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This is from page 90 of Munkres. We say $f: [0,1] \to \mathbb{R}$ is increasing if $f(x_1) \geq f(x_2)$ whenever $x_1 \geq x_2$. If $f,g : [0,1] \to \mathbb{R}$ are increasing and non-negative, show that the function $h(x,y) = f(x)g(y)$ is integrable over $[0,1]$.

My attempt: Consider that if either $f$ or $g$ are constantly zero, then the function is constant and integrable. Therefore, we assume that both are zero only at countably many or finitely many points but certainly not all. The function $h(x,y)$ is also increasing.

Take an equal partition $P$ where each cube has area $\dfrac{1}{N^2}$ was an idea but I'm not sure

dansidorkin
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    First show that for a monotone univariate function, it is integrable. Consider $U(P,f)-L(P,f)$ and note that the sum telescopes, then let $n\to\infty$. Think about how to extend this to a function on the unit square. – Andrew Jan 10 '23 at 20:59
  • So I figured it out for the monotone uni-variate function case, and I saw what you meant when you said the sum was telescoping. I ended up extending it to the unit-square, and the sum still telescopes except the area of each cube is now $\frac{1}{N^3}$. So the only thing I'm left with is $\frac{M_f(R_N)-m_f(R_1)}{N^3}$. And when $N \to \infty$ we get $U(f, P) - L(f,P) \to 0$. Let me know if that makes any sense! (In this case, I labelled my first rectangle $R_1$ and last $R_N$ for convenience). – dansidorkin Jan 10 '23 at 21:18
  • I’m not actually sure if the result still holds in space (i.e. $\mathbb R^3$, but it probably does?) However, your computations in $\mathbb R^3$ are not correct. Try it in $\mathbb R^2$ first. ETA: This result definitely generalizes to $\mathbb R^n$. I was thinking of $g(x_1,\dots,x_n)$ which is monotone increasing in each variable but does not factor into univariate functions (I think this also holds true) – Andrew Jan 10 '23 at 21:25
  • I'm working in $[0,1]^2 \times \mathbb{R}$, does that change anything? – dansidorkin Jan 10 '23 at 21:29
  • Isn’t your question to show $h$ is integrable? $h$ is defined on the plane. Anyways, the sum does not telescope so nicely (some terms do, but not all). You need to be more careful – Andrew Jan 10 '23 at 21:30
  • Yes, that $h$ is integrable, my bad I had a typo with $f$. But it's asking integrability over $[0,1]^2$ so the function is $[0,1]^2$ into $\mathbb{R}$. Basically, $h$ is monotone increasing since $f$ and $g$ are monotone increasing so the inf and sup for some cubes will be the same and it's practically all of them except the end points. – dansidorkin Jan 10 '23 at 21:33
  • I think there is some misunderstanding. If you are showing a function defined on the plane is integrable, you are working with squares. In space, you work with cubes. (Remember that on the real line you work with closed intervals). Anyways, write it out carefully for the case $N=2$, i.e. you have split up the unit square into 4 squares. You will see that not everything telescopes like you wrote. – Andrew Jan 10 '23 at 21:40
  • Yes it does! @grand_chat Thank you! – dansidorkin Jan 11 '23 at 00:14
  • @AndrewZhang I'm not really sure why all this time I was working with a subset of $\mathbb{R}^3$ instead of a plane because my initial univariate calculation was done a plane. Thoroughly confused myself! – dansidorkin Jan 11 '23 at 00:18

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