Change of sign for Brownian motion For a fixed $\epsilon > 0$, I want to show that almost surely (i.e., with probability $1$), a standard Brownian motion $W_t$ would change sign over $[0,\epsilon]$.
I thought about defining a random variable $U_t = \mbox{sign} (W_t)$ which would satisfy $(1+U_t)/2$ follows symmetric Bernoulli distribution
But I can't conclude with this. Am I on a good track?

Another way to solve this problem is with the Blumenthal zero-one law.  Let $\mathcal{F}_t = \sigma(B_s : 0 \le s \le t)$, and $\mathcal{F}_{0+} = \bigcap_{t > 0} \mathcal{F}_t$.  The Blumenthal zero-one law asserts that $\mathcal{F}_{0+}$ is almost trivial; for each $A \in \mathcal{F}_{0+}$ we have $\mathbb{P}(A) = 0$ or $1$. Consider the event $A_+$ that $B_{1/n} > 0$ for infinitely many $n$.  Show that $\mathbb{P}(A_+) \ge 1/2$, and that $A_+ \in \mathcal{F}_{0+}$.  Conclude that $\mathbb{P}(A_+) = 1$.  Show the same for $A_-$, the event that $B_{1/n} < 0$ for infinitely many $n$.  On the event $A_+ \cap A_-$, $B_t$ changes sign in every interval $[0,\epsilon]$.
