A random variable $X$ is a measurable function from a probability space $(\Omega, \mathcal{F}, \mathbb{P})$ to the reals:
$X: \Omega \to (-\infty, \infty)$
such that for every Borel set $B$,
$X^{-1}(B)=\{\omega \in \Omega:X(\omega) \in B\} \in \mathcal{F}$
shorthand notation: $\{X \in B\}=\{\omega \in \Omega:X(\omega) \in B\}$
in other words, the fact that value of $X$ belongs to a given Borel has to be an event (has to belong to the set of events $\mathcal{F}$).
Is the probability density function unique for every random variable $X$ in the probabilty space $(\Omega, \mathcal{F}, \mathbb{P})$?
I think the answer is yes, because $\mathbb{P(}{a \le X \le b})$ is defined by the probability function $\mathbb{P}$.
$a \le X \le b$ is an event belonging to $\mathcal{F}$, and $\mathbb{P}$ defines what are probabilities of events in $\mathcal{F}$.
Best Answer
No.
Let it be that $f,g:\mathbb R\to[0,\infty)$ are Borel-measurable functions with: $$\lambda\{x\in\mathbb R\mid f(x)\neq g(x)\}=0$$
where $\lambda$ denotes the Lebesgue measure.
If $f$ serves as PDF for random variable $X$, then so does $g$.
This because for each Borel-measurable set $A$: $$\mathbb P(X\in A)=\int_Afd\lambda=\int_Agd\lambda$$