Measure induced on [0, 1] by infinite tosses of biased coin 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.
 A: For $\omega \in [0,1]$, let $X_i(\omega)$ be the $i$th binary digit of $\omega$. (If $\omega$ is a dyadic rational and thus has two binary expansions, let's say we choose the expansion that ends with all 0s; it makes no difference).  Under Lebesgue measure, the $X_i$ are iid Bernoulli $1/2$ random variables, so by the strong law of large numbers, $\frac{1}{n}(X_1 + \dots + X_n) \to 1/2$ almost surely.  In other words, the set $A_{1/2} := \{\omega : \lim_{n \to \infty} \frac{1}{n} (X_1(\omega) + \dots + X_n(\omega)) = \frac{1}{2}\}$ has Lebesgue measure $1$.
If $\mu_p$ is the measure coming from a coin that comes up heads with probability $p \ne 1/2$, then under $\mu_p$, the $X_i$ are iid Bernoulli $p$.  By the strong law again, we have $\mu_p(A_p) = 1$.  But $A_p$ and $A_{1/2}$ are clearly disjoint, so Lebesgue measure and $\mu_p$ are mutually singular.
I guess the "standard technique" here is to use a zero-one type theorem from probability giving an almost sure statement about limiting behavior, and look for a place where the two measures exhibit different limiting behavior.
A: Yes. You're right; the measure is singular. One soft way to do this is to use ergodic theory. The "coin-tossing measures" are all distinct ergodic shift-invariant measures on $\{0,1\}^{\mathbb N}$ and any two ergodic invariant measures for any fixed transformation are mutually singular. The map from sequences of 0's and 1's to $[0,1]$ is 1-1 off a countable set, and so mutual singularity is preserved when the measures are transferred to $[0,1]$. 
One reference for all of this is Peter Walters' book, An Introduction to Ergodic Theory (publ. Springer Graduate Texts in Mathematics). 
A: 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).
