Problem to understand the proof of the reflexion principle of Brownian motion in wikipedia We want to prove that $$\mathbb P(\sup_{0\leq s\leq t}W_s\geq a)=2\mathbb P(W_t\geq a).$$
Here is the proof.
My mistakes are in the proof of $$\mathbb P(\sup_{0\leq s\leq t}W_s\geq a, W_t<a)=\frac{1}{2}\mathbb P(W_t\geq a).$$
So let $\tau_a=\inf\{t\geq 0\mid W_t=a\}$. A famous result says that $(X_t:=W_{t+\tau_a}-W_{\tau_a})_t$ is a Brownian motion independent of $\mathcal F_{\tau_a}$. Then, they say :
\begin{align*} \mathbb P\left(\sup_{0\leq s\leq t}W_s\geq a,W_t<a\right) =&\mathbb P\left(\sup_{0\leq s\leq t}W_s\geq a, X(t-\tau_a)<0\right) \\ &=\mathbb E[\mathbb E[\boldsymbol 1_{\{\sup_{0\leq s\leq t}W_s\geq a\}}\boldsymbol 1_{\{X(t-\tau_a)<0\}}\mid \mathcal F_{\tau_a}]]\\ &=\mathbb E[\boldsymbol 1_{\{\sup_{0\leq s\leq t}W_s\geq a\}}\mathbb E[\boldsymbol 1_{\{X(t-\tau_a)<0\}}\mid \mathcal F_{\tau_a}]]. \end{align*}
Q1) Why $\boldsymbol 1_{\sup_{0\leq s\leq t}W_s\geq a}$ is $\mathcal F_{\tau_a}-$measurable ? I agree that it's $\mathcal F_t-$measurable, but since a priori $\mathcal F_t$ is not in $\mathcal F_{\tau_a}$, I don't understand why we can take it out of the expectation.
Q2) (with several under questions)


*

*After they say that $\mathbb E[\boldsymbol 1_{\{X(t-\tau_a)\}}\mid \mathcal F_{\tau_a}]=\frac{1}{2}$, and also, I don't understand why. 

*I agree that $(X_t)$ is a brownian motion, and so, indeed, $\mathbb P\{X_t\leq 0\}=\frac{1}{2}$, but here we have $X(t-\tau_a)$ (which is not clear what is this process... why should it be a Brownian motion ?) By the way, I have the impression that $t-\tau_a$ can be negative, and so $X_{t-\tau_a}$ could be not well defined. Indeed, it's weird to say that $t>\tau_a$ since $\tau_a$ is random, so we don't know pointwise what is it. 
 A: Since $$\left\{ \sup_{0 \leq s \leq t} W_s \geq a \right\} = \{\tau_a \leq t\} \tag{1}$$ and $\tau_a$ is $\mathcal{F}_{\tau_a}$-measurable, it follows that $$\left\{\sup_{0 \leq s \leq t} W_s \geq a \right\} \in \mathcal{F}_{\tau_a}.$$ This answers your first question. Re the 2nd one: You are right that one has to be careful with these computations. I would argue as follows: Since $(X_t)_{t \geq 0}$ is independent from $\mathcal{F}_{\tau_a}$ and $\tau_a$ is $\mathcal{F}_{\tau_a}$-measurable, it holds that
$$\mathbb{E}(1_{\{\tau_a \leq t\}} 1_{\{X(t-\tau_a)<0\}} \mid \mathcal{F}_{\tau_a}) = g(\tau_a)$$ where $$g(s) := \mathbb{E}(1_{\{s \leq t\}}  1_{\{X(t-s)<0\}}),$$see the proposition in this answer. Clearly, $$g(s) = 1_{\{s \leq t\}} \mathbb{P}(X(t-s)<0) = \frac{1}{2} 1_{\{s \leq t\}},$$
and so 
$$\mathbb{E}(1_{\{\tau_a \leq t\}} 1_{\{X(t-\tau_a)<0\}} \mid \mathcal{F}_{\tau_a})= \frac{1}{2} 1_{\{\tau_a \leq t\}}.$$
Taking expectation and using $(1)$ this gives
$$\mathbb{P}(\tau_a \leq t, X(t-\tau_a) <0) = \frac{1}{2} \mathbb{P}(\sup_{s \leq t} W_s \geq a).$$
Combining this with the calculations in your question, the assertion follows.
