Moving Limit within Probability/Law of Large Numbers

Question

Edits in Bold

When I look up the strong law of large numbers it says that (Looking at Discrete Random Variables) $$P\left(\lim_{n\rightarrow\infty}\bar{X}_{n}=\mu\right)$$ That got me wondering are the following equivalent $$P\left(\lim_{n\rightarrow\infty}\bar{X}_{n}=\mu\right)=\lim_{n\rightarrow\infty}P\left(\bar{X}_{n}=\mu\right)$$

I tried to think of this question in terms or translate this question in terms of a Measure on some object (to think of it somewhat pictorially) but I didn't know how to translate the summing of more random variables to this that type of thinking.

If this is not necessarily true are there conditions that would allow it to be true?

Answer 1

As noted already, if $\overline{X}_n$ is continuous, then $P(\overline{X}_n = \mu)=0$.

If the $X_n$ are discrete, then the averages $\overline{X}_n$ will also be discrete, but the space of values that they can take become arbitrarily large as $n \to \infty$. So, $\lim_{n \to \infty}P(\overline{X}_n=\mu)=0$, even though $P(\overline{X}_n = \mu)$ may be nonzero.

Answer 2

They are not.

If $\bar{X}_n$ is continuous, then $P(\bar{X}_n=c)=0$ for any constant $c$ (including the specific case $c=\mu$). So the limit on the RHS is just $0$ and the probability on the LHS is 1 (in the case when the SLLN applies.).