Existence of the law of a random variable

Solution 1:

A "random variable" is by definition a measurable function $X$ from $(\Omega,\mathcal{F}, \mathbb{P})$ to $(\mathbb{R},\mathcal{B}(\mathbb{R}))$.

"Measurable" means that for every $B\in\mathcal B(\mathbb R)$, the inverse-image $X^{-1}(B)$ is a measurable set, i.e. is a member of $\mathcal F$. The inverse-image is defined as $$ X^{-1}(B) = \{ \omega\in\Omega : X(\omega)\in B\}. $$ Then we have $$ L_X(B) = \mathbb P\big( \{ \omega\in\Omega : X(\omega)\in B\} \big). $$ That certainly defines a non-negative function whose domain is $\mathcal B(\mathbb R)$. Next there is the problem of proving that that is actually a probability measure. It must assign measure $1$ to $\mathbb R$ and it must be countably additive. The first part is easy; let's look at the second. We have $$ L_X(B_1\cup B_2\cup B_3\cup \cdots) = \mathbb P\left( \{ \omega\in\Omega : X(\omega)\in B_1\cup B_2\cup B_3\cup \cdots\} \right). $$ If $X(\omega)\in B_1\cup B_2\cup B_3\cup \cdots $ then for some $i$, $X(\omega)\in B_i$, so $\omega \in X^{-1}(B_i)$ so $\omega \in X^{-1}(B_1)\cup X^{-1}(B_2)\cup X^{-1}(B_3)\cup\cdots$. Similarly if $\omega$ is in that latter set it's easy to show $X(\omega)\in B_1\cup B_2\cup B_3\cup \cdots $. Hence we have $$ \{ \omega\in\Omega : X(\omega)\in B_1\cup B_2\cup B_3\cup \cdots\} = \bigcup_{i=1}^\infty \{\omega\in\Omega : X(\omega) \in B_i \}. $$

If the $B$s are pairwise disjoint then $X(\omega)$ cannot be in both $B_i$ and $B_j$ for $i\ne j$, so $\omega$ cannot be in both $\{\omega\in\Omega : X(\omega) \in B_i \}$ and $\{\omega\in\Omega : X(\omega) \in B_j \}$, so those are pairwise disjoint as well. Then we can use countable additivity of $\mathbb P$ to show countable additivity of $L_X$.

"Law of $X$" is synonymous with "probability distribution of $X$".