Mod 2 normal numbers and skew products

Let E be an interval in the unit interval [0, 1). For each x ∈ [0, 1) define d_n (x) ∈ {0, 1} by d_n(x):= ∑_i=1ⁿ 1_E({2^i-1x}) (mod 2), where {t} is the fractional part of t. Then x is called a normal number mod 2 with respect to E if N^-1 ∑_n=1^N d_n (x) converges to 1/2. It is shown that for any interval E ≠ (1/6, 5/6) a.e. x is a normal number mod 2 with respect to E. For E = (1/6, 5/6) it is proved that N^-1 ∑_n=1^N d_n (x) converges a.e. and the limit equals 1/3 or 2/3 depending on x.