Is the distribution of a random variable always a probability measure? Let $\mathbb{P}X^{-1}$ denote the distribution of the random variable $X$.
For example, can I denote the distribution of a normal random variable as a probability measure?
[Math] Is the distribution of a random variable always a probability measure
measure-theoryprobability theory
Related Solutions
A (real valued) random variable is just a measurable map $X : \Omega \to \Bbb{R}$, where $(\Omega, \mathcal{F}, \Bbb{P})$ is an arbitrary probability space.
What we can then do is to consider the push-forward measure $\Bbb{P}_X = X_\ast \Bbb{P}$ of $\Bbb{P}$ by $X$. This is sometimes called the distribution of $X$. By definition, we have
$$ X_\ast \Bbb{P} (E) = \Bbb{P}(X^{-1}(E)) = \Bbb{P}(X \in E), $$
for any (measurable) $E \subset \Bbb{R}$, so that (check this) $\Bbb{P}_X$ is a probability measure on $\Bbb{R}$. Note that the last expression is the one that most mathematicians in probability theory would use.
Now - as you already stated yourself - we can associate to every (locally finite) measure $\mu$ on $\Bbb{R}$ the distribution function $F = F_\mu$ of $\mu$, given by
$$ F_\mu (x) = \mu((-\infty, x]). $$
In this way, we can also associate to the measure $\Bbb{P}_X$ the distribution function $F_X = F_{\Bbb{P}_X}$ which satisfies
$$ F_X (a) = \Bbb{P}_X ((-\infty, a]) = \Bbb{P}(X \in (-\infty, a]) = \Bbb{P}(X \leq a). $$
Sometimes, this is also called the distribution of $X$ (note that we now call the measure $\Bbb{P}_X$ and it's distribution function $F_X = F_{\Bbb{P}_X}$ the "distribution of $X$". But as each of these two objects uniquely determines the other, this is not much of a problem).
Finally, all this has not much to do with the properties of $X$ as a function (i.e. with properties like continuity of $X$, ...). To see this, note that $\Omega$ is an arbitrary probability space. Hence, it does not make sense in general to talk about continuity of $X$, for example.
There is a different notion of a continuous random variable. Here, we call $X$ a continuous random variable, if the distribution function $F_X$ is continuous. This is equivalent to the condition $\Bbb{P}(X = a) = 0$ for all $a$ (why?) and thus has nothing to do with continuity of $X$ as a function (as above, this concept does not even make sense in general).
Short summary:
1) Each real-valued random variable comes with it's own cumulative distribution function. If we place additional assumptions on $X$, then it might be the case that this distribution function is given by the one associated to Lebesgue-measure. Note that we have to restrict Lebesgue-measure to (e.g.) an interval of length $1$ to do this, because otherwise this is no probability measure.
2) As explained above, the associated CDF is given by
$$ F_X (a) = \Bbb{P}(X \leq a). $$
Yes, your constructions make perfect sense and this is in fact the systematic way to do it. And it is very easy to adjust to a more general case, where the distribution is not necessarily discrete.
Consider a probability distribution $\mu$ on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$. Let $(\Omega, \mathcal{A}, P) = (\mathbb{R},\mathcal{B}(\mathbb{R}),\mu)$ and let $X := id$. It is quite easy to see, that $P_X = \mu$, since $P(X\in A) = P(X^{-1}(A))= P(A) = \mu(A)$.
We could even generalize to random vectors. So let's say we want to construct a probability space, where $X\sim \mu$ and $Y\sim \nu$ and $X,Y$ are independent. Then we could choose $(\Omega, \mathcal{A}, P) = (\mathbb{R^2},\mathcal{B}(\mathbb{R^2}),\mu \otimes \nu)$, where $\mu \otimes \nu$ is the product measure. Here we let $X(x,y) = x$ and $Y(x,y)=y$ and it is easy to check, that $P_{(X,Y)} = \mu \otimes \nu$, which shows that $X$ and $Y$ are independent with distribution $\mu$ and $\nu$.
Much more generally the construction can be extended to infinitely many (even uncountably many) random variables. This result is known as Kolmogorovs Consistency theorem: https://en.wikipedia.org/wiki/Kolmogorov_extension_theorem
For example the theorem guarantees the existence of sequences $X_1,X_2,X_3,...$ of $i.i.d.$ variables with some distribution $\mu$.
Best Answer
Yes, by definition, the distribution of a real random variable $X$, which is a measurable function on $(\Omega,\mathcal{F},\mathbb{P})$, is the probability measure $\mathbb{P}_X$ defined on the measurable space $(\mathbb{R},B(\mathbb{R}))$ (with $B(\mathbb{R})$ the Borel sigma-algebra) by $\mathbb{P}_X(A)=\mathbb{P}(X\in A)$. So we can write that $\mathbb{P}_X=\mathbb{P}\circ X^{-1}$.
It is easy to show that it defines a probability measure on $(\mathbb{R},B(\mathbb{R}))$. It is actually a special case of a pushforward measure (see https://en.wikipedia.org/wiki/Pushforward_measure ).
Note that if $X$ is not real but $X$ has its values on a more general measurable space $(S,\mathcal{S})$, the definition is the same, and the distribution of $X$ will be a probability measure defined on $\mathcal{S}$.