I have $S_N = \Sigma^N_{n=1} X_n$, where $X_n$ is a random variable that takes either 1, -1, 0, each with probability 1/3. I know that $\frac{S_N}{\sqrt N}$ converges in distribution to a mean-zero normally distributed random variables, say $Y$.
However I am trying to come up with a set $A$, for which $\Bbb P(Y \in A)=0$, but for which $\Bbb P(\frac{S_N}{\sqrt N} \in A)= \frac{1}{2}$ for positive odd integers N, but am struggling to do so.
Any help appreciated!
First note Bora Dogan's comment - we are interested in $S_N / \sqrt{N}$, since $S_N / N$ converges in distribution to the expectation by the law of large numbers.
Consider trying $A = \left\{\frac{1}{\sqrt{N}}, \frac{3}{\sqrt{N}}, \cdots, \frac{N - 2}{\sqrt{N}}, \sqrt{N}\right\}$. The reason $\mathbb{P}\left(\frac{S_N}{\sqrt{N}} \in A\right) = 1/2$ is because for odd integers $N$, the summation is symmetric about zero. It never takes on the value zero, and for every $x > 0$, the probabilities of $S_N = x$ and $S_N = -x$ are equivalent. Note further that $\mathbb{P}(S_N > N) = 0$, so the elements of set $A$ are the only positive values that $S_N / \sqrt{N}$ may take.
But as $Y$ is normally distributed and continuous, $\mathbb{P}\left(Y \in A\right) = 0$ because $A$ is measure zero (it is countable).