This is homework so no answers please
I want to find for some $A\subset \mathbb{R}$ the limit $$\lim_{n\to \infty}\mu_{n}(A)=\lim_{n\to \infty}\frac{1}{n}E\left[\sum_{i=1}^{n}1_{A}(\frac{S_{i}}{\sqrt{n}})\right]=$$
$\sum_{k\in \mathbb{Z}}1_{A}(k)\lim_{n\to \infty}\frac{\sum_{i=1}^{n}P(\frac{S_{i}}{\sqrt{n}}=k)}{n}$
where $S_{n}=\sum_{i=1}^{n}X_{i}$ is a symmetric random walk starting at 0 and $P(X_{i}=1)=P(X_{i}=-1)=\frac{1}{2}$.
Any suggestions
Any mistakes:
One might guess: $\int_{0}^{\infty}\int_{A} P(B_{t}=x)dxdt$, where $B_{t}$ is Brownian motion.
But Donsker's invariance principle does not apply because we are not looking at an interpolation (i.e. we are not looking at $S_{\lfloor t\rfloor}+(t-\lfloor t\rfloor)(S_{\lfloor t\rfloor+1}-S_{\lfloor t\rfloor}))$
I am currently trying the LIL for random walk i.e. $$ (1-\varepsilon)\sqrt{2n\log(\log(n))}\leq S_{n}\leq (1+\varepsilon)\sqrt{2n\log(\log(n))}$$
My guess is different than yours: I claim the result of the limit is $ \int_0^1 \mathbf{P} ( B_t \in A ) dt, $ instead of integrating from $(0,\infty)$. My advice to get this result is to write the sum inside the limit as $$ n^{-1}\mathbf{E}\left[ \sum_{i=1}^n \mathbf{1}_A \left(\frac{ S_i }{ \sqrt{ n } } \right) \right] = \mathbf{E}\left[ \sum_{i=1}^n \mathbf{1}_A \left(\frac{ S_i }{ \sqrt{ n } } \right) \left(\frac{S_{i+1} - S_{i}}{\sqrt{n}} \right) \right]^2, $$ and then use invariance principle combined with Ito's isometry.