Problem with Probability Density Function definition I have a problem with the definition of probability density function (PDF).
Usually this concept is defined in terms of a given distribution function, while I would like to know if it is possible to define the concept in one shot (i.e. for both the discrete and continuous case) without passing through cdf.
Thus,...

... can we say that a PDF is any function $f : \mathbb{R} \to [0 ,1]$ that satisfies the following two basic requirements?

*

*$f \geq 0$,

*$\int f d \lambda = 1$, where $\lambda$ is the Lebesgue measure on $\mathbb{R}$.


If this is correct, does this definition encompass in one shot both the discrete and continuous case (thanks to the Lebesgue integration)?
I would say no, because condition (2) should be ill-defined for the discrete case, because it is based on the Lebesgue measure according to which every point has measure zero.

*

*Is this last intuition correct (which implies that we need to explicitly add a third condition with summation instead of integration to deal with the discrete case), or my intuition is simply wrong?

*If it is wrong, what am I missing?

As always, any feedback is enormously appreciated.
Thank you for your time.
 A: Usually you have given a probability space $(\Omega,\Sigma,\mathbb{P})$ and for a random variable $X$ you call $f$ PDF according to the Lebesuge measure $\lambda$ if 
$$ \mathbb{P}[X\in A] = \int_A f(x) \, dx $$ holds for all Borel sets $A$ and if $f \geq 0$. 
I think you are trying to characterize all possible PDF's according to the Lebesgue measure. Your definition covers almost all PDF's but your requirement $f: \mathbb{R} \to [0,1]$ is a too big restriction. There are many PDF that are point wise higher then 1. 
However, you can say:

A function $f: \mathbb{R} \to [0,\infty)$ is a PDF according to the Lebesgue measure if and only if
  $\int_{\mathbb{R}} f(x) \, dx = 1$


Now to your second question:

If this is correct, does this definition encompass in one shot both
  the discrete and continuous case (thanks to the Lebesgue integration)?

When I understand you correctly, then you are asking if this definition also includes the PDF according to the Lebesgue measure for discrete random variables. However, discrete random variables have no PDF according to the Lebesuge measure. 
Example: 
Consider a random variable with
$$\mathbb{P}[X=2]=1$$
X has no PDF according to the Lebesgue measure, since $\{2\}$ is a null set. However, if you consider the measure $(\mu, \mathcal{B})$ with $\mu(2) = 0.5$ and $\mu(3) = 0.5$ then $X$ has a PDF according to $\mu$, namely $$f(x) = \begin{cases} 2 & \text{if}\, x=2\\ 0 & else \end{cases}$$.
If you are looking for a characterization of PDF's for arbitrary measures you might find the following interesting: 


*

*The PDF of a random variable $X$ according to a measure $\mu$ is usually called Radon–Nikodym derivative and denoted as $ \frac{dX}{d\mu}$.

*The Radon–Nikodym theorem states, a random varaible $X$ has a Radon–Nikodym derivative $ \frac{dX}{d\mu}$ if and only if $\mu(A)=0$ implies $\mathbb{P}[X\in A] =0$ for all measurable sets $A$.

Heres a definition of  a PDF including discrete and continuous case:
Given a probability space $(\Omega,\Sigma,\mathbb{P})$, a measure space $(E, \mathcal{A},\mu)$ and for a random variable $X\colon \Omega \to E$ you call $f\colon E \to [0,\infty)$ PDF from $X$ according to the measure $\mu$ if 
$$ \mathbb{P}[X\in A] = \int_A f(x) \, \mu( dx) $$ holds for all $A \in \mathcal{A}$ and $f\geq 0$. 
A: Based on your comments, I'm slightly concerned that I'm not following your question.  It sounds to me like you're asking for a general way to describe all CDFs in terms of Lebesgue integration.  I think the result you're looking for is the Lebesgue Decomposition Theorem which says that any probability measure $\nu$ can be decomposed as 
$$
\nu = \nu_{ac} + \nu_d + \nu_{cs}
$$
where $\nu_{ac}$ is absolutely continuous with respect to Lebesgue measure ( and therefore has a density function, the Radon-Nikodym derivative ), $\nu_d$ is discrete ( and therefore has a "mass function" which assigns weights to a countable subset ), and $\nu_{cs}$ is continuously singular, which is the weirdest one, and says that $\nu_{cs}(x)=0$ and $\nu_{cs}(A)=\nu_{cs}(X)$ for some Borel set $A$ which has Lebesgue measure zero.  
For an example of the last one, consider the Cantor function as your CDF.  In particular, it's not enough to just add another condition for the discrete case because of this continuously singular case.
