$\frac1n \sum_{i=1}^n W_i(T_i)\to 0$ a.s. for $n\to\infty$ Let $(W_1(t))_{t\ge 0},(W_2(t))_{t\ge 0},\dots$ i.i.d. standard Wiener processes. Let be $T$ an integrable, non negative random variable, such that $\{T\le t\} \sigma\{\cup_{s\le t}W_1(s)\} $is measurable for every $t\ge0$(this is called continous stopping time). Furthermore let $T_1:=T,T_n(\omega):=T(W_n(\cdot,\omega))+T_{n-1}(\omega).$

Show $\frac1n \sum_{i=1}^n W_i(T_i)\to 0$ a.s. for $n\to\infty$ and conclude $\operatorname E(W_1(T))=0.$

I am clueless and do not even know how to set one's hand to this task. Some approaches are welcome!
 A: As it is stated right now, the first part of the problem is not correct; in general,
$$\frac{1}{n} \sum_{i=1}^n W_i(T_i)$$
does not converge to $0$. To see this let's consider the particular case that $T := 1$. Clearly, $T$ is an integrable stopping time, and by the very definition of $T_n$ we have $T_n = n$. Thus,
$$\frac{1}{n} \sum_{i=1}^n W_i(T_i) = \frac{1}{n} \sum_{i=1}^n W_i(i).$$
Note that $W_i(i)$ is Gaussian with mean $0$ and variance $i$. Since the random variables are, by assumption, independent, we find that the sum
$$S_n := \sum_{i=1}^n W_i(i)$$
is Gaussian with mean $0$ and variance
$$\sigma_n^2 = \sum_{i=1}^n i = \frac{1}{2} n(n+1).$$
In particular, $S_n$ equals in distribution $\sqrt{\sigma_n^2} S_1 \sim N(0,\sigma_n^2)$, and so
$$\mathbb{P} \left( \left| \frac{S_n}{n} \right| > 1 \right) = \mathbb{P} \left( \frac{\sigma_n^2}{n} |S_1|> 1 \right) \xrightarrow[]{n \to \infty} \mathbb{P}(|S_1|>2)$$
which shows that $S_n/n$ does not converge in probability (and hence not almost surely) to $0$.

I believe that the statement is supposed to read
$$\frac{1}{n} \sum_{i=1}^n (W_i(T_i)-W_i(T_{i-1})) \xrightarrow[]{n \to \infty} 0,$$
in fact it follows from the strong Markov property of Brownian motion that the random variables $X_i := W_i(T_i) - W_i(T_{i-1})$ are independent and identically distributed, and therefore we can apply the strong law of large numbrs to deduce that $$\frac{1}{n} \sum_{i=1}^n (W_i(T_i)-W_i(T_{i-1})) \xrightarrow[]{n \to \infty} \mathbb{E}(W_1(T)).$$ In order to show that $$\mathbb{E}(W_1(T))=0$$ I suggest that you use martingale techniques, i.e. the fact that $(W_t)_{t \geq 0}$ and $(W_t^2-t)_{t \geq 0}$ are martingales. 
