TITLE: Lebesgue Spaces and Integration by parts
QUESTION [6 upvotes]: Suppose there exists a Lebesgue Space, $L_1$ and functions functions $\phi$, $\phi'$, $f$, and $f'$ functions where
$$\phi, \phi' \in L_1$$
By rule of integration by parts,
$$uv|_a^b = \int_a^b udv + \int_a^b vdu$$
Let
$$ u = \phi, du= \phi'$$
$$ v = f, dv = f'$$
Are there any properties of Lebesgue functions that allow
$$ uv|_a^b = 0$$
Are $\phi$ and $\phi'$ convergent as integrals?
Do the unbounded limits of $\phi$ and $\phi'$ converge?

REPLY [4 votes]: Integration by parts from the point of view of Lebesgue integration is more appropriately understood in terms of functions of local finite variation. Here is a version and then I comment more on your particular problem. I hope this is helpful to you.

Theorem: Let $F$, $G$ be right--continuous functions of locally finite variation on an interval $I$ (bound or unbounded) Let $\mu_F$ and $\mu_G$ the Stieltjes-Lebesgue measures generated by $F$ and $G$ respectively. For any bounded interval $(a,b]\subset I$,
$$
\int_{(a,b]}F(t)\mu_G(dt)=F(b)G(b)-F(a)G(a)-\int_{(a,b]}G(t-)\mu_F(dt)
$$
where $G(t-)=\lim_{s\nearrow t}G(s)$.
A proof can be obtained using Fubini's theorem
\begin{aligned}
F(b)-F(a))(G(b)-G(a))&=\int_{(a,b]\times(a,b]}\mu_F\otimes\mu_G(dt,ds)\\
&=\int_{(a,b]}\Big(\int_{(a,s]}\mu_F(dt)\Big)\mu_G(s) +\int_{(a,b]}\Big(\int_{(s,b]}\mu_F(dt)\Big)\mu_G(ds)\\
&=\int_{(a,b]}\Big(\int_{(a,s]}\mu_F(dt)\Big)\mu_G(s) +\int_{(a,b]}\Big(\int_{(a,t)}\mu_G(ds)\Big)\mu_F(dt)\\
&=\int_{(a,b]} F(s)-F(a)\mu_G(s) +\int_{(a,b]}G(t-)-G(a)\mu_F(dt)\\
\end{aligned}

In the setting that you have in mind, you may have $\phi$ and $\psi$ that are absolutely continuous in an interval $[c,d]$. Then you $f=\phi'$ and $g=\phi'$ exists a.s. (with respect the Lebesgue measure in $\mathbb{R}$) and
\begin{aligned}
\phi(x)&=\phi(c)+\int^x_cf(t)\,dt,&\qquad c\leq x\leq d\\
\psi(x)&=\psi(c)+\int^x_cg(t)\, dt,& \qquad c\leq x\leq d
\end{aligned}
Then you may consider $F(x)=\phi(x)$ and $G(x)=\psi(x)$ in which case
\begin{aligned}
\mu_F((dt) &= \phi(c)\delta_c(dt) + f\mathbb{1}_{(c,d]}(t)\,dt\\
\mu_G(dt) &=\psi(c)\delta_c(dt) + g\mathbb{1}_{(c,d]}(t)\,dt
\end{aligned}
Applying the theorem above, you recover the usual integration by parts formula for  $[a,b]\subset[c,d]$.
As for the boundary condition $\phi(t)\psi(t)|^b_a=0$, there many instances in which that happens. The important thing is to notice that $\phi$ and $\psi$  should be absolutely continuous to begin with.
For infinite intervals one has to use dominated convergence (integrability conditions on the derivatives $\phi'=f$ and $\psi'=g$) type of arguments. Typical situations are when both  $|\mu_F|$ and $|\mu_F|$ are finite measures.