Showing a probability space I have recently been exposed to probability measure theory and found an exercise in my textbook which I am not able to solve.
I have to show that ($\mathbb R, \mathscr B(\mathbb R), \mu_X $) is a probability space, where $\mathscr B(\mathbb R)$ represents the Borel algebra of $\mathbb R$ and $\mu_X$ is defined as a function $\mu_X(B) = \mathbb P(X^{-1}(B)), \forall B\in \mathscr B (\mathbb R) $.
It gives a hint of using the fact that $\mathbb P$ is a probability measure and the definition of a random variable, which I would define as a function $X: \Omega \rightarrow\mathbb R $.
Can anyone help me with this and show me how to go about proving this? Thanks!
 A: In order to show that $\mu_X$ is a probability measure, we first have to check that it is indeed a measure. Since the inverse image by $X$ of the empty set is the empty set, it suffices to show the $\sigma$-additivity. Let $\left(B_i\right)_{i\in\mathbb N}$ be a collection of pairwise disjoint Borel sets. Noticing that $\left(X^{-1}B_i\right)_{i\geqslant 1}$ is also pairwise disjoint and that $\mathbb P$ is a measure, we get that 
$$
\mu_X\left(\bigcup_{i\in\mathbb N}B_i\right)=\mathbb P\left(\bigcup_{i\in\mathbb N}X^{-1}B_i\right)=\sum_{i\in\mathbb N}\mathbb P\left(X^{-1}B_i\right)=\sum_{i\in\mathbb N}\mu_X\left(B_i\right).
$$
That $\mu_X(\mathbb R)=1$ follows from $X^{-1}(\mathbb R)=\Omega$.
A: 
which I would define as a function $X: \Omega \rightarrow \mathbb R $.

To be fastidious one would say that it is a measurable function, and that means that for every Borel set $B\subseteq\mathbb R,$ the set
$$
X^{-1}(B) = \{\omega\in\Omega: X(\omega)\in B\}
$$
is a measurable subset of $\Omega.$ In defining a probability space whose underlying set is $\Omega,$ one would have specified a $\sigma$-algebra of subsets of $\Omega,$ which are its measurable subsets, and a probability measure $\mathbb P$ whose domain is that $\sigma$-algebra.
One needs to show that for every sequence $B_1,B_2,B_3,\ldots$ of pairwise disjoint Borel sets, one has
$$
\mu_X\left( \bigcup_{n=1}^\infty B_n \right) = \sum_{n=1}^\infty \mu_X(B_n).
$$
One also needs $\mu_X(\mathbb R)=1,$ and that is shown as follows: $\mu_X(\mathbb R) = \mathbb P\{\omega\in\Omega: X(\omega)\in \mathbb R\} = \mathbb P(\Omega) = 1.$
We have
$$
\mu_X\left( \bigcup_{n=1}^\infty B_n\right) = \mathbb P\left( \left\{\omega\in\Omega: X(\omega) \in \bigcup_{n=1}^\infty B_n \right\} \right). \tag 1
$$
If $i\ne j$ then $B_i\cap B_j=\varnothing,$ so for $\psi\in\Omega,$ the two propositions $X(\psi)\in B_i$ and $X(\psi)\in B_j$ cannot both be true. And therefore the two statements $\psi \in \{\omega\in\Omega: X(\omega)\in B_i)\}$ and $\psi \in \{\omega\in\Omega: X(\omega)\in B_j)\}$ cannot both be true. Thus that sequence of subsets of $\Omega$ is pairwise disjoint.
\begin{align}
& X(\psi) \in \left\{ \omega\in\Omega: X(\omega) \in \bigcup_{n=1}^n B_n \right\} \\[8pt]
\iff & X(\psi) \in\bigcup_{n=1}^\infty B_n \\[8pt]
\iff & \exists n\,\, X(\psi)\in B_n \\[8pt]
\iff & \exists n\,\, X(\psi)\in \{\omega\in\Omega: X(\omega)\in B_n\} \\[8pt]
\iff & X(\psi) \in \bigcup_n \left\{ \omega\in\Omega: X(\omega) \in B_n \right\}. \\[8pt]
\text{Therefore }  & \left\{ \omega\in\Omega: X(\omega) \in \bigcup_{n=1}^n B_n \right\} \\[8pt]
& {} \qquad {} = \bigcup_{n=1}^\infty \left\{ \omega\in\Omega: X(\omega) \in B_n \right\}. \tag 2
\end{align}
Now bring lines $(1)$ and $(2)$ together and use countable additivity of the measure $\mathbb P.$
