Why use $\mathcal{B}(\mathbb{R})$ instead of $\mathcal{P}(\mathbb{R})$ for second sigma-algebra when defining random variables? Why do we define random variables as functions $X: (\Omega, \mathcal{F}) \rightarrow (\mathbb{R}, \mathcal{B}(\mathbb{R}))$ when we can define them as $X: (\Omega, \mathcal{F}) \rightarrow (\mathbb{R}, \mathcal{P}(\mathbb{R}))$ ?
Don't we want to eventually be able to talk about the probability of $\{X \in A\}$ for all $A \subseteq \mathbb{R}$?
 A: I think what you are proposing is what probabilists did for a long time, before probability theory had the rigorous mathematical foundation it has today. So what's the problem?
The issues are non-measurable sets, sets which don't have a measure. An example is given by the Vitali-Set.
So while we would ideally want to talk about the probability of $\{X\in A\}$ for all $A \subseteq \Bbb{R}$, this shows that this is too much to ask for. This is the reason we construct the Borel $\sigma-$Algebra, where we get rid of these problematic sets.
This may be a bit of a intuitive explanation. All these notions are usually made very precise in a course on measure theory. Does this answer your question?
Edit (in response to the comment):
Let's say you had a random Variable $X \sim \mathcal{U([0,1])}$, a random variable with continuous uniform distribution on the interval $[0,1]$. Let's say that we work on the $\sigma-$Algebra $\mathcal{P}(\Bbb{R})$ and take $V$ to be the Vitali set (as linked above). What would be $\Bbb{P}(X \in V)$? By definition we would have
$$\Bbb{P}(X \in V) = \int_V \unicode{x1D7D9}_{[0,1]}(x)dx = \int_{V \cap [0,1]}1dx. $$
But this doesn't make sense, since we are trying to integrate over a non-measurable set. So it is important to make sure to work only with sets that have a well defined (Lebesgue-)measure.
