Suppose that $\{X_{ij}\}_{1\leqslant i,j\leqslant n}$ are iid random variables with $\mathbb{E}(X_{11})=0$ and $\mathrm{Var}(X_{11})=1$, does the following convergence hold: $$ \max_{1\leqslant j\leqslant n}\biggl\{\frac{1}{n^2}\sum_{i\neq i'}(X_{ij}X_{i'j})\biggr\} \to 0 \qquad \text{almost surely}? $$

Background

I am reading the AoP paper "Limit of the smallest eigenvalue of a large dimensional sample covariance matrix" by Z. Bai and Y. Yin (1993). Their Lemma 2 states a generalization of the well-known Marcinkiewicz-Zygmund strong law of large numbers to the case of multiple arrays of iid random variables.

[Lemma 2 in Bai and Yin (1993)] Let $\{\xi_{ij},i,j=1,2,\ldots\}$ be a double array of iid random variables and let $\alpha>1/2,\beta\geqslant 0$ and $M$>0 be constants. Then as $n\to\infty$, $$ \max_{j\leqslant Mn^{\beta}} \biggl|n^{-\alpha}\sum_{i=1}^n (\xi_{ij}-c)\biggr|\to0\quad \text{almost surely}, $$ if and only if $$ (i)\quad \mathbb{E}|\xi_{11}|^{(1+\beta)/\alpha}<\infty $$ $$ (ii)\quad c = \left\{ \begin{array}{ll} \mathbb{E} \,\xi_{11},& \text{if }\alpha\leqslant 1, \\ \text{any number}, &\text{if }\alpha>1. \end{array} \right. $$

By our assumptions and taking $\alpha=\beta=M=1$, $\xi_i=X_{ij}^2$ in this lemma, we have $$ \max_{j\leqslant n}\biggl|\frac{1}{n}\sum_{i,j}X_{ij}^2-1\biggr|\to0\quad \text{almost surely}. $$ This result is for square terms. I wonder if there is a similar result for the cross terms $$ \max_{1\leqslant j\leqslant n}\biggl\{\frac{1}{n^2}\sum_{i\neq i'}(X_{ij}X_{i'j})\biggr\} \to 0 \qquad \text{almost surely}? $$

Attempt

I can prove that $(1/n^2)\sum_{i\neq i'}(X_{ij}X_{i'j})\to 0\; a.s.$ for any fixed $j$. But I do not know how to deal with the problem with "$\max$".

For fixed $j$, $$ \mathrm{Var}\Bigl(\sum_{i\neq i'}X_{ij}X_{i'j}\Bigr)=2\sum_{i\neq i'}\mathrm{E}\bigl(X_{ij}^2\bigr)\cdot\mathrm{E}\bigl(X_{i'j}^2\bigr)=2(n^2-n), $$ then by Chebyshev's inequality, for any $\varepsilon>0$, $$ \Pr\biggl(\frac{1}{n^2}\sum_{i\neq i'}(X_{ij}X_{i'j})>\varepsilon\biggr) =O\Bigl(\frac{1}{n^2}\Bigr), $$ which is summable. Hence, by using the Borel-Cantelli lemma, we have $$ \frac{1}{n^2}\sum_{i\neq i'}(X_{ij}X_{i'j})\to 0\qquad \text{almost surely}.$$

If we consider $\max_{1\leqslant j\leqslant n}$, and use the trivial inequality to bound it, we have $$ \Pr\biggl(\max_{1\leqslant j\leqslant n}\biggl\{\frac{1}{n^2}\sum_{i\neq i'}(X_{ij}X_{i'j})\biggr\}> \varepsilon\biggr) \leqslant n\cdot \Pr\biggl(\frac{1}{n^2}\sum_{i\neq i'}(X_{i1}X_{i'1})>\varepsilon\biggr)=O\Bigl(\frac{1}{n}\Bigr),$$ which means $$ \max_{1\leqslant j\leqslant n}\biggl\{\frac{1}{n^2}\sum_{i\neq i'}(X_{ij}X_{i'j})\biggr\}\to 0\qquad \text{in probability}.\tag{*} $$

How can we improve the result (*) to "almost surely"?