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:

  1. [Not relevant for this question]
  2. 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.
  3. 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.
  4. 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:

  1. 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.
  2. 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?
  • 1
    $\begingroup$ In a nutshell: in Theorem 3-8 sets of two-dimensional measure $0$ are considered, whereas examples below Theorem 3-10 deal with sets of one-dimensional measure $0$. And the example in Remark 4 just shows what can happen if $h(x)$ is not integrable for $x$ belonging to a set of one-dimensional measure $0$. $\endgroup$ – user539887 Apr 21 at 9:22

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.


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.


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