Almost sure equivalence of regular conditonal probability of a progressive process in a controlled SDE equation. Consider the (canonical path) space of continuous functions $\Omega = C([0,T],\mathbb{R})$ with the Borel-sigma algebra generated by the open sets induced by the uniform topology and $\mathbb{P}$ Wiener Measure on it (and canonical process $B_t(\omega) = \omega(t))$. 
Working through the paper "Julien Claisse, Denis Talay, Xiaolu Tan:  A pseudo-Markov property for controlled diffusion processes" (https://arxiv.org/abs/1501.03939) i am stuck understanding a identity of the regular conditional probability when having a diffusion SDE (to be more precise Page 6, Equation 10): 
Say we have a solution
\begin{equation}
X_T^{t,x,u} = x_t + \int_{t}^{T} \mu(z,{X}_z^{t,x,u}, u_z) dz + \int_{t}^{T} \sigma(z,{X}_z^{t,x,u},u_z) dB_z 
\end{equation}
Then it also holds for $ t \leq s    \leq T$ by writing out (and using continuity)
\begin{equation}
X_T^{t,x,u} = X_s^{t,x,u} + \int_{s}^{T} \mu(z,{X}_z^{t,x,u}, u_z) dz + \int_{s}^{T} \sigma(z,{X}_z^{t,x,u},u_z) dB_z 
\end{equation}
Assume there exists a regular conditional probability $(\mathbb{P}_\omega)_{w \in \Omega}$ for $\mathcal{F}_T:= \bigvee_{0 \leq t\leq T } \mathcal{F}_t$ given $\mathcal{F}_s$ for $ t \leq s \leq T$. Then it holds for the stopped process $[{X}^{t,x,u}]_s := (X_{s \wedge z}^{t,x,v})_{0 \leq z \leq T} $
\begin{equation}
\mathbb{P}_{\omega} \left( [{X}^{t,x,u}]_s = [{X}^{t,x,u}]_s(\omega) \right) = 1 \qquad  for \quad \mathbb{P} \quad a.all \quad \omega \in \Omega
\end{equation}
(see also Lemma 3.2 in the referenced paper). 
However why is it true that it also holds for the progressive process $u_t$ and the concatenation process $(u_z^{t,\omega})_{0 \leq z \leq T }$ defined by
\begin{equation}
u_z^{t,\omega}( \bar{\omega}) := 
\{
\begin{array}{ll}
u_z ( \omega(y) ), \quad \quad \quad \quad \quad \quad \quad if \quad 0 \leq y \leq t  \\
u_z( \omega(t) + \bar{\omega}(y) - \bar{\omega} (t)) , \quad if \quad t \leq y \leq T\\
\end{array}
\end{equation}
that
\begin{equation}
\mathbb{P}_{\omega} ( u = u^{s,\omega}) = 1 \qquad  for \quad \mathbb{P} \quad a.all \quad \omega \in \Omega \qquad ? 
\end{equation}
Especially i am wondering of how to particularly derive the form of the additional concatenation of the above process. Can anyone light this maybe up? Probably i am missing a property of a representation of processes measurable to the sigma algebra generated by the brownian motion in this setting.
 A: I resolved it: 
Like expected it is more about basic properties of the Brownian Motion (without sigma algebra...) which yields above relation while the SDE equation's can be forgotten: 
For a fix $t \in [0,T]$ it follows from a classical result about the r.c.p ( Karatzas/Shreve: Brownian Motion and Stochastic Calculus page 307 Theorem 3.18) that:
\begin{equation}
\mathbb{P}^{\omega} ( u_t = u_t (\omega) ) =1 \quad for \quad  \mathbb{P} \quad a.all \quad  \omega \in \Omega 
\end{equation}
(it is also used in my question post for the stopped path process). 
Now consider the "Shifted to t Brownian Motion" (we know that this process again is a $ \mathbb{P} $ BM by standard estimates, or any literature about BM).
\begin{equation}
B^{\bigtriangleup t}_{(\cdot)}:= B_{t+ ( \cdot )} - B_t 
\end{equation}
and the shifting operator process $\theta _t$ on $C([0,T], \mathbb{R}$ ):
\begin{equation}
(\theta_t B )_s = := 
\{
\begin{array}{ll}
B_{s-t} , \quad \quad \quad \quad \quad \qquad \qquad if \quad 0 \leq t \leq s  \\
0 , \quad \quad \quad \quad \qquad \qquad \qquad  if \quad  0\leq s \leq t\\
\end{array}
\end{equation}
It holds now (crucial equality here)
\begin{equation}
B = [B]_t + \theta_t \cdot  B^{\bigtriangleup t} \qquad \mathbb{P} -a.s
\end{equation}
which gives above additive concatenation. 
Then we can define above concatenation process function taking the  $B^{ \bigtriangleup t }$ as input (or any other Brownian Motion independent from $\mathcal{F}_t$, like done in above concatenation process) to still have $\mathbb{P} $ equality.
From the properties of the regular conditional probability, it is pathwise equal to the $\mathbb{P}$ path, indeed above equation holds $ \mathbb{P}^{\omega} $ a.s. from which the claimed equality now follows.
