Let $\{W_t\}$ be a Brownian Motion and $T(j)=\sum_{i\leq j} \tau(i)$, where each $\tau(i)$ is a stopping time (but I don't think this matters here, we can think of $T(j)$ just as a random variable. Also we have for all $\delta>0$,$$\lim_{n\to \infty}\mathbb P(max_{j\leq n}\frac{|T(j)-j|}{n}\geq\delta)=0$$ I want to show for all $\epsilon>0,$$$\lim_{n\to \infty}\mathbb P(max_{j\leq n}\frac{|W_{T(j)}-W_{j}|}{\sqrt{n}}\geq\epsilon)=0$$.
Firstly for convenience we can rescale and show $$\lim_{n\to \infty}\mathbb P(max_{j\leq n}|W_{T(j)/n}-W_{j/n}|\geq\epsilon)=0$$ This is equivalent to $$\lim_{n\to \infty}\mathbb P(max_{j\leq n}|W_{T(j)/n}-W_{j/n}|<\epsilon)=1$$. Similary we can rewrite the assumption as ,$$\lim_{n\to \infty}\mathbb P(max_{j\leq n}\frac{|T(j)-j|}{n}<\delta)=1$$.
I think the result should follow from the sample continuity of Brownian Motion (each $W_t(\omega) \in C[0,1])$. The problem is that by sample continuity I have to choose $\delta(\omega)$ for each $\omega$. Any help is appreciated.