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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?

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  • $\begingroup$ He wrote the set $\{X<Y\}$ as a countable union of measurable sets showing that the set itself is measurable. $\endgroup$ Feb 19, 2015 at 9:43
  • $\begingroup$ @StefanHansen I am not getting why if you go over $Q$, build corresponding sets $\{ X< q\} \bigcap \{ Y > q\}$ and then join them you get the original set $\{X< Y\}$, what about the set $\{ X< \sqrt 2\} \bigcap \{ Y > \sqrt 2\}$, it is not a part of the union above, but is a part of $\{X< Y\}$. $\endgroup$
    – zesy
    Feb 19, 2015 at 10:43
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    $\begingroup$ You simply show the two inclusions. The union is clearly a subset of $\{X<Y\}$. On the other hand, if $\omega\in\{X<Y\}$, then we can find a $q\in \mathbb{Q}$ such that $X(\omega)<q<Y(\omega)$ and hence $\omega$ belongs to the union. Hence the two sets are equal. $\endgroup$ Feb 19, 2015 at 11:30
  • $\begingroup$ @StefanHansen I have used your hint to give the alternative prove for the case $\{X< Y\}$ in the answer below. Can you probably give me a hint how to show that $\{ X = Y \} \in \mathcal A$, then $\{ X \le Y \} \in \mathcal A$ follows automatically. $\endgroup$
    – zesy
    Feb 19, 2015 at 21:25
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    $\begingroup$ You could use that $\{X=Y\}=(X,Y)^{-1}(B)$ where $B=\{(x,y)\mid x=y\}\in\mathcal{B}^2$ since it's a closed set. As $(X,Y)$ is $\mathcal{A}$-$\mathcal{B}^2$-measurable it follows directly that $\{X=Y\}\in \mathcal{A}$. Alternatively, you could note that $\{X=Y\}=\{X\leq Y\}\cap\{Y\leq X\}\in\mathcal{A}$ since it's an intersection between two $\mathcal{A}$-measurable sets (per what you've just shown). $\endgroup$ Feb 20, 2015 at 6:47

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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 $\{X \ge Y\} \bigcap \{X \le Y\}=\{X=Y\} \in \mathcal A$.

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