question about meaning of indistinguishable in an example in stochastic process I have a question about example 3.4 on Page 6 of the document here:
http://stat.math.uregina.ca/~kozdron/Teaching/Regina/862Winter06/Handouts/revised_lecture1.pdf
My specific question is that I understand stochastic process $X$ and $Y$ are the same "version". However I don't quite understand why they are NOT indistinguishable. 
The definition of indistinguishable in Definition 3.3. says that they are indistinguishable when $P(X_{\alpha} = Y_{\alpha} \forall \alpha \in I) = P(\{w\in \Omega : X_{
\alpha}(w) = Y_{\alpha}(w) \forall \alpha \in I\})=1$
I think it means that the set that constitute $\Omega$ has to satistifies the condition of $P(X_{\alpha} = Y_{\alpha}$) for every single $\alpha$. So for every single $\alpha$, we have that condition of $P(X_{\alpha}(w) = Y_{\alpha}(w)) $ for all the w  that has non-zero measure in $\Omega$ I guess. Is  my understanding correct?
Then if so, could someone explains why the X and Y in Example 3.4 are not indistinguishable?
Because isn't it true that no matter what z is generated from $Z \sim N(0,1)$, we have $X_t=0$. And $Y_t$ is also zero? So we have for each $t$, the set of z that makes $X_t=0$ and $Y_t=0$ at the same time is actually all the element in set of Z? So all element z in Z in the normal distribution satisfies the $X_t(w)=Y_t(w)$ condition? so shouldn't this the Definition of 3.3 equals to 1 now?   
IF someone could explains this example, could you also explain Exercise 3.5 as to why they are not indistinguishable too? 
Thanks 
 A: In case the link ever breaks, here is a summary of the examples in question:

Example 3.4.
  Let $Z$ be an absolutely continuous random variable on a probability space $(\Omega, \mathcal{F}, P)$.
  Define the stochastic processes $X$ and $Y$ on $[0, \infty) \times \Omega$ by $X_t = 0$ for all $t \geq 0$ and
  $$
Y_t =
\begin{cases}
0, &\text{if $t \neq |Z|$}, \\
1, &\text{if $t = |Z|$}.
\end{cases}
$$
  Then $Y$ is a version of $X$ (i.e., $P(X_t = Y_t) = 1$ for all $t \geq 0$), but $X$ and $Y$ are not indistinguishable (i.e., it is not the case that $P(\text{$X_t = Y_t$ for all $t \geq 0$}) = 1$).

Proof that $X$ and $Y$ are not indistinguishable.
To show that $X$ and $Y$ are not indistinguishable, we must show that $P(\text{$X_t = Y_t$ for all $t \geq 0$}) \neq 1$.
In fact, we can show the stronger statement that the event
$$
A = \{\omega \in \Omega : \text{$X_t(\omega) = Y_t(\omega)$ for all $t \geq 0$}\}
$$
is empty. This will imply that $P(A) = 0$, so in particular $P(A) \neq 1$.
Indeed, suppose $\omega \in \Omega$.
We will show that $\omega \notin A$.
Let $t = |Z(\omega)|$, which is some number in the interval $[0, \infty)$.
Then $X_t(\omega) = 0$ (since $X_t$ is identically zero), and $Y_t(\omega) = 1$ (since $t = |Z(\omega)|$). In particular, $X_t(\omega) \neq Y_t(\omega)$, so $\omega \notin A$. Since $\omega \in \Omega$ was chosen arbitrarily, it follows that $A = \emptyset$, and we are done.

Note. This proof did not require assuming that $Z$ was an absolutely continuous random variable; this assumption was only used in the proof that $X$ and $Y$ are versions of each other.


Example 3.5. Let $\Omega = [0, 1]$, $\mathcal{F}$ the completetion of the Borel $\sigma$-algebra with respect to Lebesgue measure, and $P$ the uniform/Lebesgue measure on $\mathcal{F}$.
  Let $X$ and $Y$ be the stochastic processes on $[0, 1] \times \Omega$ be given by $X_t = 0$ for all $t \in [0, 1]$ and
  $$
Y_t(\omega) =
\begin{cases}
1, & \text{if $t = \omega$}, \\
0, & \text{otherwise} \\
\end{cases}
$$
  for all $t \in [0, 1]$ and $\omega \in \Omega$.
  Then $Y$ is a version of $X$, but $X$ and $Y$ are not indistinguishable.

Proof that $Y$ is a version of $X$.
Fix $t \in [0, 1]$. Then $\{X_t \neq Y_t\} = \{t\}$, so $P(X_t \neq Y_t) = 0$.
Proof that $X$ and $Y$ are not indistinguishable.
To show that $X$ and $Y$ are not indistinguishable, we must show that $P(\text{$X_t = Y_t$ for all $t \in [0, 1]$}) \neq 1$.
In fact, we can show the stronger statement that the event
$$
A = \{\omega \in \Omega : \text{$X_t(\omega) = Y_t(\omega)$ for all $t \in [0, 1]$}\}
$$
is empty. This will imply that $P(A) = 0$, so in particular $P(A) \neq 1$.
Indeed, suppose $\omega \in \Omega$.
We will show that $\omega \notin A$.
Let $t = \omega$, which is some number in the interval $[0, 1]$.
Then $X_t(\omega) = 0$ (since $X_t$ is identically zero), and $Y_t(\omega) = 1$ (since $t = \omega$). In particular, $X_t(\omega) \neq Y_t(\omega)$, so $\omega \notin A$. Since $\omega \in \Omega$ was chosen arbitrarily, it follows that $A = \emptyset$, and we are done.

Note. I copy-and-pasted the last proof above from the first example, making only minor changes. Both proofs use a similar "diagonalization" argument.
