Measure induced on [0, 1] by infinite tosses of biased coin

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.

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).

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).