probability - How to Calculate CDF of Random Variable Depends on $omega$

Suppose that $X$ is a random variable defined as such where $\Omega = (0,1]$

$$X(\omega)= \begin{cases} \frac{1}{3}&\, 0 < \omega \leq \frac{1}{3}\\ \omega &\, \frac{1}{3}\leq \omega\leq \frac{2}{3}\\ \omega^2&\, \frac{2}{3} < \omega\leq 1\\ \end{cases}$$

I am not sure how or if it is possible to assign a CDF (or PDF) to this Random Variable. It seem to me that this mapping is not unique? Do we assume that $\omega$ is uniform?

For $E(X)$ I calculated $$E(X)=\int_{0}^{\frac{1}{3}}\frac{1}{3}\space +\int_\frac{1}{3}^\frac{2}{3}x\space + \int_\frac{2}{3}^1x^2$$

Is this correct? Or even going down the right path? Thanks

Answer

First we have to check the continuity. As we can see X is not continuous at 2/3 so CDF of this does not exist. And for expectation you have to integrate Xf(X) not only f(X).