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Do the Lebesgue-Stieltjes integral and the Riemann integral have the same rules about the change of order of integration? I mean I know how to deal with Riemann integral, but I'm not sure if I can simply apply the same rules to the Lebesgue-Stieltjes.

Thanks.

It's double integration in $\mathbb{R}^2$, by the way.

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The Fubini–Tonelli theorems apply to arbitrary product measure spaces (not just $\mathbb R^2$) as long as the factor spaces are both $\sigma$-finite.

Basically, if $(X,\mathscr M,\mu)$ and $(Y,\mathscr N,\nu)$ are $\sigma$-finite, then we can consider the product measure space $(X\times Y,\mathscr M\otimes\mathscr N,\mu\times\nu)$. We have the following:

  • (Tonelli) If $f:X\times Y\to\mathbb [0,\infty]$ is $\mathscr M\otimes\mathscr N$-measurable, then the order of integrals can be interchanged: \begin{align*}&\int_{x\in X}\int_{y\in Y}f(x,y)\,\mathrm d\nu(y)\,\mathrm d\mu(x)=\int_{y\in Y}\int_{x\in X}f(x,y)\,\mathrm d\mu(x)\,\mathrm d\nu(y)\\=&\int_{(x,y)\in X\times Y}f(x,y)\,\mathrm d(\mu\times\nu)(x,y).\tag{*}\end{align*}
  • (Fubini) If $f:X\times Y\to\mathbb C$ is $\mathscr M\otimes\mathscr N$-measurable and $$\int_{(x,y)\in X\times Y}|f(x,y)|\,\mathrm d(\mu\times\nu)(x,y)<\infty,$$ then every term appearing in $(*)$ above is well-defined, and the equalities there all hold.
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  • $\begingroup$ Thank you so much triple_sec! But it looks like something grad students have learned. I'm an undergrad and I don't quite understand the measure space thing you metioned:( So basically the answer to my question is yes? $\endgroup$
    – larrybr
    Jul 17, 2014 at 4:23
  • $\begingroup$ @larrybr Fair enough, what I presented may well be more general than what you need. Let me simplify it. If (i) $F,G:\mathbb R\to\mathbb R$ and two non-decreasing, bounded, right-continuous functions such that $F(-\infty)=G(-\infty)=0$; (ii) $f:\mathbb R^2\to\mathbb R$ is Borel measurable; and (iii) the double Lebesgue–Stieltjes integral $$\int\int|f(x,y)|\,\mathrm dF(x)\,\mathrm dG(y)<\infty$$ is convergent, then $$\int\int f(x,y)\,\mathrm dF(x)\,\mathrm dG(y)=\int\int f(x,y)\,\mathrm dG(y)\,\mathrm dF(x)$$ is well-defined, and, yes, the order of integration is interchangeable. $\endgroup$
    – triple_sec
    Jul 17, 2014 at 4:46
  • $\begingroup$ Thanks a lot! That's really helpful! $\endgroup$
    – larrybr
    Jul 17, 2014 at 5:44

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