Problem to understand random variable : For example, $X(\omega )=1$ is really random ? I have difficulty to understand random variable. Let $(\Omega ,\mathcal F,\mathbb P)$ a probability space. Let say $\Omega =[0,1]$, $\mathbb P=m$ the lebesgue measure and $\mathcal F=\mathcal B_{[0,1]}$ the Borel $\sigma -$algebra. Let take for example, $X(\omega )=1$ or $X(\omega )=\omega $. In what is it random ? If I fix $\omega \in \Omega $, then $X(\omega )=\omega $ is know, and not unknow. 
Other example, if I want a number in $[0,1]$. So, $X(\omega )=\omega $ should work. So if I fix $\omega $, I know $X(\omega )$... why is it random ? Also, if $Y$ is a uniform r.v. in $[0,1]$ don't we have that $X(\omega )=Y(\omega )=\omega $ a.s. ? 
 A: Here's a perspective that was imparted to me by my thesis advisor: You have a function (i.e. $X$, the random variable) on a domain (i.e. $\Omega$, the sample space). It behaves just like all other functions you've ever encountered behave. The catch, though, is that you never really pick an individual $\omega$ and evaluate $X(\omega)$. Instead, Tyche, the Greek goddess of fortune, chooses an input $\omega$ for you, and she will make her choices according to the measure $\mathbb P$. Your task is to describe broadly what will happen to the outputs $X(\omega)$. 
You're quite right that if you fix a particular $\omega \in \Omega$, then nothing is really random; you're just evaluating a function. The "random" component is the part where you surrender control over which particular $\omega$ you choose back to Tyche.

Also, if Y is a uniform r.v. in [0,1] don't we have that X(ω)=Y(ω)=ω a.s. ?

Not necessarily; it depends on how $X$ and $Y$ map $\Omega$ onto the real number line. Consider this example, where the sample space is $\Omega = [0, 1]$ and the probability measure is the Borel / Lebesgue measure:
$$X(\omega) = \begin{cases} \omega, & 0 \leq \omega \leq 1 \\ 0, &\text{otherwise} \end{cases}$$
$$Y(\omega) = \begin{cases} 1-\omega, & 0 \leq \omega \leq 1 \\ 0, &\text{otherwise} \end{cases}$$
Both $X$ and $Y$ have an equally legitimate claim to the title "uniformly-distributed random variable on $[0, 1]$". Consider why; your only concern is the probability of $X$ or $Y$ being in particular regions after Tyche has chosen her input. The essential characteristic is that if you choose $a, b$ such that $0 \leq a \leq b \leq 1$, that $\mathbb P(a \leq X \leq b) = b-a$. It is also true that $\mathbb P(a \leq Y \leq b) = b-a$. Yet, it also happens to be true that $\mathbb P(X = Y) = 0$. Here's a third variable that has the same property:
$$Z(\omega) = \begin{cases} \omega, & 0 \leq \omega \leq 1 \text{ and } \omega \not \in \mathbb Q \\ 0, &\text{otherwise} \end{cases}$$
Here, $Z$ again has just as much claim to be a "uniformly distributed variable on $[0, 1]$" as $X$ or $Y$ do. This function is not terribly well-behaved, since it has infinitely many discontinuities. Yet, all we care about are the distributions after $Z$ has done its mapping.
How $X$ maps the particular sample spaces onto $\mathbb R$ is not relevant; to characterize a random variable, your only concern is to characterize statements like $\mathbb P(a \leq X \leq b)$. There are many ways to accomplish this; an equivalent one would be to successfully characterize all statements like $\mathbb P(X \leq c)$. But the point is that you don't care about how the actual inputs are structured, and you don't care about how exactly they're mapped onto the reals by $X$; you only care about things like $\mathbb P(X \in A) = \mathbb P(\omega: X(\omega) \in A)$.
