I've seen the following statement of Fubini's theorem for double integrals in a multivariable course notes:
Let $I=[a,b]×[c,d]$ and $f:I\to \mathbb{R}$ an integrable function. If foreach $x \in [a,b]$ the function $g_x(y)=(x,y)$ is integrable, then $$\int \int_I f=\int_a^b\int_c^df(x,y)dydx$$
But I've also found this more general version:
Let $A \subset \mathbb{R}^n$ and $B \subset \mathbb{R}^m$ be closed boxes (i.e product of closed intervals) and $f:A×B \to \mathbb{R}$ an integrable function. If for each $x \in A$ the function $g_x(y)=f(x,y)$ is integrable, then $$\int_{A×B}f=\int_A\int_B f$$ provided $\int_A\int_B f$ exists.
So my question is why is that extra hypothesis needed in the second version. Vould someone provide an example illustrating this necessity?
Note: both cases refer to the Riemann integral
Consider the function $f:[0,1]^2 \to \mathbb{R}$ where $f(x,y) = 0$ if $x$ or $y$ is irrational and $f(x,y) = 1/q$ if $y$ is rational and $x = p/q$ in lowest terms.
Here, $f$ is integrable over $[0,1]^2$ since the discontinuities are measure $0$ in $\mathbb{R}^2$, and it can be shown that
$$\int_{[0,1]^2} f = 0.$$
For fixed irrational $y$, the function $f(\cdot,y) = 0$ is integrable with $g(y) =\int_0^1 f(x,y) \, dx = 0$. For fixed rational $y$, the function $f(\cdot,y)$ is an integrable Thomae function where $g(y) = \int_0^1 f(x,y) \, dx = 0$ as well.
Hence, $g$ is integrable and
$$0 = \int_{[0,1]^2} f = \int_0^1 g(y) \, dy =\int_0^1 \left(\int_0^1 f(x,y) \, dx \right) \, dy $$
However, for fixed rational $x$, the function $f(x, \cdot)$ is a non-integrable Dirichlet function and $\int_0^1 f(x,y) \, dy$ does not exist as a Riemann integral.
In this case, the iterated integral
$$\int_0^1 \left(\int_0^1 f(x,y) \, dy \right) \, dx$$
does not exist.