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)?
2 Answers
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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.

$\begingroup$ Or in other word this shows that $P$ and $Q$ are mutually sigular with respect to the $\sigma$algebra $\mathcal{F}=\Vee_{t}\mathcal{F}_t$. $\endgroup$ Commented Jun 28, 2011 at 8:34
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I am afraid that (1) is not correct, since the law of the iterated logarithm states that $$B_t\approx \sqrt{2t\log\log(1/t)}$$ as $t\to0$. This yields $$\frac{B_t}{t}\to\infty$$ as $t\to0$

$\begingroup$ sorry, I want $t\to \infty$ for all the above limits. Those are typos. $\endgroup$– kennethCommented Jun 28, 2011 at 8:02