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?
The very short answer, which doesn't really help you at your level, is that the Cantor distribution is the normalized Hausdorff measure inherited by the Cantor set from the Lebesgue measure, and the normalized Lebesgue measure on an interval is the uniform distribution on that interval.
A longer answer is to consider putting the uniform distribution on each of the closed sets $C_n$ which approximate the Cantor set. These are unions of intervals, so they have a uniform distribution in the familiar sense of the phrase. (For example, the uniform distribution on $C_1$ has pdf $3/2$ on $[0,1/3]$, $3/2$ on $[2/3,1]$ and zero elsewhere.) When you send $n \to \infty$ in the sense of convergence in distribution, you get the Cantor distribution. Note that this is only meaningful as a limit of the CDFs; the PDFs go to zero off the Cantor set and blow up on the Cantor set.