Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 4 / 25J (Probability and Measure)

Work in the real $L^1$ space, consistent with the specified real step coefficients. First approximate an integrable function by a finite-valued measurable function with bounded support. Truncate its domain to $[-M,M]$ and its values to $[-M,M]$; the $L^1$ error tends to zero by integrability. Quantize those remaining values in intervals of mesh $\delta$, producing a simple function $u={\sum }_{j=1}^m{c}_j{1}_{B_j}$ with finite-measure Borel sets $B_j$ and error at most $2M\delta$ on the truncated domain. Thus $\Vert f-u{\Vert }_1$ can be made arbitrarily small.

For each $B_j$, use the allowed approximation by a finite union of intervals $A_j$, choosing the symmetric-difference errors soThese unions can be written as finite disjoint bounded intervals, and changing endpoints to the required half-open convention changes only a null set. The function $s={\sum }_j{c}_j{1}_{A_j}$ is therefore a step function of the specified form, andCombining the two approximations proves the step functions are dense in $L^1(\mathbb{R})$. In the complex version of $L^1$, the same argument applies to real and imaginary parts when complex step coefficients are allowed.

For $t\ne 0$, direct integration of each interval in the Fourier transform givesHence $|\hat{s}(t)|\le 2{\sum }_k|c_k|/|t|$, andThe finite value at zero is ${\sum }_k{c}_k(b_k-a_k)$ and does not affect this limit.

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.