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 $$P\Big(\lim_{t\to 0} \infty} \frac{W_t}{t} = 0\Big) = 1$$ This implies $$P\Big(\lim_{t\to 0} \infty} \frac{t + W_t}{t} = 1\Big) = 1. \quad (1)$$ 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, $$Q\Big(\lim_{t\to 0} \infty} \frac{t + W_t}{t} = 0\Big) = 1,$$ which implies $$P\Big(\lim_{t\to 0} \infty} \frac{t + W_t}{t} = 0\Big) = 1, \quad (2)$$ since $P$ is equivalent to $Q$. Why does this argument leads to a contradiction between (1) and (2)?
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 $$P\Big(\lim_{t\to 0} \frac{W_t}{t} = 0\Big) = 1$$ This implies $$P\Big(\lim_{t\to 0} \frac{t + W_t}{t} = 1\Big) = 1. \quad (1)$$ 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, $$Q\Big(\lim_{t\to 0} \frac{t + W_t}{t} = 0\Big) = 1,$$ which implies $$P\Big(\lim_{t\to 0} \frac{t + W_t}{t} = 0\Big) = 1, \quad (2)$$ since $P$ is equivalent to $Q$. Why does this argument leads to a contradiction between (1) and (2)?