Indistinguishable Random Variables exercise Let $I=[0,+\infty[$, $\Omega = [0,2[$ and let $Y_t=0$ for $t\in I$
We define
$$X_t =
\begin{cases}
0 \hspace{1cm} \text{if } \omega \in [0,1[\\
0 \hspace{1cm} \text{if } \omega \in [1,2[ & t\neq \omega\\
1 \hspace{1cm} \text{if } \omega \in [1,2[ & t=\omega
\end{cases}
$$
We suppose now the $\sigma$-algebra
$$ \mathcal F = \sigma(X_t, t\geq 0)
$$
According to the solution, this directly yields
$$ \mathcal F = \sigma(X_t, t\geq 0\}=\sigma(\{t\}, t\in ]1,2[)
$$
I don't understand this last equality, how did it get to the $\{t\}$ part.
A second thing is that later on, when the exercise asked to prove wether $X_t$ and $Y_t$ were indistinguishable, $i.e$
$$ \mathbb P \{\omega \in \Omega : X_t(\omega)=Y_t(\omega), t\in I\} =1
$$
The solution proceeded to check the measurability of the set above, so it got that
$$ \{\omega \in \Omega : X_t(\omega)=Y_t(\omega), t\in I\} = [0,1] \notin \mathcal F
$$
I still don't get where did these results came from.
 A: By definition $$\sigma{(X_t)} = \{ X_t^{-1}(B) | B - \text{Borel set} \}$$
If $t \notin ]1,2[$ then $X_t = 0$ and $\sigma{(X_t)} = \{ \Omega, \varnothing\}$.
If $t \in ]1,2[$ then $X_t(\omega) = I_{t = \omega}$ takes values from the set $\{ 0 ,1 \}$. Thus, for example, if $B = (\frac12, 3)$ we have $ X_t^{-1}(B) = \{ \omega: X_t(\omega) \in B \} = \{ t \} $ and if $B = (-\infty, \frac45)$ then  $ X_t^{-1}(B) = \{ \omega: X_t(\omega) \in B \} = \Omega \backslash \{ t \} $. If $B = (-\infty, 7)$ then  $ X_t^{-1}(B) = \{ \omega: X_t(\omega) \in B \} = \Omega$.
Moreover,
$$\sigma{(X_t)} = \{ {t}, \Omega \backslash \{t\}, \Omega, \varnothing\} = \sigma(\{t\}) $$
for $t \in ]1,2[$. Hence
$$ \mathcal F = \sigma(X_t, t\geq 0\}=\sigma( \cup_{t \ge 0} \sigma{(X_t)}\}=
\sigma(\{t\}, t\in ]1,2[). 
$$
Thus $\mathcal F$ is sigma-algebra, generated by dots $\{ t \}$.
Statement: $$\mathcal F = \{ A \subset \Omega | A \text{ or } \Omega \backslash A  \text{ is countable } \} $$
Proof. Put by definition $M = \{ A \subset \Omega | A \text{ or } \Omega \backslash A  \text{ is countable } \}$. As $\{t\} \in \mathcal F$ hence countable uniouns of $t$ belong to $\mathcal F$ and hence $M \subset \mathcal F$. Now it's enough to prove that $M$ is sigma-algebra. But it may be easily checked by definition.
Finally $[0,1]$ is not a countable set and hence $[0,1]$ doesn't belong to $\mathcal F = \{ A \subset \Omega | A \text{ or } \Omega \backslash A  \text{ is countable } \} $.
