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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

  1. In proving $E(|X|)=\int_0^\infty P(|X| > t)dt$, Edvin and Did used Fubini's theorem, while Ben used integration by parts;
  2. 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?

  1. 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.

  2. No idea yet about going from integration by parts to Fubini's theorem.
Did
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Tim
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1 Answers1

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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 $$ J+L=g(a)(f(b)-f(a))+f(b)(g(b)-g(a))=f(b)g(b)-f(a)g(a). $$

Did
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