### Mod 2 normal numbers and skew products

Let E be an interval in the unit interval [0,1). For each x ∈ [0,1) define dₙ(x) ∈ 0,1 by $d\u2099\left(x\right):={\sum}_{i=1}^{n}{1}_{E}\left({2}^{i-1}x\right)\left(mod2\right)$, where t is the fractional part of t. Then x is called a normal number mod 2 with respect to E if ${N}^{-1}{\sum}_{n=1}^{N}d\u2099\left(x\right)$ 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 ${N}^{-1}{\sum}_{n=1}^{N}d\u2099\left(x\right)$ converges a.e. and the limit equals 1/3 or 2/3 depending on x.