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I think that the best thing to do is prove that the upper and lower sums are equal in the limit. Since $f$ is monotonic I know that for any partition $\{x_0,\dots,x_N\}$ the upper and lower sums are given by $$U=\sum_{i=1}^Nx_i(x_i-x_{i-1})$$and$$L=\sum_{i=1}^Nx_{i-1}(x_i-x_{i-1})$$respectively. I considered showing that the the limit of $U-L$ as $N\rightarrow\infty$ is $0$, hoping that I would get some kind of telescoping situation, but that doesn't seem to be happening:$$U-L=\sum_{i=1}^N(x_i-x_{i-1})(x_i-x_{i-1})$$I can't see a nice way to show that that is going to be less than any $\epsilon$. Does this seem like the right approach? Am I missing something?

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Since the function is bounded and continuous on $[0,1]$, then it is Riemann integrable. –  Mhenni Benghorbal Jan 11 '13 at 6:48
    
Hint: because the $N$ points $x_i$ are $1/N$ units apart (modulo an off-by-one error), the product in your sum will be roughly $1/N^2$ and you'll be summing $N$ terms of that size. What will the result be, and can you formalize that? –  Steven Stadnicki Jan 11 '13 at 6:48
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@MhenniBenghorbal I know that it is. I'm meant to show that it's true from the definition of the Riemann integral. –  crf Jan 11 '13 at 6:49
    
@StevenStadnicki can I specify that that's the case about the partition? Don't I need to show that it holds for any partition? –  crf Jan 11 '13 at 6:50
    
@crf Usually, yes, you need to work with an arbitrary partition. However, if you use a partition where the points are evenly spaced $1/N$ apart, you can use the formula $\sum_{k=1}^n k = \frac{n(n+1)}{2}$ to get a nice formula for the upper and lower sums for this sort of partition. Let $N \to \infty$ and use the fact that upper sums are upper bounds for lower sums, and lower sums are lower bounds for upper sums. –  kigen Jan 11 '13 at 7:05

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up vote 2 down vote accepted

I turned my comment into an answer.

Let $P_N$ be the partition $\{0,\frac{1}{N},\frac{2}{N},...,1\}$, i.e. each point is evenly spaced with distance $1/N$. The upper and lower sums for such a partition are:

$$ U(P_N,f) = \sum_{i=1}^N \sup_{[\frac{i-1}{N},\frac{i}{N}]}x \cdot \Delta x_i = \sum_{i=1}^N \frac{i}{N} \cdot \frac{1}{N} = \frac{1}{N^2}\frac{N(N+1)}{2} = \frac{1}{2} + \frac{1}{2N}$$

$$ L(P_N,f) = \sum_{i=1}^N \inf_{[\frac{i-1}{N},\frac{i}{N}]}x \cdot \Delta x_i = \sum_{i=1}^N \frac{i-1}{N} \cdot \frac{1}{N} = \frac{1}{N^2}\frac{N(N-1)}{2} = \frac{1}{2} - \frac{1}{2N}. $$

Let $N \to \infty$. Both $U(P_N,f)$ and $L(P_N,f)$ go to $\frac{1}{2}$, from above and below respectively. Since upper sums are upper bounds for lower sums, every lower sum is bounded above by $\frac{1}{2}$. Since there are lower sums that are arbitrarily close to $\frac{1}{2}$ (take $N$ large enough) it follows that $\frac{1}{2}$ is the least upper bound for the lower sums. Simliarly we conclude that $\frac{1}{2}$ is the greatest lower bound for the upper sums. This shows that $f(x)$ is Riemann integrable on $[0,1]$ with integral $\frac{1}{2}$.

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This looks good, but I'm thinking of the case where we have a partition $P$ which includes an irrational number... Then there is no $P_N$ which is a refinement of $P$. Does that matter? –  crf Jan 11 '13 at 7:40
    
It doesn't matter. There's no need to consider refinements at all. If you're looking to just show integrability and not the value of the integral, then just subtract these two expressions: you'll get $|U-L| = \frac{1}{4N} \to 0$, so using the alternate definition of integrability (the difference of the upper and lower sums can be made arbitrarily small using partitions) we get our desired result. –  kigen Jan 11 '13 at 7:55
    
Yeah reviewed notes, saw alternate and much nicer definition of integrability, now I am happy. I actually suspected that we'd be using that $N(N+1)/2$ business; I was just worried about capturing every partition. But this is great, thank you very much! –  crf Jan 13 '13 at 5:38
    
@kigen where did you get $\frac{i-1}{N},$ from on $L(P_N, f)$? –  user104235 Nov 30 '13 at 7:13
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@user104235 $\frac{i-1}{N}=\inf x$ for $x$ in the interval $[\frac{i-1}{N},\frac{i}{N}]$. –  kigen Nov 30 '13 at 17:54

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