I am concerned with the following problem:

I am wondering if there exists any sort of integration by parts formula for a multidimensional Lebesgue-Stieltjes integral. In my case the integral is given by (1) and its domain is a multidimensional subset of $\mathbb{R}^n$. I suspect that this would involve the function $f$ instead of the corresponding measure $\mu$ according to Theorem 2 below.

I am looking for a result similar to the well known result in the case of $n=1$, i.e.: $$\int\limits_a^b g(t) \, df(t) = g(b)f(b)-g(a)f(a)-\int\limits_a^b f(t) dg(t)$$

I'd be greatful for any hint on literature that contains relevant information on such multidimensional scenarios.

So far I found plenty of literature on the one-dimensional case, i.e. $\Omega \subset \mathbb{R}$, but nothing definitive on $\Omega \subset \mathbb{R}^n$ for $n>1$.

Let's consider the following well known result from duality theory in Analysis:

Theorem 1(cf. Aliprantis Border, Infinite-Dimensional Analysis Theorem 14.14)
Let $\Omega \subset \mathbb{R}^n$ and $\Phi: C_c(\Omega) \rightarrow \mathbb{R}$ be a continuous linear functional, where $C_c(\Omega)$ is the set of all real valued continous functions on $\Omega$ with compact support. Then $\Phi$ can be expressed the follwing way: $$\Phi(g) = \int\limits_{\Omega} g(x) \, d\mu(x) \mspace{1in} \forall g \in C_c(\Omega)\tag{1}$$ where $\mu$ is a regular signed Borel measures of bounded variation, which is uniquely determined.

To further work with this measure $\mu$ I found the following interesting fact:

Let's define $$\Delta_h f(x) := \sum\limits_\delta (-1)^{\sum_{i=1}^n \delta_i} f(x-h(\delta)) \tag{2}$$ with $\delta_i$ being either $0$ or $1$, $h = (h_1,\dots,h_n)$ and $h(\delta) = (h_1 \delta_1, \dots, h_n \delta_n)$ and the sum over $delta$ meaning the sum over all possible binary vectors with length $n$. Then the following theorem holds

Theorem 2 (cf. Aliprantis Border, Infinite-Dimensional Analysis Theorem 10.50)
If $f : \mathbb{R}^n \rightarrow \mathbb{R}$ is continuous from above and satisfies $\Delta_h f(x) \geq 0$ for all $x \in \mathbb{R}^n$ and all $h \in \mathbb{R}_+^n$, then there exists a unique Borel measure $\mu$ on $\mathbb{R}^n$ satisfying (2).
Conversely, if $\mu$ is Borel measure on $\mathbb{R}^n$, then there exists a function $f : \mathbb{R}^n \rightarrow \mathbb{R}$ (unique up to translation) that is continuous from above, satisfies $\Delta_h f(x) \geq 0$ for all $x \in \mathbb{R}^n$ and all $h \in \mathbb{R}_+^n$, and satisfies (2).


I did not find any, so only list close results that perhaps might be used to prove your result.

For $\mathbb R^1$ your result is given in, e.g., Mark J. Machina, "Expected utility ..." Lemma 2, p. 314.

A more general $\mathbb R^1$ result is given in Theorem 21.67(iv) and Remarks 21.68 of Real and Abstract Analysis: A modern treatment of the theory of functions of ... E. Hewitt, K. Stromberg p. 419

Standard Integration by parts result for $\mathbb R^n$ but for smooth functions instead of Stieltjes are given in: http://www.math.ucsd.edu/~bdriver/231-02-03/Lecture_Notes/pde8.pdf pp. 110-111 (Cor 8.26 & Cor 8.27). Probably this is also true with piecewise smooth boundary and even for BV functions (i.e., what you asked). You might want to try to modify the proofs correspondingly, but I don't know if that is an easy way. N.B. in 8.26 there is probably $f$ missing in the third term. But maybe 8.27 is better anyway (and correct).

Tuo-Yeong Lee: "A multidimensional integration by parts formula for the Henstock-Kurzweil integral" (.pdf): Remark 4.11 says that it holds for the Lebesgue integral too. (BVHK functions are defined on p. 67. $\mathcal P_m$ and $\Gamma$ are defined on p. 68, $\tilde F_\alpha$ and $<s,t>$ on p. 71. $\mu_0:=$counting measure.) The notation is awful (not the fault of the author Lee but the fault of mathematics, I believe), if you are not used to the subject. So I think that this is a kind of "actual answer" to your question, but probably does not count anyway, as there is surely a better one, using the Aliprantis terminology you cite above.

Thanks for an interesting question. If you find an answer, please post it here.


Well... there's the Stokes' theorem:

$$ \int_{\Omega }^{ } \nabla (F \cdot d\Omega ) - (\nabla \cdot F )d\Omega = \oint_{d\Omega}^{ } dr \times F $$

Somehow it resembles the result you are looking for.


W.H.Young, On multiple integration by parts and the second theorem of the mean, Proc. Lond. Math. Soc. 16(1917) 273-293. See also E.W.Hobson, Functions of a Real Variable, vol 1, 3rd edition, p. 666ff.


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