TITLE: Relation between integral by parts and Fubini's theorem
QUESTION [8 upvotes]: In probability, I have seen some examples for which both Fubini's theorem and integration by parts (for Riemann-Stieltjes integrals with cdf as integrator) provide different but correct solutions. For example

In proving $E(|X|)=\int_0^\infty P(|X| > t)dt$, Edvin and Did used Fubini's theorem, while
Ben used integration by parts;
In proving $\operatorname{median}(X)$ solves $\min_{c \in \mathbb{R}} E |X-c|$, Did used Fubini's theorem, while Sivaram
used integration by parts in Edit.

So I wonder if the two are related somehow? 
For example, in some cases (especially the two examples above), can one lead to the other?

A wide guess for going from Fubini's theorem to integration by parts is: 
Integration by parts says
$$
    \begin{align} f(b)g(b) - f(a)g(a) & =  \int_a^b  g(x) \, df(x) + \int_a^b f(x) \, dg(x). \end{align}  $$
If there is some $c \in \mathbb{R}$ such that $g(c)=0$, then $$
\int_a^b g(x) \, df(x) = \int_a^b \int_c^x dg(t)  \, df(x ) $$ If
Fubini's theorem or some of its variants can apply, then for some $d
\in \mathbb{R}$, $$ \int_a^b \int_c^x dg(t)  \, df(x ) = \int_c^b
\int_d^t df(x)  \, dg(t )  = \int_c^b \int_d^x df(t)  \, dg(x )  $$
one step closer to $\int_a^b f(x) \, dg(x)$, but still far away from
integration by parts.
No idea yet about going from integration by parts to Fubini's
theorem.

REPLY [8 votes]: Fubini implies integration by parts. To see this in dimension $1$, consider  some $C^1$ functions $f$ and $g$ and the integral
$$
I=\int\limits_a^bf'(x)g(x)\mathrm dx.
$$ 
Then, the fundamental theorem of calculus shows that, for every $x$, 
$$
g(x)=g(a)+\int\limits_a^xg'(t)\mathrm dt,
$$ 
hence $I=J+K$ with
$$
J=g(a)\int\limits_a^bf'(x)\mathrm dx=g(a)(f(b)-f(a)),
$$
and, using Fubini theorem to justify the second equality sign below,
$$
K=\int\limits_a^bf'(x)\int\limits_a^xg'(t)\mathrm dt\mathrm dx=\int\limits_a^bg'(t)\int\limits_t^bf'(x)\mathrm dx\mathrm dt=\int\limits_a^bg'(t)(f(b)-f(t))\mathrm dt.
$$
One sees that $K=L-M$ with
$$
L=f(b)\int\limits_a^bg'(t)\mathrm dt=f(b)(g(b)-g(a)),\qquad
M=\int\limits_a^bg'(t)f(t)\mathrm dt.
$$
Finally, $I=(J+L)-M$ with 
$$
J+L=g(a)(f(b)-f(a))+f(b)(g(b)-g(a))=f(b)g(b)-f(a)g(a).
$$