stochastic process: determining whether process is cadlag I had asked a similar problem before and Didier Piau was very kind to help me with the answer. I have another question on the same problem setting. 
Let $T$ be a random exponentially distributed time. $P(T>t)=e^{−t}$. Define $M$ via $M_t=1$ if $t−T∈Q^+$, $M_t=0$ otherwise. Where $Q^+$ being positive rationals. let $F_t$ be a filtration generated by the process $M$.
I can see that the process described above is a martingle but i am trying to prove it is not cadlag. i am trying to apply the definition $\lim_{h\rightarrow 0} M_{t+h} =M_t$ and $\lim_{h\rightarrow 0} M_{t-h} =exists$. 
At $M_{t+h}$ i will generate a random variable $T$ and depending on whether $T>t+h$, $M_{t+h}$ will have 0 or 1. In the limit $h\rightarrow 0$ , $M_{t+h}$ does not necessarily go to $M_t$ as at $t+h$ no matter how small $h$ is a new random variable $T$ will decide the value of $M_{t+h}$ which can make $M_{t+h}$ different from $M_t$. 
Is my thought process correct to arrive at the conclusion that $M_t$ is not cadlag? It would be very kind of someone to help as I am trying to learn stochastic process on my own reading from a book.     
 A: If I understand correct, you have $M_t = 0$ for $t<T$ and $M_t = D(t-T)$ for $t\geq T$ where $D(x)$ is a Dirichlet Function, which is continuous nowhere. It means that for $t\geq T$ the process $M_t$ has a trajectory which is continuous nowhere and has no right/left limits.
Edited: Since you have some questions on the construction of this process, I try to discuss it here. As you have described, the construction is following:


*

*you pick up a random variable $T\sim\mathcal E(1)$; 

*you define
$$
M_t = \begin{cases}
1, \text{ if }t-T\in\mathbb Q^+;
\\
0, \text{ otherwise}.
\end{cases}
$$
This means that whenever you know $T$ you can construct the process $M_t$. Vice versa, if you know $M_t$ then clearly $T = \min\{\tau\geq 0:M_t = 1\}$: we can take minimum since $M_T = 1$.
If $t-T\in \mathbb Q^+$ then $M_t = 1$ but there is a sequence $t_k\geq T$ such that $M_{t_k} = 0$ but $t_k\to t$ with $k\to\infty$.
 You can also just change notation a bit and write
$$
M_t = \begin{cases}
D(t), \text{ if }t\geq T;
\\
0, \text{ if }t<T.
\end{cases}
$$
where $D(t)$ is a Dirichlet function. This immediately implies that $M_t$ is not cadlag. 
Note that here you simulate $T$ only once, if I understand your question correct. That's why you don't need to simulate two different random variables $T_1,T_2$ at times $t,t+h$. There is one-to-one correspondence between $T\in\mathbb R_{\geq0}$ and $M_t(T)$.
Let me show you an example which is more easy to visualize. Let us take
$$
M'_t = \begin{cases}1+\log{t}, \text{ if }t\geq T;
\\
0, \text{ if }t<T.
\end{cases}
$$
Here is the plot, for $T\approx 4.32$. In your case situation is almost the same - just you have $D(t)$ instead of $1+\log{t}$, but the former function is hard to visualize.

