I recently made a post on triple integrals with cylindrical/spherical coordinates, and that lead me to a breakthrough in my own understanding of double and triple integrals with variable bounds of integration in the iterated integrals. I wanted to check to make sure if this logic is correct, and if it's not, please correct me!
Here we are going to extend the evaluation of triple integrals using the triple Reimann sum definition to not only rectangular prism regions, but regions of variable shapes. First Define a function $ M(x, y, z) $ on region Q in space with bounds $\alpha\le{x}\le{\beta}, h_1(x)\le{y}\le{h_2(x)}, g_1(x,y)\le{z}\le{g_2(x,y)}$ upon which we would like to integrate (assume all h's and g's are continuous) Now construct a rectangular prism that completely surrounds Q, which we will call P, with bounds $a\le{x}\le{b}, c\le{y}\le{d}, e\le{z}\le{f}$ where $$a\le{\alpha}\le{\beta}\le{b}$$ $$c\le{h_1(x)}\le{h_2(x)}\le{d}$$ $$e\le{g_1(x,y)}\le{g_2(x,y)}\le{f}$$ Define the following function:
$$N(x, y, z)= \begin{cases} M(x, y, z), &\text{(x, y, z) ∈ Q} \\ 0, &\text{(x, y, z) ∈ P and (x, y, z) ∉ Q} \end{cases}$$
(we will prove N is Reimann integrable later)
Now, partition P and construct the triple integral on the region P:
$$\lim_{max\Delta{x} \to 0, max\Delta{y} \to 0, max\Delta{z} \to 0}\sum_{i=1}^\theta\sum_{j=1}^{\phi}\sum_{k=1}^{\rho}N(x_i, y_j, z_k)\Delta{z_k}\Delta{y_j}\Delta{x_i} = \iiint_P N(x, y, z)dzdydx$$
Use Fubini's Theorem:
$$\iiint_P{N(x, y, z)dzdydx} = \int_a^b\int_c^d\int_e^fN(x, y, z)dzdydz$$
This is where integrability comes into play. Notice, however, that when integrating upon any fixed (x,y), there are only 2 possible discontinuities right on the surface of Q. So, because we have a finite number of discontinuities for a fixed (x,y), N would be integrable on the interval [e,f]. the inside z-integral would then become:
\begin{align} \int_e^fN(x, y, z)dz & = {\int_e^{g_1(x,y)}N(x, y, z)dz} + \int_{g_1(x,y)}^{g_2(x,y)}N(x, y, z)dz + {\int_{g_2(x,y)}^fN(x, y, z)dz} \\ & = 0 + \int_{g_1(x,y)}^{g_2(x,y)}N(x, y, z)dz + 0 \\ & = \int_{g_1(x,y)}^{g_2(x,y)}M(x, y, z)dz \end{align}
The change from $N$ to $M$ is significant but can be done because on $g_1(x,y)\le{z}\le{g_2(x,y)}$, $N=M$. Now, realize that the inner-most integral is a function of x and y which we will call $W(x,y)$; also realize that because we're only dealing in 2 variables x and y, we now only have to deal with the projection of P in the xy-plane, called the rectangular region D.
\begin{align*} \int_c^d\int_e^fN(x, y, z)dzdy &= \int_c^{h_1(x)}\int_e^fN(x,y,z)dzdy + \int_{h_1(x)}^{h_2(x)}\int_e^fN(x,y,z)dydz \\ &+\int_{h_2(x,y)}^d\int_e^fN(x,y,z)dydz. \end{align*}
With the outer two iterated integrals, only y values that lie in P would be evaluated, which would thus make them equal 0. The middle iterated integral stays and the variable bounds of the z-integral can be added using the previous analysis:
$$\int_c^d\int_e^fN(x,y,z)dzdy = \int_{h_1(x)}^{h_2(x)}\int_{g_1(x,y)}^{g_2(x,y)}M(x,y,z)dzdy$$
Again, a similar logic can be applied to x (except no variables need to be fixed now): break the iterated integral up into 3 iterated integrals (with constants as bounds on the inner integrals), realize that the outer 2 iterated integral expressions are 0 because only x-values in P are evaluated, and thus only the middle integral remains with the bounds of $\alpha$ and $\beta$. Plug in the variable-bounded, doubly-iterated integral discussed earlier, and we finally have our solution:
\begin{align*} \int_a^b\int_c^d\int_e^fN(x,y,z)dzdydx &= \int_{\alpha}^{\beta}\int_{h_1(x)}^{h_2(x)}\int_{g_1(x,y)}^{g_2(x,y)}M(x,y,z)dzdydx \\ &= \iiint_Q M(x,y,z)dzdydx. \end{align*}
We have shown that regions of non-rectangular shape can be evaluated in a triple integral using functions as bounds on iterated integrals.
Proof of N being Reimann integrable:
N is always defined for every point in the solid region P, so any triple Reimann sum can be constructed in order to create a triple integral of N over P based on the fact that $N(x, y, z)$ always exists. The triple integral can be broken up into two volumes: one for the volume in Q, and one for the volume in P but outside of Q (called R here):
$$\int\int\int_P{N}dzdydx = \int\int\int_Q{N}dzdydx + \int\int\int_R{N}dzdydx$$
Because the second integral will always have 0 as the integrand based on the definition of N, the only integral left is the triple integral on Q, which we proved had variable bounds above. As long as M is continuous, that iterated integral is evaluate-able. So, N is Reimann Integrable on P, and as long as M is continuous, the triple integral can be evaluated with iterated integrals.