this is a more general question.
Consider a sequence $(X_j)_{j \in \mathbb{Z}}$ of iid real-valued random variables with mean zero and $\mathbb{E}(X_1^2) = 1$ on a probability space $(\Omega, \mathcal{A}, \mathbb{P})$, let $S_n$ be it's partial sum process and let further $Z$ be a standard normal distributed random variable.
The CLT then tells us that $$\frac{1}{\sqrt{n}} S_n \overset{\mathcal{D}}{\to} Z.$$
Now, I was wondering under what conditions such results would hold almost surely and there is an insane amount of literature discussing this topic. But I stumbled over the following result (referred as the Komlos-Major-Tusnady approximation) which says that if we take the setting from above and assume further that $\mathbb{E}(\vert X_1 \vert^p) < \infty$ for $p > 2$, then $$\frac{S_n - \mathbb{B}(n)}{n^{1/p}} \to 0 \,\,a.s.$$ where $\mathbb{B}(\cdot)$ shall denote a standard Brownian motion. As a property of the Brownian motion $\sqrt{n}^{-1}\mathbb{B}(n) \sim \mathcal{N}(0,1)$ and further, since $p > 2$ the above result would imply $$\frac{S_n - \mathbb{B}(n)}{\sqrt{n}} \to 0 \,\,a.s.$$
Wouldn't this imply, that the CLT above holds almost surely? Do I have a mistake in reasoning? Because if that would be the case, I would feel a little bit betrayed because – at least in the iid case – we would only need the additional assumption $\mathbb{E}(\vert X_1 \vert^p) < \infty$ to have the CLT hold almost surely and nobody ever mentioned this in any of our probability theory or statistics courses.
Let $Y_n=\mathbb B(n)/\sqrt n$. It is true that by the strong invariance principle, $S_n/\sqrt n-Y_n\to 0$ almost surely and that $Y_n$ has a standard normal distribution for each $n$.
It is true that in your comment, $S_n^*-Z^*\to 0$ almost surely. But you cannot infer from this that $S'_n-Z\to 0$ because the almost sure convergence is a property of the law of the whole sequence and you do not know that $(S'_n-Z)_{n\geqslant 1}$ has the same law as $(S^*_n-Z^*)_{n\geqslant 1}$. You cannot deduce convergence in probability (not even that you have something Cauchy in probability) as we do not know the law of $S'_n-S'_m$.