Let $(\Omega, \mathcal{F}, P)$ be a probability space, on which
$\mathcal{F}\_t$ is filtration satisfying general conditions. $W\_t$ is
a standard Brownian motion. By law of iterated logarithm, one has
$$ P\Big(\lim\_{t\to \infty} \frac{W\_t}{t} = 0\Big) = 1 $$
This implies
$$ P\Big(\lim\_{t\to \infty} \frac{t + W\_t}{t} = 1\Big) = 1. \quad (1)$$
We have contradiction in this below. By Girsanov theorem, there exists
a probability measure $Q$ equivalent to $P$, such that $t+ W\_t$ is a
Brownian motion w.r.t. $Q$. By law of iterated logarithm,
$$ Q\Big(\lim\_{t\to \infty} \frac{t + W\_t}{t} = 0\Big) = 1, $$
which implies
$$ P\Big(\lim\_{t\to \infty} \frac{t + W\_t}{t} = 0\Big) = 1, \quad (2)$$
since $P$ is equivalent to $Q$.
Why does this argument leads to a contradiction between (1) and (2)?



law of iterated logrithm
Girsanov's theorem tells you that on any *finite* interval $[0,T]$ you can find an equivalent probability that makes $t+B\_t$ a Brownian motion. You just gave the proof that it cannot be done on the whole real line.


