TITLE: Measure induced on [0, 1] by infinite tosses of biased coin
QUESTION [19 upvotes]: It is well-known that one can get the Lebesgue measure on [0, 1] by tossing a fair coin infinitely (countably) many times and mapping each sequence to a real number written out in binary.
I was trying to explore what happens if you follow the same procedure with a biased coin. I managed to prove that if the induced measure is absolutely continuous with respect to the Lebesgue measure, then the density must be discontinuous on at least a dense set (which includes all rational numbers in [0, 1] with a finite binary expansion).
But I suspect it's much worse than that. When I try to visualize the cumulative distribution function, it seems to have a "fractal staircase" type of shape, so I suspect the measure is, in fact, singular, but I can't prove it.
Q1. Are there any standard techniques for proving a measure is singular ?
Q2. I strongly suspect that a natural question like this has already been resolved. Any link to a textbook and/or paper would be appreciated.

REPLY [12 votes]: Just a side comment: if you pass from binary to trinary, you can still obtain the Lebesgue measure by choosing the digits $0$, $1$, $2$, with probabilities $(\frac{1}{3},\frac{1}{3},\frac{1}{3})$ (something one may call "a fair trinary coin"). Now, if you use the "biased" coin $(\frac{1}{2},0,\frac{1}{2})$, then you'll arrive exactly at the Cantor function, which puts mass $1$ to the Cantor set (which has null Lebesgue measure).