I am looking for a proof of the following fact:
Suppose that $H: \mathbb{R} \rightarrow \mathbb{R}$ is a periodic function with period $1$. Suppose further that there is no continuous function $f:\mathbb{R} \rightarrow \mathbb{R}$ with $H = f$ almost everywhere with respect to Lebesgue measure. Denote $\mathbb{T} = \{ z \in \mathbb{C}: |z|=1 \}$. Under these assumptions, there exists $\lambda \in \mathbb{R}$ such that function $R: \mathbb{T} \rightarrow \mathbb{T}$ given by $R(e^{2\pi i \theta}) = e^{2\lambda\pi i H(\theta)}$ is not equal a.e. to any continuous function on $\mathbb{T}$ (for any continuous $f:\mathbb{T} \rightarrow \mathbb{T}$ there is $m(\{ f \neq R \}) > 0$, where $m$ is Lebesgue measure on unit circle $\mathbb{T}$).
Do you know where I can find it?