There is a probability space $(\Omega,\mathcal A, P)$ and random variables $X:\Omega \to \mathbb R$ and $Y:\Omega \to \mathbb R$, show that:
$\{\omega | X(w) \le Y(w)\} = \{ X \le Y \} \in \mathcal A$
$\{\omega | X(w) < Y(w)\} = \{ X < Y \} \in \mathcal A$
$\{\omega | X(w) = Y(w)\} = \{ X = Y \} \in \mathcal A$
My solution
$X$ and $Y$ are measurable functions (w.r.t. $\sigma$ algebras $\mathcal A$ and $\mathcal B^1$), it means that $Z$ is a random variable $(*)$, i.e. measurable function w.r.t. the same $\sigma$ algebras.
Thus $\{ X \le Y \}=\{Z \le 0\}=Z^{-1}((\infty,0]) \in \mathcal A$ because $(\infty,0] \in \mathcal B^1$. The same for $\{ X < Y \}$ and $\{ X = Y \}$.
$(*)$ there is a theorem saying that if $X$ and $Y$ are random variables, and $g:\mathbb R^2 \to \mathbb R$ is a continuous function, then $g(X,Y)$ is a random variable too.
My professor wrote back, that the solution is OK, but he actually expected another reasoning: $\{X< Y\}=\bigcup_{q \in Q}(\{ X< q\} \bigcap \{ Y > q\})$... (did not specify further).
Any ideas what he meant? Does anybody knows an alternative proof which starts as specified above?
I firstly prove:
$$\{X< Y\}=\bigcup_{q \in Q}(\{ X< q\} \bigcap \{ Y > q\})$$
$\forall \omega \in \{X< Y\}, X(\omega)$ and $Y(\omega)$ are real numbers $\implies \exists q \in Q$, such that, $X(\omega) < q < Y(\omega) \implies \omega \in \bigcup_{q \in Q}(\{ X< q\} \bigcap \{ Y > q\})$.
$\forall \omega \in \bigcup_{q \in Q}(\{ X< q\} \bigcap \{ Y > q\}) \implies \exists q$, such that $X(\omega) < q < Y(\omega)$, thus $\omega \in \{X< Y\}$.
Now $\{ X< q\} \in \mathcal A$ and $\{ Y > q\} \in \mathcal A \implies \{ X< q\} \bigcap \{ Y > q\} \in \mathcal A \implies \bigcup_{q \in Q}(\{ X< q\} \bigcap \{ Y > q\}) \in \mathcal A$.
Now $\{X \ge Y\}=\{X < Y\}^c \implies \{X \ge Y\} \in \mathcal A$ because $\{X < Y\} \in \mathcal A$ as proved above. The same way can be proved $\{X \le Y\} \in \mathcal A$. So we can conclude that $\{X \ge Y\} \bigcap \{X \le Y\}=\{X=Y\} \in \mathcal A$.