We say $f:[0,1]\rightarrow\mathbb{R}$ is increasing if $f(x_1)\le f(x_2)$ whenever $x_1<x_2$. If $f,g:[0,1]\rightarrow \mathbb{R}$ are increasing and non-negative, show that the function $h(x,y)=f(x)g(y)$ is integrable over $[0,1]^2$.
I'm using the Darboux definition of integration, so I want to prove that for any $\epsilon>0$, there exists a partition $P$ of $Q$ such that $U(f,P)-L(f,P)<\epsilon$. Equivalently, there exists a partition $P$ of $Q$ such that $$\left|\sum_Rv(R)(M_R(f)-m_R(f))\right|<\epsilon,$$ where $M_R(f)$ is the supremum of the values of $f$ inside the rectangle $R$, and $m_R(f)$ is the corresponding value for infimum. (Here, $R$ ranges over all subrectangles formed by the partition $P$.)
Toward that end, I tried to take the partition $P$ to be $[0,\dfrac1n,\dfrac2n,\ldots,1]\times[0,\dfrac1n,\dfrac2n,\ldots,1]$. Since the function $h(x,y)$ is increasing in $x$ and $y$, the maximum of any square is attained at the top-right corner, and the minimum at the bottom-left corner. Therefore, $\left|\sum_Rv(R)(M_R(f)-m_R(f))\right|$ is equal to $$\dfrac{1}{n^2}\left(\sum_{i=1}^{n}f(\dfrac in)g(1)+\sum_{i=1}^{n-1}f(1)g(\dfrac in)-\sum_{i=0}^{n-1}f(\dfrac in)g(0)-\sum_{i=1}^{n-1}f(0)g(\dfrac in)\right)$$
I'm not sure if this is the right partition to take. Does this sum approach $0$ as $n\rightarrow\infty$?
Best Answer
Hint: Given the cancellations that you have already performed, along with $f(x)\leq f(1)$ and $g(y)\leq g(1)$ for all $x,y\in[0,1]$, the triangle inequality implies $$ \left\lvert\sum_R v(R)(M_R(f)-m_R(f))\right\rvert\leq\frac{4nf(1)g(1)}{n^2}. $$