TITLE: How can I show that the "binary digit maps" $b_i : [0,1) \to \{0,1\}$ are i.i.d. Bernoulli random variables?
QUESTION [6 upvotes]: In this post What is the Lebesgue measure of the set of numbers in $[0,1]$ that has two thirds of ones in their infinite base-2 expansion? we needed the fact that if we let $b_i (x) \in \{0,1\}$ for $x \in [0,1)$ denote the $i$-th digit in the infinite binary expansion of $x$, i.e.
$$
x = \sum_{n=1}^\infty b_n (x) \cdot 2^{-n},
$$
where $b_n (x) = 1$ for all but finitely many $n$ (except for $x=0$), then the $(b_n)_n$ form a family of independent Bernoulli random variables, if they are considered as maps defined on the probability space $([0,1), \mathcal{B}, \lambda)$, where $\mathcal{B}$ are the Borel sets and $\lambda$ denotes Lebesgue measure.
A quick search did not turn up any useful results, so my question is for a nice, more or less elementary, proof of this result, which can also be used for future reference.
I will "answer my own question", i.e. provide one possible proof. But I would also be interested in other proofs!

REPLY [2 votes]: The infinite binary expansion means the non-terminating binary expansion, i.e., when there exist two expansions for $x$, one a non-terminating expansion $0.\ldots 01111\cdots$ with $1$ trailing forever, and the equivalent expansion $0.\ldots 10000\cdots$ which terminates, choose the former. (You don't want the condition "$b_i(x)=1$ for all but finitely many $n$", which is satisfied for only a countable number of $x$.) With this in mind, the interval $(0,1]$ is a more natural domain for the $b_i$'s.

Let $e_1,e_2,\ldots,e_n\in\{0,1\}$.  The set $$\{b_1=e_1,b_2=e_2,\ldots,b_n=e_n\}:=\bigcap_{k=1}^n\{x\in(0,1]:b_k(x)=e_k\}\tag1$$ consists of all $x$ whose binary expansion begins $0.e_1e_2\cdots e_n$. This set has measure $2^{-n}$, since it coincides with the interval $(d, d+2^{-n}]$ whose left endpoint is the dyadic rational $d:=\sum_i^n e_i/2^i$. (Or prove by symmetry: all $2^n$ possible initial expansions $0.e_1\cdots e_n$ of $n$ elements from $\{0,1\}$ must have equal measure.) Hence
$$\lambda(b_1=e_1,b_2=e_2,\ldots,b_n=e_n)=\frac1{2^n}.\tag2$$ By representing the set $\{b_k=e_k\}$ as a disjoint union of sets of the form (1), conclude for each $k$
$$\lambda(b_k=e_k) = \frac12.\tag3$$ Combining (2) and (3) gives, for each $e_1,\ldots,e_n$,
$$\lambda(b_1=e_1,b_2=e_2,\ldots,b_n=e_n)=\prod_{k=1}^n \lambda(b_k=e_k).\tag4$$ So (4) is saying that $b_1,\ldots, b_n$ are independent, while (3) is saying each $b_k$ has Bernoulli($1/2$) distribution.