Below is a classic example of an integral expression that can be computed using the improper Riemann integral, but not using the Lebesgue integral:
$$\int\limits_0^\infty\frac{sin(x)}{x}$$
The problem with taking the Lebesgue integral here is that the Lebesgue integral is defined as $\int f^+ - \int f^-$, and $f^+$ and $f^-$ are both $\infty$ in this case, because the harmonic series diverges. The sum is conditionally convergent, and Lebesgue integrability requires absolute convergence.
To analyze it from a different angle, it seems the underlying issue here is that the Lebesgue integral works by taking the measure of the preimage of small intervals on the vertical axis, and measure doesn't very much depend on the order the points in a set appear in (countable additivity and translation invariance mean you are allowed to break up the set and shuffle the parts around quite a bit without changing the measure!). So it stands to reason that a conditionally convergent sum isn't always good enough, because a conditionally convergent sum is by definition one where the order matters.
On the other hand, something along the lines of a Riemann integral does sum things in a particular order, because the rectangles it takes the area of all happen to be lined up in a particular order along the horizontal axis. So in this case, conditionally convergent sums work.
My question is: can't we just hack the Lebesgue integral so that it does the same thing?
The following may not be defined for Lebesgue integrals:
$$\int\limits_0^\infty\frac{sin(x)}{x}$$
But what about something like this?
$$\lim_{n\to\infty} \sum_0^n \int\limits_i^{i+1} \frac{sin(x)}{x}$$
This expression seems like it would evaluate to $\pi/2$, same as the improper Riemann integral.
Obviously, going about it this way is fairly unwieldy compared to just using an improper Riemann integral or gauge integral. But does it actually work? Why couldn't something like this be used to generalize the Lebesgue integral to evaluate functions like $sin(x)/x$?
