I was wondering what are the weakest hypothesis for applying integration by parts to calculate $$\int_a^b fg \, dm,$$ where $m$ denotes the Lebesgue measure on $\mathbb R$.
Is it enough that $f$ be differentiable on $(a,b)$, $f' \in L^1(a,b)$ and $g \in L^1(a,b)$ to write
$$
\int_a^b fg \,dm = \left[f(x)\int_a^x g(t) \,dm(t)\right]_a^b – \int_a^b f'(x) \left(\int_a^x g(t) \,dm(t) \right) \,dm(x)\text{ ?}
$$
Best Answer
Proposition: Assume that $f$ is absolutely continuous on $[a,b]$ and $g\in L^1([a,b])$. For $x\in[a,b]$, denote $G(x)=\int_a^xg~dm$. Then $$\int_a^b fg~dm=f(b)G(b)-f(a)G(a)-\int_a^bf'G~dm.\tag{1}$$
Proof: Since both $f$ and $G$ are absolutely continuous on $[a,b]$, $fG$ is also absolutely continuous on $[a,b]$, and
$$(fG)'(x)=f'(x)G(x)+f(x)g(x)\quad a.e. x\in[a,b].\tag{2}$$ Integrating $(2)$ over $[a,b]$, $(1)$ follows. $\quad\square$
Remark: If $f$ is differentiable on $[a,b]$ and $f'\in L^1([a,b])$, then $f$ is absolutely continuous on $[a,b]$. See, for example, Theorem 7.21 in Real and Complex Analysis(Third Edition) by Walter Rudin.