There is some known deficiency through the Lebesgue Integral? The Integral in the Riemann sense has a lot of deficiencies, and the Lebesgue Integral can solve almost all of them.
I know that over limited intervals, Lebesgue Integral is a generalization of the Riemann Integral. But
$$\int_a^\infty\frac{\sin(x)}{x}dx$$
converges in Riemann sense, but it is not true in the Lebesgue sense. So, we could consider this like a deficiency.
But... There is some known deficiency through the Lebesgue Integral over limited intervals?
I read (in wikipedia) that in some cases, the Henstock–Kurzweil Integral is more general than the Lebesgue one. Is it a deficiency? What kind of cases are they talking about?
 A: Let $L$ the set of all Lebesgue-integrable functions. There are some deficiencies with the class $L$:
(1) There exists improper Riemann integrable functions that aren't inside $L$. For example, the important Dirichlet integral $\int_0^{\infty}\frac{\sin(x)}{x}dx$;
(2) The FTC (Fundamental Theorem of Calculus) version for class $L$ is not too efficient. For example, if $F'(x)=f(x)$ for all $x\in[a,b]$, so to write $$F(b)-F(a)=\int_a^bf(x)dx$$ we have to assume that $f$ is limited or that $F$ is absolutely continuous. The Henstock-Kurzweil integral solve this problem definitively;
(3) $L$ is not closed under product of funcitons. For example, let $f(x) = 1/\sqrt{2}$ if $x\in(0,1]$ and $f(0)=0$. So $f\cdot f\notin L$. BUT, if one of the functions is limited, then the product belongs to $L$. However, this deficiency belongs to Lebesgue, Henstock and (I think) any generalization of the Riemann integral that do not require limitation for integration;
(4) This is not a deficiency, but a observation from the pedagogical point of view. The Lebesgue theory evolves high abstraction (needs a measure theory), and generally have none connection with those integrals that we studied in calculus or basic analysis. The Lebesgue-Riesz integral may be a solution at the first contact. But the abstraction is needed to reach all of the power that the Lebesgue integral have to be applied in more general than the euclidean space, etc. Some application require abstraction, so we have to do abstraction.
(Again, apologize the bad English)
A: One thing you might consider a deficiency of the Lebesgue integral is that there are integrable functions whose derivatives are not integrable. For example (taken from Priestley's Introduction to Integration) $f(x)=x^2 \sin(x^{-2})$ is integrable on $(0,1]$ (since it extends continuously to $[0,1]$) but its derivative
$$ f'(x) = 2x\sin(x^{-2})-2x^{-1}\cos(x^{-2})$$
is not integrable on $(0,1]$. The first term is integrable, the second isn't because the integral of its absolute value diverges.
