 Solution

Write the interval as $I=(c,d]$, where $0<c<d<1$, to avoid confusing its left endpoint with the rotation parameter $\alpha$. For every sufficiently large integer $k$, define an inner and outer interval on the circle byApply part (b) simultaneously to this countable collection of intervals. The intersection $G$ of the corresponding full-measure sets still has full Lebesgue measure, and is therefore dense in the circle.

Fix an arbitrary $x$. For each sufficiently large $k$, choose $y_k\in G$ with circle distance less than $1/k$ from $x$. An irrational circle rotation preserves this distance, so for every $j\ge 0$,Averaging and using part (b) for $y_k$ givesLetting $k\to \infty$ proves the everywhere interval frequency under an irrational rotation: