TITLE: Durrett: what is a "uniform" distribution on the Cantor set?
QUESTION [5 upvotes]: On p11, Durrett:

Example 1.2.4 Uniform distribution on the Cantor set. The Cantor set $C$ is defined by removing $(1/3, 2/3)$ from $[0,1]$ and then removing the middle third of each interval that remains. We define an associated distribution function by setting $F(x) = 0$ for $x\leq 0$, $F(x) = 1$ for $x\geq 1$, $F(x) = 1/2$ for $x\in[1/3, 2/3]$, $F(x) = 1/4$ for $x\in[1/9, 2/9]$, $F(x) = 3/4$ for $x\in[7/9,8/9]$, ... There is no $f$ for which $(1.2.1)$ holds because such an $f$ would be equal to $0$ on a set of measure $1$. From the definition, it is immediate that the corresponding measure has $\mu(C^c) = 0$.


I just don't understand why the author called it a uniform distribution. In what sense? Clearly the PDF doesn't exist so the "uniformity" isn't to be interpreted in the PDF sense. So why is such a distribution called uniform?

REPLY [2 votes]: This is an example of a probability measure that is singular with respect to Lebesgue measure.  Singular measures have the odd property that their "mass" is concentrated on a set of measure zero, but the measure of every individual point is still zero.   This is in contrast to point measures which concentrate their probability on single points.
So, the interpretation of "uniform measure" in this case is tricky - you have to think about it in terms of limits (as suggested by Ian).