 Solution

Choose a step function $s$ with $\Vert f-s{\Vert }_1<\epsilon$, by part (i). The Fourier transform satisfiesBy part (ii), ${\mathrm{limsup}}_{|t|\to \infty }|\hat{f}(t)|\le \epsilon$. Since $\epsilon$ is arbitrary,This proves the Riemann-Lebesgue lemma with the transform convention in the question.