The definition is as following according to the book of John B. Walsh,
Let $(\Omega, \mathbb{F}, P)$ be a probability space. A Random Variable is a real-valued function X on $\Omega$ such that for all $x \in \mathbb{R}, \{\omega:X(\omega) \leq x\} \in \mathbb{F}$
my simple question is why the definition limits for each event of R.V. to be less than or equal a real number x? (I just want to know the intuitive of the $X(\omega) \leq x$).
In the book he argues that when you want to prove say X + Y (where X and Y both are R.V.) first we prove that $\{\omega : X(\omega) + Y(\omega) > x\} = \bigcup_{r \in \mathbb{Q}}\{\omega : X(\omega) > r, Y(\omega) > x – r\}$. Then verify that this is in F, and conclude that $x \in \mathbb{R}, {\omega:X(\omega) + Y(\omega) \leq x} \in \mathbb{F}$. Why he could go to prove that the R.V. of X and Y is greater than a real number x, then he concludes that they are less than or equal to a real number x? (This make me confuse about what is behind the definition of R.V.)
R.V. : Random Variable.
Best Answer
The random variable is not limited to values less than $x$. For instance, I can show you that the function $$ X(t) = \frac{1}{t} $$ is a measurable function on $(0, 1)$. Here's how. Let's look at
$$ \{ t \mid X(t) \le 11 \} $$ That's the set of all points in the domain for which $X(t) = 1/t$ is less than 11, which is exactly $$ A = \{t \mid \frac{1}{11} \le t < 1 \} $$
Clear enough? Now look at $A$. Is it a measurable set in the measure space $(0, 1)$? Sure. It's a half-open interval, and those are all measurable sets!
Now there was nothing special about the number 11: I could have chosen $6$ or $113$ or $-12$ (although if I'd chosen $-12$, then the set $A$ woudl have been empty. Fortunately, the empty set is measurable, too!
Now go back and look at the definition above: it says "whenever you build a set like $A$, it's measurable." It does not say that the function $X$ is bounded. Indeed, in my case, we have $$ X(0.1) = 10 \\ X(0.01) = 100 \\ X(0.001) = 1000 $$ and so on, and it's pretty clear that $X$ itself is an unbounded function on $(0, 1)$.