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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 $n\rightarrow\infty$?

Paul S.
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2 Answers2

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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 $$ \left\lvert\sum_R v(R)(M_R(f)-m_R(f))\right\rvert\leq\frac{4nf(1)g(1)}{n^2}. $$

Nick Peterson
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    Oh, I didn't think about bounding every term with $f(1)$ and $g(1)$. But that works because the number of terms is linear in $n$. Thanks, Nicholas! – Paul S. Aug 30 '13 at 14:04
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    However, we do not know that $f$ and $g$ are bounded on $I$, so how can you show that that bound can be made less than $\epsilon$ with an appropriate $n$ ? – Our Jun 15 '18 at 10:26
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    Because $f(1)$ and $g(1)$ are just fixed numbers -- they don't depend on $n$. So, when you cancel an $n$ from the numerator and denominator, you're left with some number over $n$... and it is pretty easy to make that small by making $n$ large. – Nick Peterson Jun 15 '18 at 12:23
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    @NickPeterson Ok, thank for your answer. – Our Jun 16 '18 at 05:13
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$$\sum_{R} M_R(f)v(R)=\sum_{1\leq i\leq n\\1\leq j\leq n}f(\frac{i}{n})g(\frac{j}{n})\frac{1}{n^2}\leq\sum_{0\leq i\leq n\\0\leq j\leq n}f(\frac{i}{n})g(\frac{j}{n})\frac{1}{n^2}.$$
$$\sum_{R} m_R(f)v(R)=\sum_{0\leq i\leq n-1\\0\leq j\leq n-1}f(\frac{i}{n})g(\frac{j}{n})\frac{1}{n^2}.$$

$$\sum_{R} M_R(f)v(R)-\sum_{R} m_R(f)v(R)\leq\sum_{0\leq i\leq n\\0\leq j\leq n}f(\frac{i}{n})g(\frac{j}{n})\frac{1}{n^2}-\sum_{0\leq i\leq n-1\\0\leq j\leq n-1}f(\frac{i}{n})g(\frac{j}{n})\frac{1}{n^2}\\=\frac{1}{n^2}(f(\frac{0}{n})g(\frac{n}{n})+f(\frac{1}{n})g(\frac{n}{n})+\dots+f(\frac{n}{n})g(\frac{n}{n})+f(\frac{n}{n})g(\frac{n-1}{n})+\dots+f(\frac{n}{n})g(\frac{0}{n}))\leq\frac{1}{n^2}(2n+1)f(1)g(1)\to 0(n\to\infty).$$

tchappy ha
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