Proving that brownian motion pass almost ever by zero in $]0,t[$ Consider $B=(B_0)_{t\geq 0}$ a real $\mathcal F_t$ - brownian motion starting at zero, in a probability space $(\Omega, \mathcal F, (\mathcal F_t)_{t\geq 0}, \mathbb P)$. Then, consider 
$$ \Phi_t(x) :=  \mathbb P \left\{\exists s \in ]0, t[ : B_s + x= 0\right\}$$ 
Show that $\Phi_t(0)=1 \forall t \geq 0$.
I have tried to explore the fact that $\Phi_t$ is even, since $- B$ is also a brownian motion. However, I still don't see how to prove it.
Any advice will be appreciate. Thank's in advance.
 A: Of course, what is the best approach to this question depends very much on what you know and what you do not know and, unfortunately, you are silent about this, but anyway, here is a try, using only elementary tools...
Let $X_t=\mathrm e^{-B_t-t/2}$, then $(X_t)$ is a martingale. Let $T$ denote the first hitting time of $0$ by $B$. Let $x\gt0$. Then, conditionally on $B_0=x$, $(X_{t\wedge T})$ is uniformly integrable since $X_t\leqslant1$ almost surely when $t\leqslant T$. Conditionally on $B_0=x$,  $X_0=\mathrm e^{-x}$ and $X_T=\mathrm e^{-T/2}$, hence the stopping time theorem yields $E_x[\mathrm e^{-T/2}]=\mathrm e^{-x}$. By symmetry, for every $x\ne0$, $E_x[\mathrm e^{-T/2}]=\mathrm e^{-|x|}$.
We deduce from this the case when $B_0=0$ almost surely. Consider the first hitting time $T_t$ of $0$ after $t$, then $T\leqslant T_t$ and $B_t\ne0$ almost surely hence 
$$
E_0[\mathrm e^{-T/2}]\geqslant\mathrm e^{-t/2}E_0[E_{B_t}[\mathrm e^{-T/2}]]=E_0[Y_t],\qquad Y_t=\mathrm e^{-|B_t|-t/2}.
$$
This holds for every $t\gt0$ and, when $t\to0$, $Y_t\to1$ and $|Y_t|\leqslant1$ almost surely, hence $E_0[Y_t]\to1$. Thus, $E_0[\mathrm e^{-T/2}]\geqslant1$. Since $\mathrm e^{-T/2}\leqslant1$ almost surely, $\mathrm e^{-T/2}=1$ almost surely, that is, $T=0$ almost surely.
A: Note the fact that $\limsup B_t=+\infty$ a.s. and $\liminf B_t=-\infty$ a.s. This may be proved rather easily using 0-1 law (or just look it up online). It just tells you that $B_t$ crosses 0 i.o. a.s. on $t \in [T,\infty], \forall T$ Now note that $tB_{1/t}$ is a standard Brownian motion. The result should be pretty obvious at this point. 
