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!
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.