= Solution
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 $u=\sum_{j=1}^mc_j\mathbf1_{B_j}$ with finite-measure Borel sets $B_j$ and error at most $2M\delta$ on the truncated domain. Thus $\|f-u\|_1$ 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 so
$$
\sum_j|c_j|\mu(A_j\mathbin\triangle B_j)<\varepsilon.
$$
These 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 $s=\sum_jc_j\mathbf1_{A_j}$ is therefore a <step function> of the specified form, and
$$
\|u-s\|_1\leq\sum_j|c_j|\mu(A_j\mathbin\triangle B_j)<\varepsilon.
$$
Combining the two approximations proves \b[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.
Back to article page
