Definition of probability density

Question

I know that every probability density function (pdf) must be nonnegative and integrate one on $\mathbb R$.

But the definition of pdf requires a distribution function $F$: "we say that a function $f$ is the pdf associated to $F$ if $F(x)=\int_{-\infty}^x f(t)dt$".

This makes me think if there are functions $f$ such that $f\geq 0$ and $\int_\mathbb R f(t)dt=1$ but are not a pdf of any distribution function. Probably, people gave this definition for a good reason.

Detailed question If $X$ is a real absolutely continuous random variable on $(\Omega,\mathcal A,P)$, then its probability distribution $P_X$ is absolutely continuous w.r.t. the Lebesgue measure. Then Radon-Nikodym theorem guarantees that there is a function $f$ such that $P_X((-\infty,c])=\int_{-\infty}^c f(t)dt$. This function $f$ is called probability density function, and satisfies $f\geq 0$ and $\int f(t)dt=1$. Now I ask you, is there a function $g$ such that $g\geq 0$, $\int g(t)dt=1$ and is not a probability density function of any absolutely continuous random variable? If I check that $g\geq 0$ and $\int g(t)dt=1$, I can automatically call $g$ a probability density function?

Answer 1

If a probability space $(\Omega,\mathcal{A},\mathbf{P})$ is rich enough to simulate some absolutely continuous distribution, then it can simulate every distribution on $\mathbb{R}$.

We can argue this using the following two separate results:

Theorem 1. Assume that a random variable $X$ has an absolutely continuous distribution. Then $$U = F_X(X)$$ has the uniform distribution on $[0, 1]$.

Theorem 2. (Inverse Transform Sampling) Let $U$ be a random variable having the uniform distribution on $[0, 1]$. Also, let $\mu$ be a probability measure on $\mathbb{R}$, and let $F^{-1}$ denote the inverse CDF corresponding to $\mu$. Then $$Y = F^{-1}(U)$$ has the distribution $\mu$.

As a corollary, every Borel-measurable function $f:\mathbb{R} \to [0, \infty)$ with $\int_{\mathbb{R}} f(x) \, \mathrm{d}x = 1$ is a density of some distribution, because $\mu(A) = \int_A f(x) \, \mathrm{d}x$ defines a probability measure on $\mathbb{R}$ with the density $f$.