= Solution
Outside the <countable set> of dyadic rational numbers, the <binary expansion> $x=\sum_{j\geq1}\epsilon_j(x)2^{-j}$ is unique. Multiplication by two modulo one deletes the first binary digit, so
$$
\epsilon_j(x)=\mathbf1_{[1/2,1)}(E_2^{j-1}x).
$$
Let $g=\mathbf1_{[1/2,1)}$. By part (b), the <doubling map> is a <measure-preserving transformation> and is <ergodic> for <Lebesgue measure>. Applying the <Birkhoff ergodic theorem> to this <indicator function> gives
$$
\frac1n\#\{1\leq j\leq n:\epsilon_j(x)=1\}
=\frac1n\sum_{k=0}^{n-1}g(E_2^kx)
\longrightarrow\int_0^1g(x)\,dx=\frac12
$$
for almost every $x$. The dyadic rationals have <Lebesgue measure> zero, so the choice of their two possible <binary expansions> does not affect the conclusion. Thus \b[the limiting frequency of the digit one is $1/2$ almost everywhere].
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