Total row variance and the Lindeberg condition

For a centered triangular array as in Row-wise independent centered triangular array, suppose every entry has finite second moment. Write $v_{n,k}={\mathbb{E}}_n{X}_{n,k}^2={\mathrm{Var}}_n(X_{n,k}),s_n^2={\sum }_{k=1}^{k_n}{v}_{n,k}\mathrm{.}$ Variance is defined in Moments, variance, and covariance on a probability space. Every term is finite and nonnegative, so the finite sum and its nonnegative square root s_n exist. Whenever $s_n>0$, define $L_n(\epsilon )=s_n^{-2}{\sum }_{k=1}^{k_n}{\mathbb{E}}_n\text{\hspace{0.17em}⁣}\left[X_{n,k}^2{\mathbf{1}}_{\{\mathrm{\mid }{X}_{n,k}\mathrm{\mid }>\epsilon s_n\}}\right](\epsilon >0)\mathrm{.}$ The Lindeberg condition is $L_n(\epsilon )\to 0$ for every fixed $\epsilon >0$, with $s_n>0$ for all n under consideration. The event is measurable because absolute value is continuous and the entries are measurable; products are measurable by Arithmetic and lattice operations preserve measurability whenever they are defined. Its nonnegative integrand is bounded by $X_{n,k}^2$, so its expectation exists and is finite by Monotonicity and nonnegative homogeneity of the nonnegative integral and Expectation of a nonnegative or integrable random variable. In particular $0\le L_n(\epsilon )\le 1$.

For $Y_{n,k}=X_{n,k}\mathrm{/}{s}_n$, scalar homogeneity gives ${\sum }_k{\mathbb{E}}_n{Y}_{n,k}^2=1$. The equality $\mathrm{\mid }{Y}_{n,k}\mathrm{\mid }>\epsilon ⟺\mathrm{\mid }{X}_{n,k}\mathrm{\mid }>\epsilon s_n$ gives $L_n(\epsilon )={\sum }_k{\mathbb{E}}_n[Y_{n,k}^2{\mathbf{1}}_{\{\mathrm{\mid }{Y}_{n,k}\mathrm{\mid }>\epsilon \}}]\mathrm{.}$ Thus the normalized and unnormalized conditions are exactly equivalent, not merely asymptotic. The cutoff uses strict inequality; equality at the threshold is excluded. A zero-variance row is allowed in the initial variance definition, but its normalization and Lindeberg expression above are undefined. No independence or choice is needed to define these quantities.