Let $g$ be a function such that $\underline{\int_{y\in B}}f(x,y)\leq g(x)\leq \overline{\int_{y\in B}}f(x,y)$ for all $x\in A$. Show that if $f$ 
(a) Exercise 1 of that section is: Let $f,g: Q\to \mathbb{R}$ be bounded function such that $f(x)\leq g(x)$ for $x\in Q$. Show that $\underline{\int_{Q}}f\leq \underline{\int_{Q}}g$ and $\overline{\int_{Q}}f\leq \overline{\int_{Q}}f$, 
I think of using the exercise to conclude that $\underline{\int_{x\in A}}\underline{\int_{y\in B}}f(x,y)\leq \underline{\int_{x\in A}}g(x)$ and $\overline{\int_{x\in A}}g(x)\leq \overline{\int_{x\in A}}\overline{\int_{y\in B}}f(x,y)$ for all $x\in A$, But fubini's theorem tells me that $\int_{Q}f=\int_{x\in A}\underline{\int_{y\in B}}f(x,y)=\int_{x\in A}\overline{\int_{y\in B}}f(x,y)=\underline{\int_{x\in A}}\underline{\int_{y\in B}}f(x,y)\overline{\int_{x\in A}}\overline{\int_{y\in B}}f(x,y)$, with which $\overline{\int_{x\in A}}g(x)=\int_{Q}f=\underline{\int_{x\in A}}g(x)$, with this I could not conclude that $g$ is integrable over $A$?
I need help for (b) and (c), could someone help me please? Thank you.
 A: Part (a) clarification
Given your first inequality for $g$ and the result of the exercise, we have
$$\tag{*}\underline{\int}_{x \in A}\,\underline{\int}_{y\in B}f \leqslant \underline{\int}_{x\in A}g \leqslant \overline{\int}_{x\in A}g  \leqslant \overline{\int}_{x \in A}\,\overline{\int}_{y\in B}f$$
If $f$ is integrable over $Q$ then by Fubini's theorem we have
$$\int_Q f = \int_{x \in A}\,\underline{\int}_{y\in B}f= \underline{\int}_{x \in A}\,\underline{\int}_{y\in B}f, \\ \int_Q f = \int_{x \in A}\,\overline{\int}_{y\in B}f= \overline{\int}_{x \in A}\,\overline{\int}_{y\in B}f$$
Using this result along with (*) we have
$$\int_Q f\leqslant \underline{\int}_{x\in A}g \leqslant \overline{\int}_{x\in A}g  \leqslant \int_Qf,$$
which proves both that $g$ is integrable over $A$ (since the lower and upper integrals are equal) and that
$$\int_Qf = \int_Ag$$
Part (b)
Consider  the example
$$
  f(x,y)=\begin{cases}
    0, & x \notin \mathbb{Q}\\
    0, & x \in \mathbb{Q}, \, y \notin \mathbb{Q}\\
    1/q, & x \in \mathbb{Q}, \, y = p/q \text{ in lowest terms}  \end{cases}$$
We have $\int_0^1 f(x,y) \, dy = 0$, since if $x$ is irrational $f(x,y) = 0$, and if $x$ is rational then $y \mapsto f(x,y)$ is the Thomae function whose integral is zero. Hence the iterated integral $\int_0^1 \left(\int_0^1 f(x,y) \, dy \right) \, dx$ exists.
However, if $y$ is rational, then $x \mapsto f(x,y)$ is a Dirichlet function and the Riemann integral $\int_0^1 f(x,y) \, dx$ does not exist. 
For (c) see the answer given by David C. Ulrich here
