# Law of Random Variable is Always a Measure?

I have a conceptual question regarding the law of a random variable in probability theory.

Suppose we begin with a probability space $$(\Omega,F,P)$$. We also endow a measurable space of reals with its Borel $$\sigma$$-algebra, $$(\mathbb{R},B(\mathbb{R}))$$.

Given $$X$$ is a real-valued random variable, we define the distribution measure of law of $$X$$ as

$$P_X(B)=P(X^{-1}(B))=(P\circ X^{-1})(B)-P(\omega:X(\omega)\in B)=P(X\in B).$$

Then, the law of $$X$$ is entirely characterized by its cumulative distribution function such that:

$$F_X(x)=P_X((\infty,x])=P(X\leq x).$$

My understanding is:

$$P:F\rightarrow [0,1]$$.
$$P_X:B(\mathbb{R})\rightarrow[0,1].$$

Hence, the law of $$X$$ is a probability density on the measruable space $$(\mathbb{R},B(\mathbb{R})).$$

(1) Does this mean any two random variables that are from distinct distributions have their own probability distribution on the space $$(\mathbb{R},B(\mathbb{R}))$$? In other words, is there distinct laws for each distinct random variable?

(2) A random variable is just a measurable function from one measurable space to another, say $$(E,A),(F,B)$$. Is it possible for the law of $$X$$ not be a non-negative and $$\sigma$$-finite measure defined on $$(F,B)$$. In other words, does the way we define and characterize the law of $$X$$ in probability have to do with some "nice" properties of reals or the Borel sigma algebra of reals?

(1) Yes, by definition $$X\stackrel{d}{=} Y$$ ($$X$$ is equal to $$Y$$ in distribution) iff the laws of $$X$$ and $$Y$$ coincide, that is $$P_X = P_Y$$.
But it is possible that $$X \neq Y$$ yet $$P_X = P_Y$$. For example, if $$X\sim\mathcal{N}(0,1)$$, then $$-X \sim \mathcal{N}(0,1)$$ as well but $$X \neq -X$$.
(2) A random variable is by definition a measurable function $$X:(\Omega, \mathcal{F}) \to (E, \mathcal{E})$$ where $$(\Omega, \mathcal{F}, P)$$ is a probability space. In this case, the law of $$X$$ is a measure (this is a special case of image measure)
$$P_X: \mathcal{E} \to [0,1]$$
and in particular, $$P_X$$ is positive and finite, so also $$\sigma$$-finite.