Show that Px$(B(s)\ge0 $ for all 0 $\le s \le t$ and B(t) $\in$ M) = Px(B(t) \in M)$-$P-x$(B(t) \in M)$,where B(s) is brownian motion,x>0.

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I want to show

Px$(B(s)\ge0 $ for all 0 $\le s \le t$ and B(t) $\in$ M) = Px(B(t) \in M)$-$P-x$(B(t) \in M)$

where,x>0,M is measurable set in [0,$\infty$).

The difficulty for me is how to handle the left side of the equation.I know the distribution of hitting time.But here it is not only about hitting time.

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This is called the reflection principle (often named after Désiré André) and is most probably explained in your notes: one starts from the decomposition $$ [B_t\in M]=[B_t\in M,T_0\lt t]\cup[B_t\in M,T_0\gt t]. $$ where $T_0=\inf\{s\gt0\mid B_s=0\}$ is the first hitting time of $0$. Once one has identified the first and the third event above, one is left with the crucial remark that, under $P_x$, the process $B'$ defined by $$ B_s'=\left\{\begin{array}{cll}B_s&\text{if}&s\leqslant T_0\\-B_s&\text{if}&s\gt T_0\end{array}\right. $$ is distributed like $(B_s)$ and that $\inf\{s\gt0\mid B_s'=0\}=T_0$. Furthermore, for every $x\gt0$, $M\subseteq[0,\infty)$ hence the events $[B_0=x,B_t\in M,T_0\lt t]$ and $[B_0'=x,B_t'\in-M]$ coincide. It follows that $$ P_x(B_t\in M,T_0\lt t)=P_x(B_t'\in-M)=P_x(B_t\in-M), $$ which ends the proof of the formula in your question.