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In an occasion, I'd like to use Fubini's Theorem to swap the order of integration of a countour integral with an integral with respect to a measure (show that $\int_{\gamma} \int_{\Omega} f(x,z) d\mu(x) dz = \int_{\Omega} \int_{\gamma} f(x,z) dz d\mu(x)$). For that, I need to take two measure spaces, $(\Omega, \mathcal{F}, \mu)$ and $(\mathbb{C}, \mathcal{B}(\mathbb{C}), c)$, and construct the product measure in the appropriate way.

Looking at the definition of the (Riemann) contour integral, the first thought would be to proceed as is done with the Riemann-Stieltjes integrals to properly choose the measure $c$ so that $\int_{\gamma} f(z) dz = \int_{\tilde{\gamma}} f dc$ (where $\tilde{\gamma}$ is some subset of $\mathbb{C}$ associated to the countour $\gamma$. It seems it would be something like $c=m+im$, where $m$ is Lebesgue measure on $\mathcal{B}(\mathbb{R})$, but that wouldn't be a measure in the usual sense.

How can I make a correspondence between contour integrals and Lebesgue integrals and thus be able to apply Fubini's Theorem? It seems possible, since the Riemann sums for the contour are just the "simple functions" (as $c$ is not a measure) $\sum_{k=1}^n f(\gamma(t_k)) c(\Delta t_k)$...

Thank you very much and sorry, this is the first time I deal with complex-valued measurable functions.

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If you break everything down into real and imaginary parts, with $f=u+iv$, $dz=dx + i \, dy$, you get that the integrand is $(u+iv) (dx + i \, dy) = u \, dx - v \, dy + i(v\, dx + u \, dy)$. To make sense of $dx$ integrals in the context of measure theory, you have to interpret $dx$ as a signed Borel measure (aka Borel charge). For a subarc $I$ of $\gamma$ from $z_1 = x_1 + iy_1$ to $z_2 = x_2 + iy_2$ (in the orientation given by $\gamma$) you want $\int_I \, dx = x_2 - x_1$. If $\gamma$ is rectifiable, then this extends to a finite signed Borel measure on $\gamma$, by the same argument that a continuous function of bounded variation on a real interval corresponds to a signed Borel measure. Obviously, the construction for $dy$ is exactly the same.

An alternative is to use a parametrization $\gamma(t) = x(t) + i y(t)$, in which case $dx$ and $dy$ are the signed Borel measures $x'(t) dt$ and $y'(t) dt$ on some interval on the real line.

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  • $\begingroup$ In this case, if $t \in [a,b]$ is the parameter for $\gamma$, we'd just have that $\int_{\gamma} f(z) dz = \int_{[a,b]} (u \circ \gamma) + i (v \circ \gamma) dx - \int_{[a,b]} (v \circ \gamma) +i (u \circ \gamma) dy$, right? $\endgroup$ – José Siqueira Nov 23 '13 at 17:56
  • $\begingroup$ Yes, if you interpret it correctly. (The notation could be a little confusing because $\int_{[a,b]} \ldots dx$ looks almost like an integral with respect to Lebesgue measure.) $\endgroup$ – Lukas Geyer Nov 23 '13 at 18:05
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    $\begingroup$ Indeed, but I meant $dx=x'(t)dt$ and $dy=y'(t)dt$, as you've mentioned. So this is basically writing down the contour integral in terms of the parametrization and using the equivalence of Riemann integration of a continuous map over a compact interval and Lebesgue integration, as long as $\gamma$ is continuously differentiable, right? $\endgroup$ – José Siqueira Nov 23 '13 at 18:15
  • $\begingroup$ Exactly. And it works in slightly more general form (e.g., piecewise continuously differentiable), as long as you know that the integral agrees with the Lebesgue integral. $\endgroup$ – Lukas Geyer Nov 23 '13 at 18:28
  • $\begingroup$ Alright, thank you very much! I'll later try to construct the product measure(s) and justify the commutation of the integrals. $\endgroup$ – José Siqueira Nov 23 '13 at 18:40

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