if $\int_Q f$ exists then $\int_{y \in B} f(x, y)$ exists for each $x \in A − D$, where $D$ is of measure zero. Let $f : Q = A \times B \to \Bbb R$ be bounded, where $A, B$ are respectively rectangles in $\Bbb R^l$ and $\Bbb R^k$. Show that if $\int_Q f$ exists then
$\int_{y \in B} f(x, y)$ exists for each $x \in A − D$, where $D$ is of measure zero.
Require Hints to solve the problem.
 A: If you are allowed to use Fubini theorem, then it is much easier:
\begin{align*}
\infty>\int_{A\times B}|f|=\int_{A}\int_{B}|f(x,y)|d\mu(y)d\mu(x),
\end{align*}
so $F(x)=\displaystyle\int_{B}|f(x,y)|d\mu(y)$ is a (measurable) function that $\displaystyle\int_{A}|F(x)|d\mu(x)<\infty$, a standard fact from real analysis says that $F(x)<\infty$ a.e. $A$-$\mu(dx)$, in other words, $\displaystyle
\int_{B}f(x,y)d\mu(y)$ exists a.e. $A$-$\mu(dx)$.
I will include the standard fact: Suppose $\displaystyle\int_{X}|f|d\mu<\infty$ then $|f|<\infty$ $\mu$-a.e. Write $S=\{x\in X: |f(x)|=\infty\}$, we are to show $\mu(S)=0$. Now $S=\displaystyle\bigcap_{n}S_{n}:=\bigcap_{n}\{x\in X:|f(x)|\geq n\}$, we have $n\mu(S_{n})\leq n\displaystyle\int_{X}\chi_{S_{n}}d\mu\leq\int_{X}|f|d\mu$, so $\mu(S)\leq\dfrac{1}{n}\displaystyle\int_{X}|f|d\mu$, now taking $n\rightarrow\infty$.
A: Hints: (if you are working with Riemann integration)
(1) For $f$ bounded and Riemann integrable on $Q = A \times B$ show that 
$$\int_Q f = \int_{x \in A} \underline{\int}_{y \in B}f(x,y) = \int_{x \in A}\overline{\int}_{y \in B}f(x,y)$$
where $\underline{\int}$ and $\overline{\int}$ denote lower and upper Darboux integrals.
(2) Since it must hold that
$$ \int_{x \in A} \left( \overline{\int}_{y \in B}f(x,y)-  \underline{\int}_{y \in B}f(x,y)  \right) = 0,$$
what can you conclude about where the upper and lower integrals are equal and consequently the existence of the integral $\int_{y \in B} f(x,y)$ for fixed $x$?
