Let $X_n\in \mathbb{R}^{p\times n}$ be a random matrix whose entries are i.i.d. $\mathcal{N}(0,1)$. Define $S_n = \frac{1}{n}X_n X_n^\top$. If $p/n\to y\in (0,1)$, the well-known Bai-Yin theorem yields that with probability 1, we have $$\lim_{n\to +\infty}\lambda_{\mathrm{min}}(S_n)=(1-\sqrt{y})^2\,.$$

In light of this, can we say that there exists $N$ such that for all $n>N$, we have $\lambda_{\mathrm{min}}(S_n)>(1-\sqrt{y})^2/2$ almost surely? This argument was used in Hastie et al. multiple times. For example, on page 38, in the paragraph right after equation (110), the authors wrote "$\lambda_{\mathrm{min}}(\Sigma) \ge c$, and the Bai-Yin theorem (Bai and Yin, 1993), which implies that the smallest eigenvalue of $Z^\top Z/n$ is almost surely larger than $(1 −\sqrt{\gamma})^2/2$ for sufficiently large n".

Let $(\Omega,\mathcal{F},P)$ be the underlying probability space. Define $f_n(\omega)=\lambda_{\mathrm{min}}(S_n)$, where $\omega\in \Omega$. I think that to get $f_n(\omega)>(1-\sqrt{y})^2/2$ for all sufficiently large $n$, we need a stronger type of convergence, i.e., uniform convergence for almost all $\omega$ (i.e., all $\omega$ except a zero probability set), rather than just almost sure convergence (pointwise convergence for almost all $\omega$). In other words, the $N$ should work for almost all $\omega$, rather than that the choice of $N$ depends on $\omega$.

My first question is:

**Is the argument in Hastie et al. correct?**

I have another simple example. Define $X_n(\omega)=n\omega^n$, where $\omega\in \Omega=[0,1]$ and $P$ is the Lebesgue measure. Almost surely we have $X_n\to 0$. However, it is not true that $X_n(\omega)<1$ holds almost surely for all sufficiently large $n$.

I have another minor question:

**What is the underlying probability space in the expression $\Pr[\lim_{n\to +\infty}\lambda_{\mathrm{min}}(S_n)=(1-\sqrt{y})^2]$?**

I thought that it might be an infinite product probability space of a univariate Gaussian probability measure $\mathcal{N}(0,1)$. Consider an infinite 2D array $A_{ij}$ whose entries are i.i.d. $\mathcal{N}(0,1)$. Then $X_n$ is just a 2D subarray $\{A_{ij}\mid 1\le i\le p,1\le j\le n\}$. Am I right?

Thank you!