Let $\mathbb{T}$ be the unit circle and $\lambda$ be the Lebesgue measure on $\mathbb{T}$. Let $A_n := e^{2\pi i[1/2^{2n},1/2^{2n+1}]}$, $n\ge 1$. Define a function $f$ on the set of all the Lebesgue measurable set $E$ of $\mathbb{T}$ by $$ f(E) = \min (\sum_n \frac{\lambda(E\cap A_n)}{\lambda(A_n)} ,\sum_n \frac{\lambda(A_n- A_n\cap E)}{\lambda(A_n)} ). $$
Let $U_t $ be a swift operator defined by $$E_t:=U_t ( e^{2\pi ix}) := e^{2\pi i (x+t)} $$ for every $e^{2\pi ix} \in E$ for every $E\subset \mathbb{T}$.
Now, if we let $E$ be such that $\lambda(E\cap A_n)/\lambda(A_n)=\frac{1}{2}$, we have $f(E)=\infty$. However, as we all know, the density of almost all the point of $E$ is $0$ or $1$.
Can we find a $E\subset\mathbb{T}$ such that
$$\int_0^1 f(E_t)dt =\infty?$$