Let $X_1,X_2,...$ be iid random variables $P(X_i=1)=\frac{1}{2}=P(X_i=-1)$ and let $S_n=X_1+...+X_n$ be simple random walk. Let $M_{2n}$ be the first time of hitting maximum in the walk of $2n$ steps. One of the arcsine laws for simple random walks states precisely that for $0<a<b<1$ we have $\lim_{n \to \infty}P(a\leq \frac{M_{2n}}{2n}\leq b)=\int_a^b\frac{1}{\pi\sqrt{x(1-x)}}dx$ (in Durret for instance).
I am trying to prove analogue of this theorem for Wiener process using Donsker's invariance principle and in particular I want to show that $\frac{M_{2n}}{2n}$ converges in distribution to arcsine distribution (then I will be done with the proof). Since arcsine distribution have density $\frac{1}{\pi\sqrt{x(1-x)}}$ it suffices to show that $F_n(t)$ converges to $F(t)$ pointwise, where $F_n$ and $F$ are cdf's of $\frac{M_{2n}}{2n}$ and arcsine distribution respectively. I thought it would be possible and easy since $\lim_{n \to \infty}P(a\leq \frac{M_{2n}}{2n}\leq b)=\int_a^b\frac{1}{\pi\sqrt{x(1-x)}}dx$, however I have a problem because $$F_n(t)=P(\frac{M_{2n}}{2n}\leq t)=P(0\leq \frac{M_{2n}}{2n}\leq t)$$ and limit does not work for $a=0$ (in the statement od theorem for random walks). If I could interchange order of limits like that $$\lim_{n \to \infty}(F_n(t)-F_n(0))=\lim_{n\to \infty}\lim_{a \to 0}P(a\leq \frac{M_{2n}}{2n}\leq t)=\lim_{a \to 0}\lim_{n\to \infty}P(a\leq \frac{M_{2n}}{2n}\leq t)=\int_a^b \frac{1}{\pi\sqrt{x(1-x)}}dx$$ I would be done since I believe $F_n(0)$ converges to $0$. But I don't know how to justify interchange of limits, I don't even know if it is possible.
I know there are plenty ways to prove this differently but I want this way or similarly short different way. Any suggestion is welcomed though.