I am realising that the Lebesgue integral is not omnipotent. Let's consider the function $f(x)=x^2\sin (1/x)$ defined on $[-1,1]$ with $f(0)=0$. $f'(x)=2x\sin(1/x)-\cos(1/x)$, so $f$ is a.e. differentiable on $[-1,1]$. $f'$ is not Lebesgue integrable, however, because the integral of its abosolute value over $[0,1]$ is unbounded. I feel it is sensible to define $\int _{[-1,1]} f'(x)dx=f(1)-f(-1)=2\sin 1$. More precisely, define the integral in the following way (quoted from here)
The Cauchy integral is defined as follows:
Let $f$ be a mapping of an interval $I \subset \mathbf{R}$ into a Banach space $F$. We say that a continuous mapping $g$ of $I$ into $F$ is a primitive of $f$ in $I$ if there exists a denumerable set $D > \subset I$ such that, for any $\xi \in I - D$, $g$ is differentiable at $\xi$ and $g'(\xi) =f(\xi)$ .
If $g$ is any primitive of a regulated function $f$, the difference $g(\beta) - g(\alpha)$, for any two points of $I$, is independent of the particular primitive $g$ which is considered, owing to (8.7.1); it is written $\int_\alpha^\beta f(x) dx$, and called the integral of $f$ between $\alpha$ and $\beta$. (A map $f$ is called regulated provided that there exist one-sided limits at every point of $I$).
We can show that the value $g(\beta) - g(\alpha)$ is independent of the choice of $g$.
However, of course, being Cauchy integrable does NOT mean being Lebesgue integrable.
In the end, I am thinking about this: apparently, the Lebesgue integral could be extended to a larger class of functions. What are some sensible ways of doing this? Of course, we can integrate any function in the vector space spanned by Lebesgue integrable and Cauchy integrable functions.