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 to the Cantor function, which puts mass $1$ to the Cantor set (which has null Lebesgue measure).

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).

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 to the Cantor function (https://en.wikipedia.org/wiki/Cantor_function), which puts mass 1 to the Cantor set (which has null Lebesgue measure).

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 to the Cantor function, which puts mass $1$ to the Cantor set (which has null Lebesgue measure).