Let $\{X_{nj}\}$, $n=1,2,...$, $j=1,2,...,k_n$, be a double array of random variables and for each $n$, $X_{n1},\ldots,X_{nk_n}$ are independent. Define $S_n=\sum_{j=1}^{k_n}X_{nj}$ and
$$\begin{array}{ll}
\mathscr{E}(X_{nj})=\alpha_{nj}, & \mathscr{E}(S_n)=\sum_{j=1}^{k_n}\alpha_{nj}=\alpha_n; \\
\sigma^2(X_{nj})=\sigma^2_{nj}, & \sigma^2(S_n)=\sum_{j=1}^{k_n}\sigma^2_{nj}=s^2_n. \\
\end{array}$$Suppose $\alpha_{nj}=0$ for all $n$ and $j$, and $s^2_n=1$. In order that as $n\rightarrow\infty$ the two conclusions below both hold:
(1) $S_n$ converges in distribution to $\Phi$.
(2) $\{X_{nj}\}$ is uniformly asymptotically negligible (UAN);
it is necessary and sufficient that for each $\eta>0$, we have $$\underset{n\rightarrow\infty}{\lim}\sum_{j=1}^{k_n}\mathscr{E}\left[X_{nj}^2\,I\left(|X_{nj}|>\eta\right)\right]=0$$
$\bullet$ Proof.
(Lindeberg: $\Longleftarrow$) The proof of $$\forall\,\eta>0,\;\underset{n\rightarrow\infty}{\lim}\sum_{j=1}^{k_n}\mathscr{E}\left[X_{nj}^2\,I\left(|X_{nj}|>\eta\right)\right]=0\implies S_n\overset{d}{\rightarrow}\Phi$$has already shown in Linderberg-Feller's Central Limit Theorem (short version). Here, we prove the UAN part. Suppose that the Lindeberg's condition holds. Then for all $\eta>0$, $$\begin{array}{l}\underset{n\rightarrow\infty}{\lim}\underset{1\leq j\leq k_n}{\max}\mathscr{P}\{|X_{nj}|>\eta\}\\ \leq\underset{n\rightarrow\infty}{\lim}\underset{1\leq j\leq k_n}{\max}\frac{1}{\eta^2}\int_{|x|>\eta}x^2\,dF_{nj}(x)\quad\left(\because\,|x|>\eta\,\Rightarrow\,\frac{x^2}{\eta^2}>1\right)\\ \leq\underset{n\rightarrow\infty}{\lim}\frac{1}{\eta^2}\sum_{j=1}^{k_n}\int_{|x|>\eta}x^2\,dF_{nj}(x)\rightarrow0\mbox{ as }n\rightarrow\infty. \end{array}$$Thus, $\{X_{nj}\}$ is uniformly asymptotically negligible (UAN).
(Feller: $\Longrightarrow$) Suppose (1) $S_n\overset{d}{\rightarrow}\Phi$ and (2) $\{X_{nj}\}$ is UAN. First we introduce a theorem that represents the definition of UAN as the characteristic function.
[Theorem] Let $\{X_{nj}\}$, $n=1,2,...$, $j=1,2,...,k_n$, be a double array of random variables and $\{f_{nj}(t)\}$ be their ch.f.'s. $$\forall\,\varepsilon>0,\,\underset{n\rightarrow\infty}{\lim}\max_{1\leq j\leq k_n}\mathscr{P}\{|X_{nj}|>\varepsilon\}=0\iff\forall\,t\in\mathbb{R},\,\underset{n\rightarrow\infty}{\lim}\max_{1\leq j\leq k_n}\left|f_{nj}(t)-1\right|=0.$$For more details, please see Uniformly Asymptotically Negligible (2): Connect to the Characteristic Function.
Back to the proof. By (1) and (2), we have for all $t\in\mathbb{R}$, $$\begin{array}{ll}\underset{n\rightarrow\infty}{\lim}\max_{1\leq j\leq k_n}\left|f_{nj}(t)-1\right|=0,&\mbox{and}\\ \underset{n\rightarrow\infty}{\lim}\prod_{j=1}^{k_n}f_{nj}(t)=\exp{\{-t^2/2\}}&\Rightarrow\,\underset{n\rightarrow\infty}{\lim}\sum_{j=1}^{k_n}\log{f_{nj}(t)}=-\frac{t^2}{2}.\end{array}$$By Taylor's expansion, let $$\log{f_{nj}(t)}=(f_{nj}(t)-1)+\Lambda|f_{nj}(t)-1|^2,\;|\Lambda|\leq\frac{1}{2}.$$Thus, $$\sum_{j=1}^{k_n}\log{f_{nj}(t)}=\sum_{j=1}^{k_n}(f_{nj}(t)-1)+\Lambda\sum_{j=1}^{k_n}|f_{nj}(t)-1|^2.$$Consider the second part, $$\begin{array}{rl}\sum_{j=1}^{k_n}|f_{nj}(t)-1|^2&\leq\underset{1\leq j\leq k_n}{\max}|f_{nj}(t)-1|\sum_{j=1}^{k_n}|f_{nj}(t)-1|\\
&=\underset{1\leq j\leq k_n}{\max}|f_{nj}(t)-1|\sum_{j=1}^{k_n}\left|\int_\mathbb{R}(e^{itx}-1)\,dF_{nj}(x)\right|\\
&=\underset{1\leq j\leq k_n}{\max}|f_{nj}(t)-1|\sum_{j=1}^{k_n}\left|\int_\mathbb{R}\left(itx+\theta\frac{t^2x^2}{2}\right)\,dF_{nj}(x)\right|\;\left(|\theta|\leq1\right)\\
&\leq\underset{1\leq j\leq k_n}{\max}|f_{nj}(t)-1|\cdot\frac{t^2}{2}\sum_{j=1}^{k_n}\int_\mathbb{R}x^2\,dF_{nj}(x)\\
&=\frac{t^2}{2}\underset{1\leq j\leq k_n}{\max}|f_{nj}(t)-1|\qquad\qquad\left(\because\,s^2_n=1\right)\\
&\rightarrow0\mbox{ as }n\rightarrow\infty \end{array}$$by $\{X_{nj}\}$ is UAN. Thus the first part should be $$\underset{n\rightarrow\infty}{\lim}\sum_{j=1}^{k_n}(f_{nj}(t)-1)=-\frac{t^2}{2}.$$Taking the real parts and multiply $-1$, we have $$\underset{n\rightarrow\infty}{\lim}\sum_{j=1}^{k_n}\int_\mathbb{R}(1-\cos{tx})\,dF_{nj}(x)=\frac{t^2}{2}.$$Hence for each $\eta>0$, we split the integral into two parts and transpose one of them $$\begin{array}{l}
\underset{n\rightarrow\infty}{\overline{\lim}}\left|\frac{t^2}{2}-\sum_{j=1}^{k_n}\int_{|x|\leq\eta}(1-\cos{tx})\,dF_{nj}(x)\right|\\ =\underset{n\rightarrow\infty}{\overline{\lim}}\left|\sum_{j=1}^{k_n}\int_{|x|>\eta}(1-\cos{tx})\,dF_{nj}(x)\right|\\ \leq\underset{n\rightarrow\infty}{\overline{\lim}}\sum_{j=1}^{k_n}\int_{|x|>\eta}2\,dF_{nj}(x)\\ \leq\underset{n\rightarrow\infty}{\overline{\lim}}\sum_{j=1}^{k_n}\frac{2}{\eta^2}\int_{|x|>\eta}x^2\,dF_{nj}(x)\quad\left(\because\,|x|>\eta\,\Rightarrow\,\frac{x^2}{\eta^2}>1\right)\\ =\frac{2}{\eta^2}\qquad\left(\because\,s^2_n=1\right) \end{array}$$Since for every real number $t$, we have $0\leq1-\cos{t}\leq t^2/2$, we have $$0\leq\underset{n\rightarrow\infty}{\overline{\lim}}\left\{\frac{t^2}{2}-\sum_{j=1}^{k_n}\frac{t^2}{2}\int_{|x|\leq\eta}x^2\,dF_{nj}(x)\right\}\leq\frac{2}{\eta^2}.$$Multiply $\frac{2}{t^2}$, the inequality turns to $$0\leq\underset{n\rightarrow\infty}{\overline{\lim}}\left\{1-\sum_{j=1}^{k_n}\int_{|x|\leq\eta}x^2\,dF_{nj}(x)\right\}\leq\frac{4}{t^2\eta^2}\quad\forall\,t\in\mathbb{R}.$$As $t\rightarrow\infty$, we have $$\underset{n\rightarrow\infty}{\lim}\left\{1-\sum_{j=1}^{k_n}\int_{|x|\leq\eta}x^2\,dF_{nj}(x)\right\}=\underset{n\rightarrow\infty}{\lim}\sum_{j=1}^{k_n}\int_{|x|>\eta}x^2\,dF_{nj}(x)=0.$$Thus the Lindeberg's condition holds.
$\Box$
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