Solution
= Solution
Choose a <step function> $s$ with $\|f-s\|_1<\varepsilon$, by part (i). The <Fourier transform> satisfies
$$
|\widehat f(t)-\widehat s(t)|
\leq\int|f(x)-s(x)|\,dx<\varepsilon.
$$
By part (ii), $\limsup_{|t|\to\infty}|\widehat f(t)|\leq\varepsilon$. Since $\varepsilon$ is arbitrary,
$$
\boxed{\widehat f(t)\to0\quad\text{as }|t|\to\infty.}
$$
This proves the <Riemann-Lebesgue lemma> with the transform convention in the question.