If $f:\mathbb{R}\rightarrow \mathbb{R}$ and $g:\mathbb{R}\rightarrow \mathbb{R}$ are continuous functions, prove that $h:\mathbb{R}^2\rightarrow \mathbb{R}$ such that $$h(x,y)=f(x)+g(y)$$ is continuous and $\displaystyle \iint_A h(x,y)dxdy=(d-c)\int_a^bf(t)dt+(b-a)\int_c^dg(t)dt$, whatever the rectangle $A=[a,b]\times [c,d]$.
Can you help me with this exercise?
On
\begin{eqnarray} \iint_A h(x,y) ~{\rm d}y{\rm d}y &=& \int_c^d\int_a^b h(x,y) ~{\rm d}x{\rm d}y \\ &=& \int_c^d\int_a^b f(x) ~{\rm d}x{\rm d}y + \int_c^d\int_a^b g(y) ~{\rm d}x{\rm d}y \\ &=& \int_a^b\int_c^d f(x) ~{\rm d}y{\rm d}x + \int_c^d\int_a^b g(y) ~{\rm d}x{\rm d}y \\ &=& \int_a^bf(x)\color{blue}{\left(\int_c^d {\rm d}y\right)}{\rm d}x + \int_c^dg(y) \color{red}{\left(\int_a^b {\rm d}x\right)}{\rm d}y \\ &=& \int_a^bf(x)\color{blue}{\left(d-c\right)}{\rm d}x + \int_c^dg(y) \color{red}{\left(b-a\right)}{\rm d}y \\ &=& (d-c)\int_a^bf(x){\rm d}x + (b-a)\int_c^dg(y) {\rm d}y \end{eqnarray}
This is straightforward - I think.
Rewrite the integrand as $f(x) + g(y)$. Now the magic happens:
\begin{align*} \int_A h \ dx \ dy &= \int_c^d \int_a^b f(x) + g(y) \ dx \ dy \\ &= \int_c^d \int_a^b f(x) \ dx \ dy + \int_c^d \int_a^b g(y) \ dx \ dy \\ &= (d-c) \int_a^b f(x) \ dx + \int_a^b \int_c^d g(y) \ dy \ dx \\ &= (d-c) \int_a^b f(x) \ dx + (b-a) \int_c^d g(y) \ dy \end{align*}
What happened? As $f$ is independent of $y$, the outer integral in that term is just $\int_c^d dy$ which we can easily do. A similar argument holds for $g$ since it's independent of $x$. Notice I changed the order. A simple version of Fubini justifies the interchange of integration order when integration over a rectangle/box.