Why $\mathbb P\{\forall t, X_t=Y_t\}=1$ doesn't really make sense?

Question

Let $(\Omega ,\mathcal F,\mathbb P)$ a probability space and $(X_t)_t$ and $(Y_t)_t$ two stochastic process. In a book I'm reading it's written that *$(X_t)$ and $(Y_t)$ are not distinguishable if there is a set $N$ of measure $0$ s.t. $X_t(\omega )=Y_t(\omega )$ for all $t$ and all $\omega \notin N$. And it's written : Even if the set $\{\forall t, X_t=Y_t\}$ is not necessarily an event, we will abuse the notation and write $$\mathbb P\{\forall t, X_t=Y_t\}=1.$$

1. I don't understand the last sentence, why $\{\forall t, X_t=Y_t\}$ is not an event ?
2. By the way, what is the difference between $$\forall t, \mathbb P\{X_t=Y_t\}=1\quad \text{and}\quad \mathbb P\{\forall t, X_t=Y_t\}=1 \ \ ?$$ To me it looks to mean more or less the same thing.

Answer 1

1) $$\{\forall t\;X_t=Y_t\}=\{\omega\in\Omega\mid\forall t\;X_t(\omega)=Y_t(\omega)\}=\bigcap_t\{X_t=Y_t\}$$

Here $\{X_t=Y_t\}$ is an event for every $t$, but if the index set of $t$ is uncountable then it is not excluded that this intersection is not an event. This because a $\sigma$-algebra is not necessarily closed under intersections that are not countable.

2)

LHS states that $P(X_t=Y_t)=1$ for every $t$ but this does not imply that also $P\left(\bigcap_t\{X_t=Y_t\}\right)=1$ which is the RHS. It is even quite will possible that no $\omega\in\Omega$ exists with $X_t(\omega)=Y_t(\omega)$ for every $t$. In that case $\bigcap_t\{X_t=Y_t\}=\varnothing$ and consequently $P\left(\bigcap_t\{X_t=Y_t\}\right)=0$.

RHS does imply LHS. This because the intersection is a subset of $\{X_t=Y_t\}$ for every $t$.

Answer 2

1) The set $\{\forall t, X_t=Y_t\}$ can be not measurable (a $\sigma -$algebra is a priori not stable by uncountable intersection.)

2) Let $([0,1], \mathcal B([0,1]), m)$ where $\mathcal B(\mathbb R)$ is the Borel $\sigma -$ algebra and $m$ the Lebesgue measure. Let $D$ be the diagonal of $[0,1]\times [0,1]$ and for all $t\in [0,1]$ and all $\omega \in [0,1]$ $$X_t(\omega )=0\quad \text{and}\quad Y_t(\omega )=\boldsymbol 1_D(t,\omega ).$$ Then for fixed $t$, $$\mathbb P\{X_t=Y_t\}=1,$$ but $$\mathbb P\{\forall t, X_t=Y_t\}=0.$$