What exactly is a probability distribution, a Radon-Nikodym derivative of a CDF function?

Question

Let's call a function $F: \mathbb{R} \rightarrow \mathbb{R}$ a CDF if it satisfies the following properties:

1. $\lim_{x\to -\infty} F(x) = 0$
2. $\lim_{x\to\infty} F(x) = 1$
3. $F(a) \le F(b)$ for all $a<b$ (never decreasing)
4. $F$ is right-continuous.

(I'm pretty sure that this is the standard way to define a CDF, but please tell me if I've missed a property.)

If we have a continuous CDF $F$, there there is some $f$ such that integrating $f$ gives $F$, and $f$ is a continuous distribution. Ditto for sums and discrete $f$.

However, what is the way to represent both of these at once (or even a distribution mixed between continuous and discrete)? What I am reading is that this has to do with the Radon-Nikodym derivative, which looks like it turns into the regular derivative in nice cases but gives the generality to define $f$ for any $F$ satistfying the CDF properties. However, I am struggling with the idea of absolute continuity. That seems to break down when the distribution is discrete, so perhaps the Radon-Nikodym derivative isn't quite the special sauce I seek.

In the continuous case, I would write $\int_{- \infty}^b f_X(x)dx = \int_{- \infty}^b dF_X(x) = F_X(b)$. Though I am not totally sure what the "$dF_X(x)$" part means (is that related to Radon-Nikodym?), the left and right sides of the equation are equal.

There must be a way to represent the distribution $f$ for a general CDF $F$, I figure, even if $f$ is defined implicitly by (Lebesgue) integration.

I am mostly interested in the univariate case (for now) to deal with a zero-inflated distribution. There's a discrete component to my distribution (the 0s), but then the rest is continuous.

Thanks.

Answer 1

What you are asking, actually, is whether every continuous probability distribution has a density. (That's the name for your $f$). The answer is negative. You can find here some useful information.