How do I prove I can integrate over a triangle with two different parametrizations? This seems like a simple thing that has been eluding me. Consider the two integrals:
$I_1 = \int_t^{t'} ds \int_s^{t'} ds' \; g(s',s)$
and
$I_2 = \int_t^{t'} ds' \int_t^{s'} ds \; g(s',s)$
where g(s',s) is any function of the variables.
It seems to me that the two integrals should be the same, as I'm running over the same triangle in the $(s,s')$ space, but I cannot find the appropriate change of variables to prove it.
Is that true? What is the change of variables?
 A: We can use the following trick: define
$$G(s', s) = \begin{cases} g(s', s) & (s', s) \in D \\ 0 & \text{otherwise}, \end{cases}$$
where $D$ is the triangular region that we want to integrate over.
Thus $I_1 = \int_t^{t'} ds \int_t^{t'} ds' \; G(s',s).$ Similarly, $I_2 = \int_t^{t'} ds' \int_t^{t'} ds \; G(s',s).$
Now, Fubini's theorem says that these two integrals are equal.
Source: http://ksuweb.kennesaw.edu/~plaval/math2203/doubleintgen.pdf
A: Of course, this is conceptually the same as Fubini's theorem, i.e. that the two-dimensional integral over a region $D$ is independent of which direction you slice it (so long as the boundary curves are graphs over the axis perpendicular to the slice direction).
The usual way of reducing this "stronger" notion of Fubini to the classical, rectangular definition is to introduce an indicator function
$$\chi(s,s') = \begin{cases}1, & s \geq t, s' \leq t', s'-s \geq 0, \\0, & \mathrm{otherwise},\end{cases}$$
so that
\begin{align*}
I_1 &= \int_t^{t'} ds \int_s^{t'} ds' \; g(s',s)\\
&= \int_t^{t'} ds \int_s^{t'} ds' \; g(s',s) \chi(s,s')\\
&= \int_t^{t'} ds \int_t^{t'} ds' \; g(s',s) \chi(s,s')\\
&= \int_t^{t'} ds' \int_t^{t'} ds \; g(s',s) \chi(s,s')\\
&= \int_t^{t'} ds' \int_t^{s'} ds \; g(s',s) \chi(s,s')\\
&= \int_t^{t'} ds' \int_t^{s'} ds \; g(s',s) = I_2.
\end{align*}
The nontrivial steps here are Fubini's theorem (in the middle step) and the insertion and removal of $\chi$, which requires carefully checking that for all points in the domain of integration, $\chi(s,s') = 1$. To be fully rigorous you'd also need to prove that $g\chi$ is integrable.
