# "Absolutely continuous random variable" vs. "Continuous random variable"?

In the book Limit Theorems of Probability Theory by Valentin V. Petrov, I saw a distinction between the definitions of a distribution being "continuous" and "absolutely continuous", which is stated as follows:

1. "...The distribution of the random variable $$X$$ is said to be continuous if $$P\left(X \in B\right)=0$$ for any finite or countable set $$B$$ of points of the real line. It is said to be absolutely continuous if $$P\left(X \in B\right)=0$$ for all Borel sets $$B$$ of Lebesgue measure zero..."

The concept I am familiar with is:

1. "If a random variable has a continuous Cumulative Distribution Function, then it is absolutely continuous."

My questions are: are the two descriptions about "absolute continuity" in (1) and (2) talking about the same thing? If yes, how can I translate one explanation into the other one?

• The standard example of a continuous but not absolutely continuous distribution is discussed at stats.stackexchange.com/questions/229556/…, where it is graphed and code is supplied to sample from it.
– whuber
Aug 16, 2017 at 19:34

The descriptions differ: only the first one $$(*)$$ is correct. This answer explains how and why.

### Continuous distributions

A "continuous" distribution $$F$$ is continuous in the usual sense of a continuous function. One definition (usually the first one people encounter in their education) is that for each $$x$$ and for any number $$\epsilon\gt 0$$ there exists a $$\delta$$ (depending on $$x$$ and $$\epsilon$$) for which the values of $$F$$ on the $$\delta$$-neighborhood of $$x$$ vary by no more than $$\epsilon$$ from $$F(x)$$.

It is a short step from this to demonstrating that when a continuous $$F$$ is the distribution of a random variable $$X$$, then $$\Pr(X=x)=0$$ for any number $$x$$. After, all, the continuity definition implies you can shrink $$\delta$$ to make $$\Pr(X\in (x-\delta, x+\delta))$$ as small as any $$\epsilon \gt 0$$ and since (1) this probability is no less than $$\Pr(X=x)$$ and (2) $$\epsilon$$ can be arbitrarily small, it follows that $$\Pr(X=x)=0$$. The countable additivity of probability extends this result to any finite or countable set $$B$$.

### Absolutely continuous distributions

All distribution functions $$F$$ define positive, finite measures $$\mu_F$$ determined by

$$\mu_F((a,b]) = F(b) - F(a).$$

Absolute continuity is a concept of measure theory. One measure $$\mu_F$$ is absolutely continuous with respect to another measure $$\lambda$$ (both defined on the same sigma algebra) when, for every measurable set $$E$$, $$\lambda(E)=0$$ implies $$\mu_F(E)=0$$. In other words, relative to $$\lambda$$, there are no "small" (measure zero) sets to which $$\mu_F$$ assigns "large" (nonzero) probability.

We will be taking $$\lambda$$ to be the usual Lebesgue measure, for which $$\lambda((a,b]) = b-a$$ is the length of an interval. The second half of $$(*)$$ states that the probability measure $$\mu_F(B)=\Pr(X\in B)$$ is absolutely continuous with respect to Lebesgue measure.

Absolute continuity is related to differentiability. The derivative of one measure with respect to another (at some point $$x$$) is an intuitive concept: take a set of measurable neighborhoods of $$x$$ that shrink down to $$x$$ and compare the two measures in those neighborhoods. If they always approach the same limit, no matter what sequence of neighborhoods is chosen, then that limit is the derivative. (There's a technical issue: you need to constrain those neighborhoods so they don't have "pathological" shapes. That can be done by requiring each neighborhood to occupy a non-negligible portion of the region in which it lies.)

Differentiation in this sense is precisely what the question at What is the definition of probability on a continuous distribution? is addressing.

Let's write $$D_\lambda(\mu_F)$$ for the derivative of $$\mu_F$$ with respect to $$\lambda$$. The relevant theorem--it's a measure-theoretic version of the Fundamental Theorem of Calculus--asserts

$$\mu_F$$ is absolutely continuous with respect to $$\lambda$$ if and only if $$\mu_F(E) = \int_E \left(D_\lambda \mu_F\right)(x)\,\mathrm{d}\lambda$$ for every measurable set $$E$$. [Rudin, Theorem 8.6]

In other words, absolute continuity (of $$\mu_F$$ with respect to $$\lambda$$) is equivalent to the existence of a density function $$D_\lambda(\mu_F)$$.

### Summary

1. A distribution $$F$$ is continuous when $$F$$ is continuous as a function: intuitively, it has no "jumps."

2. A distribution $$F$$ is absolutely continuous when it has a density function (with respect to Lebesgue measure).

That the two kinds of continuity are not equivalent is demonstrated by examples, such as the one recounted at https://stats.stackexchange.com/a/229561/919. This is the famous Cantor function. For this function, $$F$$ is almost everywhere horizontal (as its graph makes plain), whence $$D_\lambda(\mu_F)$$ is almost everywhere zero, and therefore $$\int_{\mathbb{R}} D_\lambda(\mu_F)(x)d\lambda = \int_{\mathbb{R}}0 d\lambda = 0$$. This obviously does not give the correct value of $$1$$ (according to the axiom of total probability).