This post is quite long, but the problem stated carries no computational burden.
Consider the equally spaced partition $t_{i}^n=\frac{i}{n}$ with $i=0,...,n$ of the interval $[0,1]$ into $n$ sub-interval of length $\delta=\frac{1}{n}$.
For each $n$ let $B_{i,n}$ with $i=1,...,n$ be a triangular array of Bernoulli variates such that $\mathbb{P}\left[B_{i,n}=1\right]=p_n$ and $\mathbb{P}\left[B_{i,n}=0\right]=1-p_n$. The $B$'s could be independent or not, it is irrelevant.
Suppose that $Y_t$ and $X_t$ with $t\in[0,1]$ are two continuous-time stochastic processes over the time window $[0,1]$ such that, when sample on the time grid $t_i^n$ it holds that
$$ \left\{ \begin{array}{lll} X_{0}&=&Y_0\\ X_{t_i^n}&=&Y_{t_i^n}\,(1-B_{i,n})+B_{i,n}\,X_{t_{i-1}^n}\quad (1), \end{array} \right. $$
and this is true irrespectively of the choice of $n\in\mathbb{N}$.
The idea is very simple. At $t=0$ the process $X$ is equal to $Y$. Later, at each time instant $t_{i}^n$ there are two possibilities:
1) if the Bernoulli is "activated", that is if $B_{i,n}=1$, then the $X$ does not update and repeats itself, that is $X_{t_i^n}=X_{t_{i-1}^n}$.
2) if the Bernoulli is not "activated", that is if $B_{i,n}=0$, then the $X$ is updated to the corresponding value of $Y$, that is $X_{t_i^n}=Y_{t_{i}^n}$.
If we write down the first two iterations of the process at the frequency $n=\frac{1}{\delta}$ we get
$$ \left\{ \begin{array}{lll} X_{0}&=&Y_0\\ X_{\delta}&=&Y_{\delta}\,(1-B_{1,n})+B_{1,n}\,Y_{0}\\ X_{2\,\delta}&=&Y_{2\,\delta}\,(1-B_{2,n})+Y_{\delta}\,B_{2,n}\,(1-B_{1,n})+Y_{0}\,B_{2,n}\,B_{1,n}. \end{array} \right. $$
So clearly $X_{\delta}$ may have as values either $Y_0$ or $Y_{\delta}$, while $X_{2\,\delta}$ may have as values or $Y_0$ or $Y_{\delta}$ or $Y_{2\,\delta}$.
My concern is related to what happen if we change the sampling frequency, for example taking $\Delta=\frac{1}{m}=2\,\delta=\frac{2}{n}>\delta$. Now the dynamics of $X$ sampled in the new time grid $t_i^m=\frac{1}{m}$ cannot be
$$ \left\{ \begin{array}{lll} X_{0}&=&Y_0\\ X_{t_i^m}&=&Y_{t_i^m}\,(1-B_{i,m})+B_{i,m}\,X_{t_{i-1}^m}. \end{array} \right. $$
since this would imply, taking just the first iteration, that
$$ \left\{ \begin{array}{lll} X_{0}&=&Y_0\\ X_{\Delta}&=&Y_{\Delta}\,(1-B_{1,m})+B_{1,m}\,Y_{0}\\ \end{array} \right. $$ so that $X_{\Delta}$ may have as value only $Y_{\Delta}$ or $Y_0$, but since $\Delta=2\,\delta$ this contradicts what we have found at the higher frequency $n>m$.
So my final question is: is the process in (1) ill-defined? Or, in other words, is it possible that the process $X$ in (1) is obtained by sampling a continuous-time process on a finite partition?
My guess is that, if my reasoning is correct, the process in (1) can exist only in discrete time and so cannot be a sampling of a continuous-time process.