It can be proven that
Suppose $f:[a,b]\times[c,d]\to\mathbf R$ is bounded and integrable. For any fixed $y$, suppose $$A(y)=\int_a^bf(x,y)\,\mathrm dx$$ exists. If $\int_c^dA(y)\,\mathrm dy$ exists, then it is equal to the integral of $f$.
The question is why do we need the assumption of the existence of $\int_c^dA(y)\,\mathrm dy$. We (my friends and I) tried finding examples using combinations of the usual "something-if-rational-zero-otherwise" trick but none of them work. Is there such an example?
EDIT
Just to clarify, my question is the following. Suppose $f:[a,b]\times[c,d]\to\mathbf R$ is bounded and integrable. Suppose $A(y)=\int_a^bf(x,y)\,\mathrm dx$ exists for all $y$. Does $\int_c^dA(y)\,\mathrm dy$ necessarily exist?
Suppose $f$ is bounded and integrable on $[a,b]\times[c,d]$. I assume that integrability here pertains to the existence of the multivariate Riemann integral, which I will distinguish from iterated integrals by denoting as
$$\int_{[a,b]\times[c,d]}f(x,y) \,d(x,y)$$
For an example where $A(y) = \int_a^bf(x,y) \, dx$ does not exist as a Riemann integral for every $y \in [c,d]$, take $f: [0,1]\times[0,1] \to \mathbb{R}$ where
$$f(x,y)= \begin{cases} 0, & x \not\in \mathbb{Q}\\0, & x \in \mathbb{Q}, \,\,y \not\in\mathbb{Q}\\1/q,& x \in \mathbb{Q}, \,\, y=p/q,\,(p,q) =1 \end{cases}$$
Here we have $f(x,y)$ equal to zero if either $x$ or $y$ is irrational, but non-zero and equal to $1/q$ if both $x$ and $y$ are rational with $y = p/q$ in lowest terms. It can be shown that $f$ has a multivariate Riemann integral by extending the proof for the univariate case Thomae's function. If $y$ is any irrational number in $[0,1]$ then the Riemann integral $A(y)$ obviously exists and takes the value $0$. However, if $y=p/q$ is a rational number in $[0,1]$, then
$$f(x,y) = \begin{cases} 1/q, & x\in \mathbb{Q}\\0,& x\notin\mathbb{Q} \end{cases}$$
In this case, $f(\cdot,y)$ is a Dirichlet-type function that is everywhere discontinuous on $[0,1]$ and is not Riemann integrable. Hence, $A(y)$ fails to exist for rational values of $y$.
This is actually the worst that can happen, and we have the following:
Theorem. If $f$ is bounded and integrable in the sense of the multivariate Riemann integral on $[a,b]\times[c,d]$, then the Riemann integral $A(y)= \int_a^b f(x,y) \, dx$ will always exist except possibly on a set of Lebesgue measure $0$.
Proof:
Since $f$ is bounded, for any $y \in [c,d]$ there always exist upper and lower Darboux integrals
$$\overline{\int_a^b} f(x,y) \, dx, \quad \underline{\int_a}^bf(x,y) \,dx,$$
and there is a version of Fubini's theorem for Riemann integrals (Munkres, Analysis on Manifolds, Theorem 12.2) that states
$$\int_{[a,b]\times[c,d]}f(x,y) \,d(x,y)= \int_c^d\left(\overline{\int_a^b} f(x,y) \, dx\right)\, dy = \int_c^d \left(\underline{\int_a}^bf(x,y) \,dx\right) \, dy$$
Hence,
$$\tag{*}\int_c^d \left(\overline{\int_a^b} f(x,y) \, dx- \underline{\int_a}^bf(x,y) \,dx\right) \, dy = 0$$
Since the integrand in (*) is a nonnegative difference of upper and lower integrals, it must be zero for almost every $y \in [c,d]$ in order that the integral vanish. In other words, there is a set $S$ of Lebesgue measure zero such that for all $y \in [c,d] - S$, we have
$$\overline{\int_a^b} f(x,y) \, dx= \underline{\int_a}^bf(x,y) \,dx$$
As the upper and lower integrals are equal, it follows that $f(\cdot,y)$ is Riemann integrable on $[a,b]$ for all $y \in [c,d]-S$.