Covering a set of intervals
You're right to look for a reduction from the subset-sum problem. The subset problem is to find a subset $S&#39;$ of $S$ such that $\Sigma_{s&#39; \in S&#39;} s&#39; = t$. But this can be reduced to your interval cover problem with intervals $\{(0, s) : s \in S\}$, $k = t$, $L = 1$. Therefore the interval cover problem is NP-hard.

PS Why don't you make all the intervals, input and output, half-open at the same end? That way you eliminate the finite set of points at which the sum of indicator functions is different.