Let A be a rectangle in $R^k$ , let B be a rectangle in $R^n$, let Q = A x B. Let f: Q $\rightarrow R$ be a bounded function. Show that if $\int_Q {f}$ exists, then $\int_{y \in B} {f(x,y)} $ (denoted by I) exits for x $\in$ A - D, where D is a set of measure zero in $R^k$.
Attempt at a solution : By Fubini's theorem we now that $\int_Q {f}$ = $\int_{x \in A} \underline\int_{y \in B} {f(x,y)}$ = $\int_{x \in A} \overline\int_{y \in B} {f(x,y)}$ From above we now that $\int_{y \in B} {f(x,y)}$ exists since f isintegrable over Q. We need to show that it also exists for x $\in$ A - D. Let P = $(P_A, P_B)$ be a partition for Q, We know that $m_{R_A * R_B} \leq$$m_{R_A} \leq m_{R_{A-D}} $ and V(A-D) = V(A) - V(D) and V(D) = 0. We can use this to show that L(f,P) $\leq L(I, P_A) \leq L(I, P_{A-D}) $ and similarly for the uper sums. and Since all three integrals will by definition exist between the inequalities at the extreme ends and since the partitions are arbitrary we can conclude that I exists for A- D, and is equal to I for A. This is just a rough draft at an attempt at the proof, I'm not sure if its even the right idea, please let me know in particular about my argument regarding V(A-D) = V(A) and whether showing that I is integrable over A-D is the same as showing that I exists for A-D. Thank you