Let $$X(\omega ):=\mathbb{1}_{B}(\omega )$$ be a random variable on a probability space $\left(\Omega=[0,1], \mathcal{F}, P\right)$ where $B=\{[0,1] \bigcap \mathbb{Q}\}$.
Does X($\omega$) admit a density?
I understand that it must hold $$P\{X\in A\}=\mathbb{E} \left[f \ \mathbb{1}_{A}\right],\ \forall A $$ in order $\forall X$ random variable to have a density.
It seems to me that in this case $$P\{X\in A\}= \begin{cases} 1\ \ if\ 0\in A \\ 0 \ \ if\ 0\not\in A \end{cases}.$$Am I correct up to here?
Then for $f$ simple function, the density could not exist because $f$ would always have measure $0$ on $B$ defined above, while $P\{X\in B\}=1$ as a consequence of $0 \in B$. Does it make sense? Then what about a general $f$?