# Same $\sigma$-field of different sets

I have a probability measure space with random variables $$X,Y:\Omega \to \mathbb{R}$$

Question is how can is show that the sets : $$A = \{\omega \in \Omega:X(\omega) =Y(\omega)\}$$,

$$B = \{\omega \in \Omega:X(\omega) >Y(\omega)\}$$

$$\Gamma = \{\omega \in \Omega:e^{X(\omega)} =Y(\omega)\}$$ belong in $$\sigma$$-field $$\mathcal{F}$$

My effort is the below but I am not sure about $$\Gamma$$

For every $$t \in \mathbb{R}$$ we have : $$\bigcap_{n=1}^{\infty} \{ \omega: X(\omega)\leq t\}= \{ \omega:\sup X (\omega) \leq t\} =\{\omega: X_{n} (\omega) \leq t \}$$ for every $$n \in \mathbb{N} = \bigcap_{n=1}^{\infty} \{\omega : X_{n}(\omega) \leq t \}$$.

But $$X_{n},n \in \mathbb{N}$$ is a random variable and the class $$\mathcal{F}$$ is a $$\sigma$$ - field then : $$\bigcap_{n=1}^{\infty} \{ \omega : X_{n}(\omega) \leq t \} \in \mathcal{F}$$

The same for $$\{\omega:Y(\omega)>t \}= \{\omega:X_{n}(\omega)>t \}$$ for every $$n=\mathbb{N}=\bigcap_{n=1}^{\infty}\{ \omega : X_{n}(\omega) >t = \bigcap_{n=1}^{\infty} \{\omega : X_{n}(\omega) \leq t \}^{c} \in \mathcal{F}\}$$ and then $$\{\omega :Y(\omega) \leq t \} \in \mathcal{F}$$ for every $$t \in \mathbb{R}$$

For $$\{\omega:Y(\omega)= e^{t} \}= \{\omega:X_{n}(\omega) =e^t \}$$ for every $$n=\mathbb{N}=\bigcap_{n=1}^{\infty}\{ \omega : X_{n}(\omega) \geq e^t = \bigcap_{n=1}^{\infty} \{\omega : X_{n}(\omega) \leq e^t \}^{c} \in \mathcal{F}\}$$ and then $$\{\omega :Y(\omega) \leq e^t \} \in \mathcal{F}$$

• You have used symbols that you did not define. What is your $X_n$? $e^{X}$ is also a random variable so there is no need to prove separately that $\Gamma \in \mathcal F$. – Kavi Rama Murthy Jan 4 at 5:22

I understand that you have a a probability measure space $$(\Omega, \mathcal{F}, P)$$ and random variables $$X,Y:\Omega \to \mathbb{R}$$.

You want to prove that the sets : $$A = \{\omega \in \Omega:X(\omega) =Y(\omega)\}$$ $$B = \{\omega \in \Omega:X(\omega) >Y(\omega)\}$$ $$\Gamma = \{\omega \in \Omega:e^{X(\omega)} =Y(\omega)\}$$ belong in $$\sigma$$-field $$\mathcal{F}$$.

There is a simple way prove those results. Let us see it.

Proof: Note that $$X-Y$$ is a random variable, so $$(X-Y)^{-1}(\{0\}) \in \mathcal{F}$$. Since $$A = \{\omega \in \Omega:X(\omega) =Y(\omega)\} = (X-Y)^{-1}(\{0\})$$ we have that $$A\in \mathcal{F}$$.

In a similar way, since $$X-Y$$ is a random variable, so $$(X-Y)^{-1}((0,+\infty))\in \mathcal{F}$$. Since $$B = \{\omega \in \Omega:X(\omega) >Y(\omega)\} = (X-Y)^{-1}((0,+\infty))$$ we have that $$B\in \mathcal{F}$$.

Finally, note that, since $$X$$ is a random variable, so is $$e^X$$. So just apply the result for $$A$$ to have that $$\Gamma \in \mathcal{F}$$.

• Where $(e^X -Y)^{-1}\in \mathcal{F}$ – Mr.Podilatis Jan 4 at 16:13
• @Mr.Podilatis , If you don't want to simply reduce the case for $\Gamma$ to the first case, you can prove it directly as follows. Since $X$ is a random variable, $e^X$ is a random variable. Since $Y$ is a random variable, $e^X-Y$ is a random variable. Since $e^X-Y$ is a random variable, $(e^X-Y)^{-1}(\{0\}) \in \mathcal{F}$. But, $$\Gamma = \{\omega \in \Omega:e^{X(\omega)} =Y(\omega)\} = (e^X-Y)^{-1}(\{0\})$$ So $\Gamma \in \mathcal{F}$. – Ramiro Jan 4 at 16:23
• I appreciate it mate thank you – Mr.Podilatis Jan 4 at 16:24