Integrability: Neither improper Riemann nor Lebesgue but Henstock-Kurzweil
Solution 1:
This is a blatant cheat, but anyway, here goes:
Take $f(x) = \frac{\sin{x}}{x}$, which is well-known to be improperly Riemann integrable, but not Lebesgue integrable.
Take the characteristic function $g$ of $[0,1] \cap \mathbb{Q}$ which is Lebesgue integrable but not improperly Riemann integrable.
The KH-integral integrates both, hence it integrates $h(x) = f(x) + g(x)$.
Clearly, $h(x)$ cannot be either, improperly Riemann integrable or Lebesgue integrable, because this would force $g$ or $f$ to have a property it doesn't have.