I am working through the UC Davis analysis notes, and was referred to the appendix for the theorem 3.62, which is not proven:
If $f, g:[a, b] \to \mathbb{R}$ are absolutely continuous, then $$ \int_a^b fg'dx = f(b)g(b) - f(a)g(a) - \int_a^bf'gdx $$ where $f', g'$ denote the pointwise a.e. derivatives of $f, g$.
I wanted to prove it myself, so I first showed that $fg$ is absolutely continuous. This part I am confident in. But then I was not sure how to go past that without using the FTC for Lebesgue Integrals, which I have not been exposed to yet in my study so I don't think I should use it.
How can I prove this result from first principles without using FTC for Lebesgue Integrals, after proving that $fg$ is absolutely continuous?
I tried researching online and found proofs like this one online, but they just say
...$fg$ is absolutely continous, and furthermore we have $$ f(b)g(b) - f(a)g(a) = \int_a^b(fg)'dx = \int_a^b[fg' + f'g]dx $$
which isn't really a justification to me. How does this Chain Rule actually follow?
There is no way to prove this statement without essentially using or reproving the FTC for Lebesgue integrals of absolutely continuous functions. Indeed, notice that if you set $f=1$, you get exactly the FTC for $g$, so the FTC follows immediately from this result.