A confusion about a problem in Zorich 11.4.3 Problem 1.c In the book of Mathematical Analysis II by Zorich, at page 135, it is asked that 

Show by example that if the values of the function $F(x)$ that occurs
  in Fubini’s theorem, which in the theorem were subjected to the
  conditions $\underline J (x) ≤ F(x) ≤ \bar J(x)$ at all points where
  $\underline J(x) < \bar J(x)$, are simply set equal to zero at those
  points, the resulting function may turn out to be nonintegrable.
  (Consider, for example, the function $f(x,y)$ on $\mathbb{R}^2$  equal
  to $1$ if the point $(x,y)$ is not rational and to $1 − 1/q$ at the
  point $(p/q, m/n)$, both fractions being in lowest terms.)

I couldn't understand what the author is asking. I mean what does he mean by "set equal to zero" ? Plus, what kind of grammatical structure does the author using in "[...] at those points, the resulting function may turn out to be nonintegrable. [...]"
Reference:

and $$F(x) = \int_y f(x,y) dy$$
 A: I took a look at the book by Zorich, to confirm that these are Riemann integrals under consideration and to clarify the notation as ,
$$F(x) = \int_0^1 f(x,y) \, dy, \\ \overline{J}(x) = \overline{\int}_0^1 f(x,y) \,dy, \\ \underline{J}(x) = \underline{\int}_0^1 f(x,y) \,dy,$$
with $\overline{J}(x)$ and $\underline{J}(x)$ denoting upper and lower Darboux integrals, which always exist when $f$ is bounded.
The question is poorly worded. At points where $\underline{J}(x) < \overline{J}(x)$ with strict inequality, the Riemann integral $F(x)$ does not exist, and it is meaningless to write $\underline{J}(x) \leqslant F(x) \leqslant \overline{J}(x)$.  If $F(x)$ exists then of course $\underline{J}(x) = F(x) = \overline{J}(x)$.  
What is really being asked is to produce an example where $F(x)$ exists at points where $\underline{J}(x) = \overline{J}(x)$, $F(x)$ fails to exist at points where $\underline{J}(x) <  \overline{J}(x)$ and is defined to be $F(x) = 0$,  yet $F$ is not Riemann integrable over $[0,1]$.
With the suggested function $f$, if $x \in[0,1]$ is irrational then $f(x,y) =1 $ for all $y \in [0,1]$. With $x$ thus fixed, the function $y \mapsto f(x,y)$ is Riemann integrable over $[0,1]$ with
$$\overline{J}(x) = \underline{J}(x) = F(x) = \int_0^1 f(x,y) \, dy = 1.$$
So, in this case, nothing unusual happens.
On the other hand, if $x \in (0,1]$ is rational then it is of the form $p/q$ in lowest terms and 
$$f(p/q,y) = \begin{cases}1, \qquad\qquad\qquad \text{if } y \in [0,1]  \text{ is irrational} \\ 1 - 1/q < 1, \quad \text{  if } y \in [0,1]  \text{ is rational}\end{cases}.$$
This is a Dirichlet-type function which is not Riemann integrable and $F(x)$ is undefined. However, the upper and lower integrtals are, respectively, $\overline{J}(p/q) = 1$ and $\underline{J}(p/q) = 1-  1/q.$ 
Hence, if $x \in (0,1]$ is rational then we have the case mentioned in the question where 
$$\tag{*}\underline{J}(x) < \overline{J}(x)$$.
By defining $F(x) = 0$ at points where (*) holds, we now have a function where $F(x) = 1$ if $x$ is irrational and $F(x) = 0$ when $x \in (0,1]$ is rational. Regardless of how $F(0)$ is defined, this Dirichlet function is not Riemann integrable.
