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Consider two definite integrals:

\begin{equation} I_1 = \int_{R_1} f(x) \:dx\qquad I_2 = \int_{R_2} g(y) \:dy \end{equation}

Then, \begin{equation} I_1 \cdot I_2 = \left[ \int_{R_1} f(x) \:dx \right] \left[ \int_{R_2} g(y) \:dy\right] \end{equation}

Which under certain conditions becomes: \begin{equation} I_1 I_2 = \int_{R_1} \int_{R_2} f(x) g(y) \:dx \:dy \end{equation} I'm unsure whether I have to force the following, but here $f(x), g(y)$ are Real Valued Functions continuous on $R_1$ and $R_2$ respectively.

My question is: What theorem(s) is required in order for this property to hold? i.e. how do we know when we can combine and separate multiple integrals?

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This is just linearity of the integral. It's automatic as long as the product $I_1\cdot I_2$ itself is meaningful (i.e. we need $I_1, I_2, I_1I_2$ to exist in some sense, so we shouldn't have e.g. $I_1=0, I_2 = \infty$)

Edit: After thinking a little more, it occurs to me that we also need the "inner" integral in your iterated integral to be finite.

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  • $\begingroup$ Cheers @Brian. But does linearity hold for all integrals? and if not, under what conditions does it not hold? Also, what is the exact definition of 'meaningful' here? Or is it simply the examples you have? with $-\infty$ included? $\endgroup$ Commented Aug 12, 2019 at 0:16
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    $\begingroup$ Yes, linearity holds by definition for all integrals. Even further, if you proposed a new type of "integral" for which linearity doesn't hold, you'll find resistance in the mathematics community to even calling it an integral. $\endgroup$ Commented Aug 12, 2019 at 0:20
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    $\begingroup$ "Meaningful" here means that the integrals themselves should exist as extended real numbers (i.e. real or $\pm \infty$) and we should be able to take the product of their values (which is true as long as one isn't $0$ while the other is $\pm \infty$) $\endgroup$ Commented Aug 12, 2019 at 0:22
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    $\begingroup$ In order for the equation $$\int_R [f(x) + g(x)]\,dx = \int_R f(x)\,dx + \int_R g(x)\,dx$$ to make sense, we need all the respective integrals to exist and we need the sum on the right-hand side to be meaningful. The easiest way to force this is for both $f(x)$ and $g(x)$ to be "integrable over $R$," which is exactly the prototypical scenario in which we define the integrals. $\endgroup$ Commented Aug 12, 2019 at 0:31
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    $\begingroup$ @KevinNivek Note that I had to add one more condition in order for the iterated integral to make sense. Basically, everything works when all your integrals are finite, but I missed a situation when $I_1I_2$ might make sense in the extended reals while the iterated integral itself is nonsense. $\endgroup$ Commented Aug 12, 2019 at 1:25

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