Tricky a.e. limit question, studying the $ \text{a.e.-lim}_{t \rightarrow \infty} \frac {1}{t} \int_0 ^ t W_s ~ds$ Could someone give some advice in order to study  $$\underset{t\to\infty}{\operatorname{a.e.-lim}} \frac {1}{t} \int_0 ^ t W_s ~ds,$$ where $(W_t)_{t\geq 0}$ is a standard brownian motion starting at zero ? 
Thank's in advance.
 A: I'll turn my comment above into an answer.  Call $X_t = \frac{1}{t} \int_0^t W_s ds$.  We claim that the family $\{ X_t\}$ is not tight and therefore cannot converge in distribution (and hence cannot converge almost surely).  This is easy to see, since $X_t$ is Gaussian with mean zero and variance of order $t$.  Therefore, for any compact set $K$ $P ( X_t \in K) \to 0$ as $t \to \infty$.
As Did suggested, a little more work gets a stronger result.  Clearly, for any $M$ the event $A = \{\limsup_{t \to \infty} X(t) > M \}$ is a tail event in the sigma algebra of the Brownian motion $W_s$.  Therefore $P(A) \in \{0,1 \}$.  An explicit computation using the normality above gives that $P(X(t) > M) \to \frac{1}{2}$ as $t \to \infty$ and (the reverse) Fatou's lemma then gives that $P(A) \geq \frac{1}{2}$, so $P(A) = 1$.  Therefore $P(\{\limsup_{t \to \infty} X(t) = \infty \}) = 1$.  Symmetry gives that $P(\{\liminf_{t \to \infty} X(t) = - \infty \}) = 1$.  As a result, the limit almost surely does not exist.
