Let $X_n$ be a sequence of i.i.d. random variables where $X_1$ has distribution function $F$. Fix $t \in \Bbb R$. Define $Y_n(t) := \boldsymbol{1}_{X_n \le t}$. Certainly, the $Y_n(t)$ are again i.i.d.
Then, $\Bbb P[Y_1(t) = 0] = \Bbb P [\{X_1 > t \}]$ and $\Bbb P[Y_1(t) = 1] = \Bbb P [\{X_1 \le t \}]$. Thus, $\Bbb E[Y_1] = \Bbb P [\{X_n \le t \}]$ and $\Bbb V[Y_1] = \Bbb E[Y_1^2] - \Bbb E[Y_1]^2 = (1 - \Bbb P [\{X_n \le t \}]) \Bbb P [\{X_n \le t \}]$.
Do I see this correctly, that by the Law of large numbers, ${1 \over n}S_n(t) \to \Bbb E[Y_1] \Bbb P$-almost surely, where $S_n(t) = \sum_{k=1}^n Y_n(t)$?
Next define $S_n^{*}(t) := {1 \over \sqrt{n}}(S_n(t) - \Bbb E[S_n(t)])$. Now I am to use the central limit theorem on $S_n^{*}(t)$ and in my task it says that this can be done by choosing a certain $t \in \Bbb R$.
First part is correct. Also $S_n^{*}$ converges in distribution to normal with mean $0$ and variance $(1-F(t))F(t)$ for any $t$. There is no need to choose a particular $t$.