# 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 $$A \subseteq \mathbb{R}$$?

• Writing both $X\in A$ and $A\in\Bbb R$ in the same formula cannot be right; if you are going to do probability or measure theory (and even if not), understanding the difference between elements and subsets is crucial. Also, like all such operators (think $\sum$ and such), $\forall$ should come before the expression where the variable it introduces is used. Commented Jun 12, 2021 at 9:44
• Small error thanks Commented Jun 12, 2021 at 11:04

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.

• So I am doing a course on measure-theoretic probability. To me, this set $\{ X \in A \}$. where $A \in \mathbb{R}$ is "measurable" because the measure $\mathbb{P}$ is applied to sets $A \in \mathcal{F}$. So as long as $\{ X \in A \} \in \mathcal{F}$, it will have a well-defined probability right? Commented Jun 12, 2021 at 7:55
• I added something in my answer, does this make sense to you? Commented Jun 12, 2021 at 8:06
• This is a good answer. I'll accept this Commented Jun 12, 2021 at 8:18