Let $X$ be a random variable with c.d.f $F_X$ and probability law $\mathbb{P}^X$ on $\mathbb{R}$. Let $D$ be the countable set of discontinuities of $F$ and suppose that $D$ is non-empty. This implies that for some points $x$ we have $$\mathbb{P}^X(\{x\})=\mathbb{P}[X=x]>0.$$
What is a correct way of defining a density function $f_X$ for $X$? Ie we need a function $f_X$ such that
$$\int_{-\infty}^\infty f_X(x)d\mathbb{P}^X(x)=1,$$
and that resembles both a p.d.f and a p.m.f.
I think when $D$ is finite or has all isolated points we can define this function easily. However, what if $D$ contains an accumulation point? Or even worse, what if $D$ is dense?
Best Answer
The simple answer is you don't, or rather you can't. Not all random variables have a PDF. In fact a real valued random variables has a density if and only if its CDF is absolutely continuous (this is a stronger condition than continuity).
Distributions are a generalisation of functions, and you can consider every measure as a distribution. A distribution $T$ is a linear function $C_{\infty}(\mathbb R) \to \mathbb R$ (which satisfies some continuity conditions, but that's not important for this discussion). For example the Dirac delta distribution can be defined by $\delta_0(f) = f(0)$.
All probability measures can be considered as a distribution, if $\mu$ is a probability measure then we can turn it into a distribution by defining $$\mu (f) = \int_{\mathbb R} f(x) \mu(\text{d} x). $$
CDFs are always just normal functions, the function $F(x) = \mu((-\infty, x])$ is always well defined after all. What we can't guarantee is that the CDF will have a corresponding density