TITLE: Existence of the law of a random variable
QUESTION [15 upvotes]: Here is the definition of the law of a random variable.

Let $X$ be a random variable on $(\Omega,\mathcal{F}, \mathbb{P})$. Then, the law of $X$, denoted by $L_{X}$, is a probability measure on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$ such that for all $B \in \mathcal{B}$, $L_X(B) = \mathbb{P}(X \in B)$, where $\mathcal{B}$ denotes the Borel set of $\mathbb{R}$.

I understand the definition, but what I am wondering is that, how do we know such $L_X$ as in this definition exist? 
Can we construct one explicitly? or do we know, by some theorem, that at least one $L_X$ exists?

REPLY [26 votes]: 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 $X$".
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