the quadratic bound gives $\mathrm{\mid }r(t{Y}_{n,k})\mathrm{\mid }\le 4t^2{Y}_{n,k}^2$. Therefore $\mathrm{\mid }{\sum }_k\mathbb{E}r(t{Y}_{n,k})\mathrm{\mid }\le \mathrm{\mid }t{\mathrm{\mid }}^3\epsilon \mathrm{/}3+4t^2{L}_n(\epsilon )$. First take limsup in n, then let the arbitrary positive epsilon decrease to zero. This proves ${\sum }_k{w}_{n,k}\to -t^2\mathrm{/}2$, without interchanging an unbounded number of unquantified little-o terms.

All hypotheses of [F4] were verified in step 1.1, so ${\prod }_k(1+w_{n,k})-\exp ({\sum }_k{w}_{n,k})\to 0$. Step 2.1 makes its limit $e^{-t^2\mathrm{/}2}$. By [F5] this product is the row-sum transform, and by [F6] its limit is the transform of N(0,1), continuous at zero. [F7] proves the result. At t=0 all factors and the limit are exactly one. AC is inherited in [F6]–[F7]; rows on different probability spaces cause no difficulty because only their laws are compared.