Why does $\lim \limits_{n\to \infty}P(A_n=p)=0$ for $A_n = S_n/n$ where $S_n$ is the number of success in $n$ Bernoulli trials?

Question

I had thought that for $n$ Bernoulli trials with probability $p$ of success for each trial (all independent), as $n \to \infty, A_n = S_n/n\to p$. However my book states that the limit $\lim \limits_{n\to \infty}P(A_n=p)=0$, but also states that per the law of large number, a Bernoulli trial $P(|S_n/n-p|\leq \epsilon) \to 1$ as $n \to \infty$. Is $P(A_n = p)$ and $P(A_n \to p)$ different things?

Answer 1

Yes, they are different.

Suppose $p$ is irrational. Then for every fixed $n$, the even $A_n=p$ would never happen (because $A_n(\omega)$ must be one of $0,\frac1n,\frac2n,\dots,\frac{n-1}n,1$ for each $\omega$). Similarly if $p\in\mathbb{Q}\cap(0,1)$ with least denominator $m$, then $A_n=p$ would never happen for $n$ which are not a multiple of $m$. So $\lim_{n\to\infty}\mathbb{P}(A_n=p)=0$ because you are taking a limit of zeros.

On the other hand, the event $\lim A_n=p$ is for all $\varepsilon>0$ there is an $N\in\mathbb{N}$ such that $\lvert A_n-p\rvert<\varepsilon$. The strong law of large numbers says $\mathbb{P}(\lim A_n=p)=1$.

In short, you can't interchange the order of limit and $\mathbb{P}$.