I am trying to understand the intuition of the CLT. If I have a sequence of R.V $\{X_i\}_{i=1}^{\infty}$
where $X_i\overset{iid}{\sim} f(x;\mu,\theta)$ where $E[X_i]=\mu$ and $Var[X_i]=\theta$
it is my understanding that by the SLLN there exists a set $N\subset\Omega$ with $P(N)=0$ such that for all $\omega \in \Omega-N$ the following is true: $$\overline{X_n}(\omega)\rightarrow \mu$$
I understand this from a real analysis perspective that $\overline{X_n}$ converges to a constant function a.s. that is a degenerate R.V over the set $\Omega-N$. With the WLLN implying something similar where the set where convergence does not occur eventually has a small probability as n gets larges and converging to 0.
Now the central limit theorem implies
$$\sqrt{\frac{n}{\theta}}\bigg[\overline{X_n}-\mu\bigg]\rightarrow Z$$
where $Z\sim N(0,1)$ that is the standardized sample mean converges to a RV(function) whose distribution is a standard normal. If I am wrong so far please correct me?
Why does the sample mean converge to a degenerate RV in probability and almost surely but then the standardized version converge to a non-degenerate random variable? The only thing I can come up with is that unlike convergence a.s, convergence in probability does not guarantee a defined set where this always holds so we can have $\omega \in \Omega $ where will not always guarantee convergence then this provides a non-degenerate RV where the central mass is at $E[X]=\mu$