Let $(\Omega, \mathscr{F}, \cal{P})$ be a complete probability space. Consider a real-valued random process (field) $X(\omega, s),\omega \in \Omega,s\in{\mathbb{R}^N}$ with (everywhere) continuous sample paths, as well as the generated sub $\sigma$-algebra $\mathscr{Y}\triangleq \sigma \{Y:\Omega \rightarrow{\mathbb{R}^M} \} \subseteq \mathscr{F}$.


The question is simple: Does the *projected* random process


\begin{equation}
Z(\omega,s)\triangleq \mathbb{E} \{ X(\cdot,s) | \mathscr{Y}\}(\omega)
\end{equation}


admit an everywhere sample path continuous modification as well?


We may assume that $X$ is ${\cal L}\_1$-dominated, uniformly in $s$, that is, we may assume existence of a ${\cal L}\_1$ random variable $B:\Omega \rightarrow{\mathbb{R}}$, such that $\sup\_{s}|X(\omega,s)|\le B(\omega)$, for all $\omega\in\Omega$.


I think the answer is negative, but counterexamples maybe?



Is the Conditional Expectation of a Sample Path Contiuous (SPC) Stochastic Process a SPC Process itself?
The answer is positive: under the assumption of "$X$ is continuous and $L^1$-dominated", $Z=\{Z(t)\stackrel{\text{def}}{=}\mathsf{E}[X(t)\mid \mathscr{Y}],0\le t\le 1\}$
admit an everywhere sample path continuous modification. 


To prove it use the fact: Suppose that $Y=\{Y(t),0\le t\le 1\}$ is a stochastic process, then the necessary and sufficient conditions of $Y$ admit an everywhere sample path continuous modification are following:
 i) $Y$ is stochastic continuous, i.e.,
$$ \lim\_{s\to t}\mathsf{P}(|Y(s)-Y(t)|>\varepsilon)=0,\quad \forall \varepsilon>0,\quad \forall t\in[0,1]
$$
ii) $Y$ is sample path uniformly continuous on $\mathbb{Q}\_1\stackrel{\text{def}}{=}\{p/q\in[0,1]: p,q\in\mathbb{Z}\_+\}$, i.e., denoting $\omega(\delta,Y)\stackrel{\text{def}}{=}\sup\{|Y(s)-Y(t)|: |s-t|\le \delta, s,t\in\mathbb{Q}\_1\} $ , the following limit is hold,
$$\lim\_{\delta\downarrow0}\mathsf{P}(\omega(\delta,Y)>\varepsilon)=0,\qquad \forall \varepsilon>0.\quad(\text{equivalently, } \mathsf{P}(\lim\_{\delta\downarrow0}\omega(\delta,Y)=0)=1.)
$$


Using above conclusion and X is $L^1$-deminated, we get
\begin{gather} 
\lim\_{s\to t}\mathsf{E}|X(s)-X(t)|=0, \qquad \forall t\in [0,1].\tag{1}\\
\omega(\delta,X)\le 2B\in L^1,\qquad \lim\_{\delta\downarrow0}\mathsf{E}[\omega(\delta,X)]=0.
\end{gather}
Now form (1) and $\mathsf{E}|Z(s)-Z(t)|\le \mathsf{E}|X(s)-X(t)|$ we have
$$\lim\_{s\to t}\mathsf{E}|Z(s)-Z(t)|=0, \qquad \forall t\in [0,1].\tag{2}$$ 
Meanwile, 
\begin{align} |Z(s)-Z(t)|&\le\mathsf{E}[|X(t)-X(s)|\mid\mathscr{Y}]\le\mathsf{E}[\omega(\delta,X)\mid\mathscr{Y}].\\
\omega(\delta,Z)&\le \mathsf{E}[\omega(\delta,X)\mid\mathscr{Y}].\\
\lim\_{\delta\downarrow0}\mathsf{E}[\omega(\delta,Z)]&\le \lim\_{\delta\downarrow0}\mathsf{E}[\omega(\delta,X)]=0 \tag{3}
\end{align}
Using (2),(3) we can conclude that $Z=\{\mathsf{E}[X(t)\mid \mathscr{Y}],0\le t\le 1\}$ admit an everywhere sample path continuous modification. 


