Why $\mathbb P(X_t=Y_t\text{ for all }t)=1$ if $\mathbb P(X_t=Y_t)=1$ for all $t$? Let $X_t$ and $Y_t$ two continuous stochastic process on $(\Omega ,\mathcal F,\mathbb P)$ s.t. $X_t=Y_t$ a.s. for all $t$. Why $\mathbb P\{X_t=Y_t\text{ for all }t\}=1$ ? 
The solution goes as : We have that $X_r=Y_r$ a.s. for all $r\in\mathbb Q$. Therefore, $\mathbb P\{\sup_{s\in\mathbb R}|X_s-Y_s|>0\}=0$. The claim follow.
I really don't understand the argument. And for me $\mathbb P(X_t=Y_t)=1$ for all $t$ is really the same as $\mathbb P(X_t=Y_t\text{ for all }t)=1$. I don't see the difference.
 A: The only thing you can say for $t$ fixed is $$\mathbb P\{X_t=Y_t\text{ for all }t\}\leq \mathbb P\{X_t=Y_t\}.$$
You should pay more attention to the answers I gave you here. It's the same problem here. 
What you must show is that there is $N$ with $\mathbb P(N)=0$ s.t. for all $\omega \notin \Omega $ and all $t$, $X_t(\omega )=Y_t(\omega ).$


*

*If $X_t=Y_t$ a.s. for all $t$, then $X_q=Y_q$ a.s. for all $q\in\mathbb Q$. Therefore, for all $q\in\mathbb Q$, there is $N_q$ with $\mathbb P(N_q)=0$ s.t. for all $\omega \notin N_q$ $X_q(\omega )=Y_q(\omega )$. 

*Set $N=\bigcup_{q\in\mathbb Q}N_q$. It's s.t. $\mathbb P(N)=0$ and for all $\omega \notin N$ and all $q\in\mathbb Q$, $X_q(\omega )=Y_q(\omega )$.

*Let $t\in \mathbb R$ and let $(q_n)_n$ a sequence of $\mathbb Q$ s.t. $q_n\to t$. If $\omega \notin N$, then, by continuity
$$X_t=\lim_{n\to \infty }X_{q_n}=\lim_{n\to \infty }Y_{q_n}=Y_t.$$
Therefore $\mathbb P\{X_t=Y_t\text{ for all $t$}\}=1$. 

For a counter example for the equality : Let $\mathbb P$ the Lebesgue measure on $\Omega =[0,1]$. Let $t\in [0,1]$, $X_t=0$ for all $\omega $ and $Y_t(\omega )=\boldsymbol 1_{\{t\}}(\omega )$.
Then $\mathbb P\{X_t=Y_t\}=1$ for all $t$, but $$\mathbb P\{X_t=Y_t\text{ for all }t\}=0,$$
since there is no $\omega $ s.t. $X_t(\omega )=Y_t(\omega )$ for all $t$ because for $\omega \in \Omega $ fixed, when $t=\omega $ then $X_t(\omega )=0$ but $Y_t(\omega )=1$. 
A: Since $X_t, Y_t$ are continuous, we have of course
$$
\mathbb P\left\{\sup_{s\in\mathbb R}|X_s-Y_s|=0\right\}
=
\mathbb P\left\{\sup_{s\in\mathbb Q}|X_s-Y_s|=0\right\}
=
\mathbb P\left(\bigcap_{s\in\mathbb Q}\{|X_s-Y_s|=0\}\right)
$$
a countable intersection of measurable sets of measure $1$.
