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 $f$

Question

(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.

Answer 1

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