Let $A$ be a rectangle in $\mathbb{R^k}$; let $B$ be a rectangle in $\mathbb{R^n}$; let $Q=A\times B$. Let $f: Q\to \mathbb{R}$ be a bounded function. Show that if $\int_Q f$ exists, then
$$\int_{y\in B}f(x,y)$$ exists for $x\in A-D$, where $D$ is a set of measure zero in $\mathbb{R^k}$.
I'm having difficulty proving this.
I think I might have to use these theorems in the proof.
My work:
Let $D$ be a set of measure zero in $\mathbb{R^k}$, and $x\in A-D$.
Since $f$ is integrable on $Q$, $f$ is continuous except in a measure zero set, say $E$.
Claim: Let $B$ be a rectangle in $\mathbb{R^n}$. If $D$ is a set of measure zero in $\mathbb{R^k}$, then $D\times B$ is a set of measure zero in $\mathbb{R^{k+n}}$.
Proof of Claim: Let $\epsilon\gt 0$ be given.Since $D$ is a set of measure zero in $\mathbb{R^k}$, there exists a countable set of rectangles $\{Q_i\}$ whose union covers $D$ and the sum of the volumes is less than $\epsilon/v(B)$, where $v(B)$ is the volume of $B$. Now consider the countable set of rectangles $Q_i \times B$. Then the union of all of these rectangles cover $D\times B$, and $\sum v(Q_i\times B)=\sum v(Q_i)\times v(B)\lt \epsilon$. QED.
Hence by the claim above, $E \cup (D\times B)$ is a set of measure zero in $\mathbb{R^{k+n}}$. Hence, $f$ is continuous almost everywhere when $x\in A-D$.
Now for a fixed $x\in A-D$, $f$ is continuous almost everywhere in $\mathbb{R^{k+n}}$. However, to complete the proof of this problem, I need to show that $f(x,y)$, for a fixed $x\in A-D$ is continuous almost everywhere in $\mathbb{R^n}$, but I don't know how to show this part.
I would greatly appreciate any hints, suggestions or solutions.
