Suppose we have functions $f_1:\Bbb R^m\rightarrow \Bbb R$ and $f_2:\Bbb R^n\rightarrow \Bbb R$ for some natural number $m,n$, and both of $f_1,f_2$ are bounded and non-negative.Now define $f:\Bbb R^m\times\Bbb R^n\rightarrow \Bbb R$ by $f(x,y)=f_1(x)f_2(y)$.
Now consider $A=A_1\times A_2\subset \Bbb R^m\times \Bbb R^n$ which is a closed bounded rectangle. I need to prove that if $f_1,f_2$ are integrable over $A_1,A_2$ respectively, then $f$ is integrable (over $A$).
In fact, I want to prove that: to be explicit, the integral over $A$ of $f$ is given by:
$$\int_A f(x)dx=(\int_{A_1}f_1(x_1)dx_1 )\cdot (\int_{A_2}f_2(x_2)dx_2)$$
I need a Riemann-Darboux proof here. So I am thinking of partition the set $A_1\times A_2$ into a "product partition". But then I feel a bit lost here.
Could someone please prove it? Or any help including providing some reference would be appreciate. Thanks in advance!
Let $P = P_1 \times P_2$ be a partition of $A_1 \times A_2$ where $P_1$ and $P_2$ are partitions of $A_1$ and $A_2$, respectively. Since $f_1$ and $f_2$ are nonnegative, for any subrectangle $R = R_1 \times R_2$ of $P$ we have
$$\inf_{(x_1,x_2) \in R}f_1(x_1)f_2(x_2) = \inf_{x_1 \in R_1}f_1(x_1) \inf_{x_2 \in R_2}f_2(x_2), \\ \sup_{(x_1,x_2) \in R}f_1(x_1)f_2(x_2) = \sup_{x_1 \in R_1}f_1(x_1) \sup_{x_2 \in R_2}f_2(x_2), $$
and lower and upper sums take the form
$$L(P,f)= L(P_1,f_1)L(P_2,f_2), \\ U(P,f)= U(P_1,f_1)U(P_2,f_2). $$
Thus
$$U(P,f) - L(P,f) = U(P_1,f_1)U(P_2,f_2) - L(P_1,f_1)L(P_2,f_2) \\ = [U(P_1,f_1) - L(P_1,f_1)]\, U(P_2,f_2) + [U(P_2,f_2) - L(P_2,f_2)]\, L(P_1,f_1).$$
Let $M_1$ and $M_2$ be upper bounds for $f_1$ and $f_2$, respectively.
It follows that $$U(P_2,f_2) \leqslant M_2 \, \text{vol}(A_2)\,\, , \\ L(P_1,f_1) \leqslant M_1 \, \text{vol}(A_1), \\ U(P,f) - L(P,f) \leqslant M_2[U(P_1,f_1) - L(P_1,f_1)] + M_1 [U(P_2,f_2) - L(P_2,f_2)]. $$
Since $f_1$ and $f_2$ are integrable, for any $\epsilon > 0$ there exist partitions $P_1$ and $P_2$ such that $U(P_1,f_1) - L(P_1,f_1) < \epsilon/(2M_2)$ and $U(P_2,f_2) - L(P_2,f_2) < \epsilon/(2M_1)$.
Whence, $f$ is integrable since
$$U(P,f) - L(P,f) < \epsilon.$$
With $f$ integrable, Fubini's theorem guarantees that
$$\tag{*} \int_{A_1 \times A_2} f = \int_{A_1} f_1 \, \int_{A_2} f_2$$
Proving (*) directly would follow the same steps as the proof of Fubini's theorem with some simplifications along the way for this specific form of $f$.
Procceding, given a subrectangle $R_1 \times R_2$, we have with $x_1 \in R_1$ fixed on the RHS
$$\inf_{(x_1,x_2) \in R_1 \times R_2}f_1(x_1) f_2(x_2) \leqslant f_1(x_1) \inf_{x_2 \in R_2}f_2(x_2),$$
and
$$\sum_{R_2}\inf_{(x_1,x_2) \in R_1 \times R_2}f_1(x_1)f_2(x_2) \, \text{vol}(R_2) \leqslant f_1(x_1)L(f_2, P_2) \leqslant f_1(x_1) \int_{A_2} f_2 $$
Take the infimum of both sides over $R_1$, multiply by $\text{vol}(R_1)$ and sum to get
$$\tag{1}L(P,f) = \sum_{R_1}\sum_{R_2}\inf_{(x_1,x_2) \in R_1 \times R_2}f_1(x_1)f_2(x_2) \, \text{vol}(R_1)\,\text{vol}(R_2) \\ \leqslant L(P_1,f_1) \int_{A_2} f_2 \\ \leqslant \int_{A_1} f_1 \int_{A_2} f_2 $$
Similarly we can show
$$\tag{2}\int_{A_1} f_1 \int_{A_2} f_2 \leqslant U(P,f).$$
Since (1) and (2) are true for any partition $P$ we have
$$\int_{A_1 \times A_2} f = \sup_P L(P,f ) \leqslant \int_{A_1} f_1 \int_{A_2} f_2 \leqslant \inf_P U(P,f) = \int_{A_1 \times A_2} f $$