# Proving $\int_A f(x,y)\ dxdy = \left(\int_a^b \phi(x)\ dx\right)\left(\int_c^d \psi(y)\ dy\right)$

Let $\phi:[a,b]\to\mathbb{R}$ and $\psi:[c,d]\to\mathbb{R}$ be integrable functions. The function $f:[a,b]\times[c,d]\to\mathbb{R}$, defined in the rectangle $A=[a,b]\times[c,d]$ by $f(x,y) = \phi(x)\psi(y)$ is integrable and $\int_A f(x,y)\ dxdy = \left(\int_a^b \phi(x)\ dx\right)\left(\int_c^d \psi(y)\ dy\right)$

My book has the converse theorem:

Let $f:A_1\times A_2\to \mathbb{R}$ integrable in the product of blocks $A_1\subset\mathbb{R}^m$ and $A_2\subset\mathbb{R}^n$. For all $x\in A_1$, let $f_x:A_2\to\mathbb{R}$ defined by $f_x(y) = f(x,y)$ and let

$$\phi(x) = \underline{\int_{A_2}} f_x(y)\ dy, \ \ \psi(x) = \overline{\int_{A_2}} f_x(y)\ dy$$

The functions $\phi, \psi: A_1\to\mathbb{R}$ are integrable, with

$$\int_{A_1}\phi(x) \ dx = \int_{A_1}\psi(x) \ dx = \int_{A_1\times A_2} f(x,y)\ dxdy$$

that is,

$$\int_{A_1\times A_2} f(x,y) \ dxdy = \int_{A_1}dx \left(\underline{\int_{A_2}}f(x,y)\ dy\right) = \int_{A_1}dx \left(\overline{\int_{A_2}}f(x,y)\ dy\right)$$

By the theorem, our $f_x(y) = \phi(x)\psi(y)$ and $f_y(x) = \psi(x)\phi(y)$ with the observation that $f_x(y)$ is for fixed $x$ and variable $y$, and $f_y(x)$ the converse. And since $\phi, \psi$ are integrable,

$$\overline{\int_{A_2}}f(x,y)\ dy = \underline{\int_{A_2}}f(x,y)\ dy = \int_{A_2}f(x,y)\ dy = \int_{A_2}f_x(y)\ dy = \int_{A_2} \phi(x)\psi(x)\ dy$$

So by the formula in the theorem:

$$\int_{[a,b]\times [c,d]} f(x,y) \ dxdy = \int_{A_1}dx \left(\underline{\int_{A_2}}f(x,y)\ dy\right) = \int_{[a,b]}dx \left(\int_{[c,d]} \phi(x)\psi(y)\ dy\right) = \mbox{ what???}$$

So I'm not getting the desired $\left(\int_a^b \phi(x)\ dx\right)\left(\int_c^d \psi(y)\ dy\right)$

I think that you are confusing yourself by having multiple meanings associated to $\phi$ and $\psi$. To avoid this lets assume that $f(x,y) = g(x)h(y)$. Now lets use the theorem: Define: \begin{equation} \phi(x) = \int_{A_2}g(x)h(y)dy = g(x) \int_{A_2} h(y)dy. \end{equation} Now we have that (last line in your theorem): \begin{equation} \int_{A_1 \times A_2}f(x,y) dx dy = \int_{A_1} \phi(x) dx = \int_{A_1}\left(g(x) \int_{A_2} h(y)dy\right)dx = \int_{A_2} h(y)dy \int_{A_1}g(x)dx. \end{equation}