Why do continuous random variables have densities?

Question

In a basic probability course, there seems to be an implicit assumption that any continuous random variable has a density function, but I don't really see why this should be. My intuition is that there's some probability space $(\Omega, \Sigma, P)$ and a measurable function $X : \Omega \to \mathbb{R}$, and you'd like to know the probability that $X$ takes on a value in some measurable $E \subseteq \mathbb{R}$. The natural thing to do is look at the push-forward measure, $X_*P(E) = P(X^{-1}(E))$, but in any introductory probability book they make some claim that there's a density function $f$ such that this probability is $\int_E f(x) dx$, which seems to mean $X_*P$ is absolutely continuous with respect to Lebesgue measure, but I don't really see why this should be true in general.

I suppose this should mean that if $E \subseteq \mathbb{R}$ has zero Lebesgue measure, then $X^{-1}(E)$ has has zero $P$-measure, but again I don't see why this needs to be true for an arbitrary measurable function $X$.

Is this just an assumption that introductory books make, or is there something else going on that I'm not seeing?

Answer 1

You might want to read the Wikipedia section on this. In short, you are right: a continuous distribution need not be absolutely continuous with respect to Lebesgue measure (the Cantor distribution is given as an example in my previous link).

Most of your post is beyond the scope of an introductory course in probability, where most continuous distributions that are studied are ones with densities.

Answer 2

You're right, a "continuous" random variable need not have a density. The actual story is that if $\mu$ is any Borel measure on $\Bbb R$ with $\mu(\Bbb R)=1$ there is a random variable $X$ such that $$P(X\in E)=\mu(E).$$Now $X$ has a density if and only if $\mu$ is "absolutely continuous".

So the actual story gets into measure theory, which is why they can't tell the actual story in an elementary class. Here's a simple "probabilistic" construction of a random variable with no density function:

Say $d_0,d_1,d_2,\dots$ are iid random variables with $P(d_j=0)=P(d_j=2)=1/2$. Let $$X=\sum_{j=0}^\infty d_j3^{-j}.$$

Then $X$ might be described as a random variable "uniformly distributed on the Cantor set". It's a continuous random variable, in that $P(X=a)=0$ for any fixed number $a$, but it does not have a density function. (The distribution of $X$ is the "Cantor distribution" mentioned in the other answer.) $X$ is an example of the sort of thing they don't want to talk about in the courses you mention.