Let $(B_t)$ a standard Brownian motion (i.e. $B_t\sim\mathcal N(0,t)$). Let $a\geq 0$. Prove that $$p\left\{\max_{0\leq u\leq t} B_u\geq a\right\}=2p\{B_t\geq a\}.$$
The proof goes like this : Set $$\tau=\begin{cases}\inf\{u\geq 0\mid b_u=a\}&\text{if}\ \{u\geq 0\mid b_u=a\}\neq\emptyset\\ +\infty &\text{if}\ \{u\geq 0\mid b_u=a\}=\emptyset\end{cases}.$$ Let $$\tilde B_t=\begin{cases}B_t&\text{if }t<\tau\\ a-(B_t-a)&\text{if }t\geq\tau\end{cases}.$$
We have that \begin{align*} p\left\{\max_{0\leq u\leq t}B_u\geq a,B_t\geq a\right\}&\underset{(1)}{=}p\left\{\max_{0\leq u\leq t}\tilde B_u\geq a,\tilde B_t\leq a\right\}\\ &\underset{(2)}{=}p\left\{\max_{0\leq u\leq t}B_u\geq a,B_t\leq a\right\}. \end{align*}
Therefore \begin{align*} p\left\{\max_{0\leq u\leq t}B_u\geq a\right\}&\underset{(3)}{=} p\left\{\max_{0\leq u\leq t}B_u\geq a,B_t\geq a\right\}+p\left\{\max_{0\leq u\leq t}B_u\geq a,B_t\leq a\right\}\\ &=2p\left\{\max_{0\leq u\leq t}B_u\geq a,B_t\geq a\right\}\\ &\underset{(4)}{=}2p\{B_t\geq a\}. \end{align*}
I'm sincerely sorry, but I don't understand $(1)$, $(2)$, $(3)$ and $(4)$. Any explanation is welcome.
Thank you :-)