Integration by parts, as stated in W. Rudin's Principles of Mathematical Analysis, Theorem 6.22, goes as follows:
Suppose F and G are differentiable functions in $[a,b]$, $F'=f\in \mathcal{R}$, and $G'= g\in \mathcal{R}$. Then $\int_a^bF(x)g(x)dx = F(b)G(b) - F(a)G(a) - \int_a^bf(x)G(x)dx$.
The proof is to put $H(x)=F(x)G(x)$ and apply the fundamental theorem of calculus to H and its derivative. The fundamental theorem of calculus is stated as:
If $f \in \mathcal{R}$ on $[a,b]$ and if there is a differentiable function $F$ on $[a,b]$ s.t. $F'=f$, then $\int_a^bf(x)dx=F(b)-F(a)$.
We also have, from an earlier theorem (6.13), that:
If $f \in \mathcal{R}$ and $g \in \mathcal{R}$ on $[a,b]$, then $fg \in \mathcal{R}$.
Rudin notes that "$H' \in \mathcal{R}$, by Theorem 6.13" (above). Is that theorem really enough to prove this? It doesn't seem to be. After all, if $H(x)=F(x)G(x)$ then $H'(x)=F(x)g(x)+f(x)G(x)$, and we only have $f(x)g(x) \in \mathcal{R}$, while $F(x)g(x) \neq f(x)g(x)$. Where do we get integrability of $F(x)g(x)$ and $f(x)G(x)$?
I'd be grateful for any pointers. Thanks!
