Is there a version of Fubini's theorem for improper Riemann integrals? Here's an example of what such a version might look like.

If $f:\mathbb{R}^n\rightarrow\mathbb{R}$ is bounded and non-negative over the set $E\subseteq\mathbb{R}^n$ and if the improper Riemann integral $I:=\int_E f\left(\mathbf{x}\right)\ \mathrm{d}\mathbf{x}$ exists and is finite, then the following iterated integral is well defined and evaluates recursively to $I$:

$$ \int_{-\infty}^\infty\int_{-\infty}^\infty\cdots\int_{-\infty}^\infty f\left(x_1,x_2,\dots,x_n\right)\mathbb{1}_E\left(x_1,x_2,\dots,x_n\right)\ \mathrm{d}x_n\cdots\mathrm{d}x_2\mathrm{d}x_1 $$

So, for almost every $\left(x_1,\dots,x_{n-1}\right)\in\mathbb{R}^{n-1}$, the function

$$ x_n \mapsto f\left(x_1,\dots,x_{n-1},x_n\right)\mathbb{1}_E\left(x_1,\dots,x_{n-1},x_n\right) $$

is non-negative, bounded and Riemann integrable on every closed interval $\left[a,b\right]\subseteq\mathbb{R}$ ($a<b$) and the function $g:\mathbb{R}^{n-1}\rightarrow\mathbb{R}$

$$ g\left(x_1,\dots,x_{n-1}\right):=\lim_{r\rightarrow\infty}\int_{-r}^r f\left(x_1,\dots,x_{n-1},x_n\right)\mathbb{1}_E\left(x_1,\dots,x_{n-1},x_n\right)\ \mathrm{d}x_n $$

is again bounded and non-negative over $E'$ = the projection of $E$ on the first $n-1$ coordinates, and the improper integral $\int_{E'}f\left(\mathbf{x}'\right)\ \mathrm{d}\mathbf{x}'$ exists and equals $I$.

  • 3
    $\begingroup$ This is certainly true for the Lebesgue integral, it is called Tonelli's theorem. Since the Riemann and Lebesgue integrals agree whenever both are defined, it therefore follows for the Riemann integral as well. The only detail is to show that for almost every slice, your function is continuous almost everywhere so that the Riemann integral makes sense. $\endgroup$ – nullUser Dec 9 '13 at 21:28

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