About the remark 3 about Fubini's Theorem in "Calculus on Manifolds" by Michael Spivak on p.60

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enter image description here I am reading "Calculus on Manifolds" by Michael Spivak.

On p.60, there is a remark(remark 3) about Fubini's theorem in this book.

Spivak wrote:

Since $\int_A \mathcal{L}$ remains unchanged if $\mathcal{L}$ is redefined at a finite number of points, we can still write $\int_{A \times B} f = \int_A (\int_B f(x, y) dy) dx$, provided that $\int_B f(x, y) dy$ is defined arbitrarily, say as $0$, when it does not exist.

What is the meaning of "$\int_B f(x, y) dy$ is defined as $0$"?

I think the meaning of $\int_B f(x, y) dy$ is already defined (and in this case $\int_B f(x, y) dy$ does not exist), so we cannot define arbitrarily the value of $\int_B f(x, y) dy$.

Instead, for some $a \in A$, we can redefine the values of $f$ like $f(a, y) = 0$ for all $y \in B$, then $\int_B f(a, y) dy = 0$ .

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What's the meaning of the symbol $\int_B f(x,y) \, dy$? It means we're fixing some $x \in A$, and considering the function $y \mapsto f_x(y):= f(x,y)$ from $B \to \Bbb{R}$, and if this function $f_x(\cdot) = f(x,\cdot)$ is integrable on $B$, we get a number \begin{align} \int_B f_x \equiv \int_B f(x,y) \, dy \in \Bbb{R} \end{align} (where $\equiv$ means same thing, but in different notation). So, we now have a function \begin{align} \phi: D_{\phi} :=\{x \in A| \, \, \text{$f_x:B \to \Bbb{R}$ is integrable on $B$}\} &\to \Bbb{R} \end{align} defined by the rule \begin{align} \phi(x) := \int_B f_x = \int_B f(x,y) \, dy \end{align} What Spivak is saying is that one possible extension of $\phi$ is the map $\Phi:A \to \Bbb{R}$ defined by \begin{align} \Phi(x) &:= \begin{cases} \phi(x) & \text{ if $x \in D_{\phi}$}\\ 0 & \text{otherwise} \end{cases} \\\\ &= \begin{cases} \int_B f(x,y) \, dy & \text{if $f_x(\cdot) = f(x,\cdot)$ is integrable on $B$} \\ 0 & \text{otherwise} \end{cases} \end{align}

Pretty much the point of this remark is to try to give meaning to the equation $\int_{A \times B}f = \int_A \int_B f(x,y) \, dy \, dx$, in the unfortunate circumstance that the RHS strictly speaking doesn't make sense (because of the lack of regularity).