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I was wondering

  1. if integration by parts can be generalized to general integrals in measure theory? By general integrals, I mean the integral that is defined for a measurable function $f: \text{ general measure space } (X, \mathcal{F}, \mu) \rightarrow (\mathbb{R} \text{ or } \mathbb{C}, \mathcal{B})$.
  2. if there are some conditions for it to be true?

Thanks and regards!

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2 Answers 2

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If $f, g$ are normalised of bounded variation, that is they are of bounded variation, right continuous and $f(-\infty) = 0$ and at least one of them is continuous then:

$$\int_{(x_0, x_1]} f \, dg + \int_{(x_0, x_1]} g \, df = f(x_1)g(x_1) - f(x_0)g(x_0).$$

where the integrals are Lebesgue-Stieltjes integrals.

You have many more similar results. Are you looking for something like this or something more general?

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  • $\begingroup$ Thanks! I would like to learn about both, although I was more interested in general. $\endgroup$
    – Tim
    Apr 9, 2011 at 13:00
  • $\begingroup$ @Tim: This is usually in a measure theory book, for example in Folland - Real Analysis or Exercise 13.13 (solutions are online) to Measures, Integrals and Martingales - Schilling (I prefer this book to learn the material from as it is easier to understand). $\endgroup$
    – JT_NL
    Apr 9, 2011 at 14:08
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Alright, so this question is a little bit old, but I'd been looking for a similar result for a long time and I found it the other day. Daniel Stroock has a complete proof of the divergence theorem in two of his books, "A Concise Introduction to the Theory of Integration" and "Essentials of Integration Theory for Analysis." As it turns out, the result in the first book is somewhat more general (fewer assumptions on f), but the proof is simpler in the second. The result states:


Again let $G$ be a smooth region in $\mathbb{R}^N$ and $U$ and open neighborhood of $\bar{G}$. If $F:U\rightarrow\mathbb{R}^N$ is continuously differentiable and either $G$ is bounded or $F\equiv 0$ off a compact subset of $U$, then

$\int\limits_G div F(x) dx = \int\limits_{\partial G} (F(x),n(x))_{\mathbb{R}^N} \lambda_{\partial G}(dx)$


In order to get integration by parts, let $F=fg$ for $f:\mathbb{R}^N\rightarrow \mathbb{R}^N$ and $g:\mathbb{R}^N\rightarrow\mathbb{R}$ and then apply the chain rule.

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