I'm reading a proof of Multivariate CLT using Lindeberg Theorem.
Let $X_n = (X_{ni},... ,X_{nk})$ be independent random vectors all having the same distribution. Suppose that $E[X_{nu}]<\infty$; let the vector of means be $c=(c_1,..., c_k)$, where $c_u=E[X_{nu}],$ and let the covariance matrix be $\Sigma = [\sigma_{uv}],$ where $\sigma_{uv}=E[(X_{nu} — c_u)(X_{nv} — c_v)].$ Put $Sn=X_{1}+\cdots X_{n}.$ Under these assumptions, the distribution of the random vector $(S_n — nc)/\sqrt{n}$ converges weakly to the centered normal distribution with covariance matrix $\Sigma$.
The proof is as follows:
Let $Y =(Y_1,...,Y_{n})$ be a normally distributed random with $0$ means and covariance matrix $\Sigma.$ For given $t=(t_1,...,t_k)$ let $Z_n=\displaystyle\sum_{u=1}^{k}t_u(X_{nu}-c_{u})$ and $Z=\displaystyle\sum_{u=1}^{k}t_uY_u.$ Then it suffices prove that $n^{1/2}\displaystyle\sum_{j=1}^{n}Z_j$ converges in distribution to $Z$ (for arbitrary $t$). But this is an instant consequence of the Lindeberg-Levy theorem.
I'm stuck following this proof. I'm not sure if Lindeberg condition is satisfied, i.e. $$\displaystyle\lim_{n\rightarrow\infty}\displaystyle\sum_{k=1}^{n}\frac{1}{s_n}\int_{\{|Z_k/\sqrt{n}|>\epsilon s_{n}\}}\frac{|Z_k|^2}{n} dP=0.$$
My idea is that $\{|Z_k/\sqrt{n}|>\epsilon s_{n}\}$ decreases to $\emptyset$; that's the reason of the integral converges to $0,$ but what about of the convergence or divergence of $s_{n}$ and the sum that tends to infinity?
Any kind of help is thanked in advanced.