Spivak "Calculus on Manifolds" remarks about Fubini Theorem On Spivak "Calculus on Manifolds" p.58 he provides a general version of Fubini theorem. Which I present below (I omit parts that are not important to the question):

Let $A\subset \mathbb{R}^n$ and  $B\subset \mathbb{R}^n$ be closed retangles, and let $f:A\times B \rightarrow \mathbb{R}$ be integrable. For $x\in A$ and $y\in B$, and let 
  \begin{equation}
\mathscr{L}(x) = \mathbf{L} \int_B f(x, y) dy
\end{equation}
  denote the lower integral of $f$ on $y\in B$. Then $\mathscr{L}$ is integrable on $A$ and: 
  \begin{equation}
\int_{A\times B} f = \int_A \mathscr{L} = \int_A \left(\mathbf{L} \int_B f(x, y) dy \right) dx
\end{equation}

He gives the following remarks:

  
*
  
*[Not relevant for this question]
  
*In practive it is often the case that $h(x) = \int_B f(x, y) dy$ is integrable, so that:
  \begin{equation}
\int_{A\times B} f = \int_{A} \left(\int_B f(x, y) dy\right) dx
\end{equation}
  can be applied. This certainly occurs if $f$ is continuous.
  
*The worst irregularity commonly encontered is that $h(x)$ is not integrable for a finite number of $x\in A$. In this case, $\mathscr{L}(x) = \int_B f(x, y) dy$ for all but these finitely many $x$. Since $\int_A\mathscr{L}$ remains unchanged if $\mathscr{L}$ is redefined at a finite number of points we can still write $\int_{A\times B} f = \int_{A} \left(\int_B f(x, y) dy\right) dx$, provided that $\left(\int_B f(x, y) dy\right)$ is defined arbitrarily, say as 0, when it does not exist. 
  
*Let $f:[0, 1] \times [0, 1]\rightarrow \mathbb{R}$ be defined by:
  \begin{equation}
f(x, y) =
\begin{cases}
1 & \text{if }x\text{ is irrational} \\
1 & \text{if }x\text{ is rational and }y\text{ is irrational}\\
1-\frac{1}{q} &  \text{if }x = p/q\text{ in lowest terms and }y\text{ is rational}
\end{cases}
\end{equation}
  Then $f$ is integrable and $\int_{[0, 1]\times[0, 1]} f = 1$. Now $\int_{0}^1 f(x, y) dy = 1$ if $x$ is irrational and does not exist if $x$ is rational. Therefore $h$ is not integrable if $h(x) = \int_{0}^1 f(x, y) dy$ is set equal to zero when the integral does not exist.
  


My question are: 


*

*On remark 3, why does it need to be a finite number of points? If we had any set with measure 0, wouldn't that be enough to just define $h(x) = 0$ on those points, and in this case, the equality:
\begin{equation}
\int_A \mathscr{L}(x) = \int_A  h(x)
\end{equation}
would still hold.

*On remark 4, why $\int_0^1 f(x, y) dy$ does not exist for $x$ rational? The set of discontinuities in this case has measure 0 and by theorem 3-8 of the same book this would be enough to guarantee the existance of this integral, wouldn't it?

 A: For (2) if $x$ is a fixed rational, then the function $g_x: [0,1] \to \mathbb{R}$  where $g_x(y) = f(x,y)$ is given by
$$g_x(y) = \begin{cases}1, & y \in [0,1]\setminus\mathbb{Q} \\ 1- 1/q \neq 1, & y \in \mathbb{Q}\cap[0,1] \end{cases}$$
This is an everywhere discontinuous Dirichlet function and is not Riemann integrable. Note that $g_x$ is discontinuous at any irrational point $\xi$, since by the density of the rationals, there is a sequence of rationals $r_n \to \xi$ such that $g_x(r_n) \not\to g_x(\xi)$.  Similarly $g_x$ is discontinuous at every rational point.
A: Seeing as there is already an answer for your question (2), I'll address (1). Spivak already proved that $\mathscr{L}$ is integrable on $A$. The key thing is if you redefine $\mathscr{L}$ at finitely many points of $A$, then it is still integrable on $A$, and
\begin{equation}
\int_A \mathscr{L}_{\text{new}} = \int_A \mathscr{L}_{\text{old}}. 
\end{equation} 
However, if you redefine an integrable function at infinitely many points (even if it only has measure zero) the result may not be integrable, as the following example shows:
Consider $\phi: [0,1] \to \mathbb{R}$, $\phi(x) = 0$, which is clearly integrable, and its modification $\psi: [0,1] \to \mathbb{R}$, 
\begin{equation}
\psi(x) = 
\begin{cases}
1 & \text{if $x \in \mathbb{Q}$} \\
0 & \text{if $x \notin \mathbb{Q}$}
\end{cases}
\end{equation}
Notice that $\psi$ is obtained from $\phi$ by redefining it on a subset of the rationals, which are countable and hence have measure zero. However, $\psi$ is nowhere continuous, and hence not integrable (I'm using the fact that Riemann-integrability is equivalent to set of discontinuities having measure zero). So it doesn't make sense to say
\begin{equation}
\int_0^1 \phi = \int_0^1 \psi
\end{equation}
If however, you can prove that even after redefining $\mathscr{L}$ on a set of measure zero, it remains integrable, then yes the equation you wrote is true.
