Why $\mathbb P(\forall s\in [a,b], B_s\neq 0\mid B_a=x)=\mathbb P(\forall s\in [0,b-a], B_s\neq -x\mid B_0=0)$? I want to compute $$\mathbb P(\forall s\in (a,b), B_s\neq 0),$$
but I already don't see why $$\{\forall s\in (a,b), B_s\neq 0 \}$$
is $\mathcal F_b$ measurable. Also, I know that if $t&gt;s$ then $$\mathbb P(B_t\in A\mid B_s=x)=\mathbb P(B_{t-s}\in A\mid B_0=x).$$
By why from this we can get $$\mathbb P(\forall t\in (a,b), B_t\neq 0\mid B_a=x)=\mathbb P(\forall t\in (0,b-a), B_t\neq -x\mid B_0=0) \ \ ?$$
First of all, mind that we are not dealing with classical conditional probabilities, that is, $$\mathbb{P}(A \mid B_s=x) \neq \frac{\mathbb{P}(A \cap \{B_s=x\})}{\mathbb{P}(B_s=x)}.\tag{1}$$

This is pretty obvious from the fact that $\mathbb{P}(B_s=x)=0$ for all $x \in \mathbb{R}$, $s&gt;0$. The expression on the left-hand side of $(1)$ is defined as follows: It follows from properties of the conditional expectation that there exists a measurable mapping $g:\mathbb{R} \to \mathbb{R}$ such that $$\mathbb{P}(A \mid B_s) := \mathbb{P}(A \mid \sigma(B_s))=g(B_s) \quad \text{a.s.}$$  Now one defines $$\mathbb{P}(A \mid B_s = x) := g(x).$$ (This definition is not specific for Brownian motion; more generally, this is the way to define $\mathbb{P}(A \mid X)$ for an event $A$ and some random variable $X$).



Let's return to your original problem. Fix some Borel set $A$ and some $a&lt;b$. Recall that the restarted process

$$W_t := B_{t+a}-B_a, \qquad t \geq 0,$$

is a Brownian motion which is independent of $(B_t)_{t \leq a}$. Clearly,

$$B_t(\omega) \neq 0 \, \, \text{for all $t \in (a,b)$} \iff W_t(\omega) \neq -B_a(\omega) \, \, \text{for all $t \in (0,b-a)$}.$$

If we set $\tau_x := \inf\{t&gt;0; W_t=x\}$, we can write this equivalently as

$$B_t(\omega) \neq 0 \, \, \text{for all $t \in (a,b)$} \iff \tau_{-B_a(\omega)}(\omega)\geq b-a.$$

Hence,

$$\mathbb{P}\left(B_t \neq 0 \,  \, \text{for all $t \in (a,b)$} \mid B_a \right) = \mathbb{P}(\tau_{-B_a} \geq b-a \mid B_a).$$

Since $\omega \mapsto \tau_{-x}(\omega)$ is independent of $\mathcal{F}_a^B:=\sigma(B_s; s \leq a)$ for all $x \in \mathbb{R}$ and $B_a$ is $\mathcal{F}_a^B$-measurable, it follows that

$$\mathbb{P}(\tau_{B_a} \geq b-a \mid B_a) = g(B_a)$$

where

$$g(x) = \mathbb{P}(\tau_{-x} \geq b-a); \tag{2}$$

see e.g. the result which I stated in the first part of this answer. Combining the previous two identites, we get
$$\mathbb{P}\left(B_t \neq 0 \,  \, \text{for all $t \in (a,b)$} \mid B_a \right)=g(B_a)$$

for $g$ defined in $(2)$. Hence,

$$\mathbb{P}\left(B_t \neq 0 \,  \, \text{for all $t \in (a,b)$} \mid B_a=x \right)=g(x), \tag{3}$$

as I explained earlier. By definition,

\begin{align*} g(x) = \mathbb{P}(\tau_{-x} \geq b-a) &amp;= \mathbb{P} \left(W_t \neq -x \, \, \text{for all $t \in (0,b-a)$}\right) \end{align*}

Using the fact taht both $(B_t)_{t \geq 0}$ and $(W_t)_{t \geq 0}$ are Brownian motions, which means in particular that they are continuous with probability $1$ and have the same finite dimensional distributions, we conclude that

\begin{align*} g(x) = \mathbb{P} \left( \bigcap_{q \in \mathbb{Q} \cap (a,b)} \{W_q\neq-x\} \right) &amp;= \mathbb{P} \left( \bigcap_{q \in \mathbb{Q} \cap (a,b)} \{B_q\neq-x\} \right) \\ &amp;= \mathbb{P} \left(B_t \neq -x \, \, \text{for all $t \in (0,b-a)$}\right)\end{align*}

Plugging this into $(3)$ proves the assertion.
I can answer your second question: 

$$\mathbb \{\forall t\in (a,b), B_t\neq 0\mid B_a=x\} = 
\bigcap_{t \in (a, b)}\{B_t\neq 0\mid B_a=x\} =
\bigcap_{t \in (a, b)}\{B_t \in \{0\}^c\mid B_a=x\}.$$
Now, by translation invariance of $B_t$ we get that the law of $B_t$ started at $B_a=x$ is equal to the law of $B_{t-s}$ started at $B_0=x$, so
$$
\mathbf P \left( \cap_{t \in (a, b)}\{B_t \in \{0\}^c\}\mid B_a=x\right) =
\mathbf P \left(\cap_{t \in (a, b)}\{B_{t-a} \in \{0\}^c\}\mid B_0=x\right) $$
This last set is equal to
$$\bigcap_{t \in (0, b-a)}\{B_t \in \{0\}^c\mid B_0=x\} \\=
\{\forall t\in (0,b-a), B_t\neq 0\mid B_0=x\} \\= 
\{\forall t\in (0,b-a), B_t - x\neq -x\mid B_0-x=0\} $$

Now, the process $B_t - x$ with $B_t$ started at $B_0 = x$ has the same distribution as another brownian motion $W_t$ started at $W_0 = 0$, so the set above has a probability equal to

$$\mathbf P(\forall t\in (0,b-a), W_t \neq -x\mid W_0 =0). $$