# Prove that $f(x)=x$ is Riemann-integrable on $[0,1]$ with $\int_0^1xdx=\frac{1}{2}$

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 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 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 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 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. – proximal Jan 11 at 7:05
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## 2 Answers

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 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. – proximal Jan 11 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 at 5:38

Using what you have, observe that $$|U - L| \leq \sum_{i=1}^N |x_i - x_{i-1}|^2 \leq \sup_i |x_i - x_{i-1}| \sum_{i=1}^N x_i - x_{i-1}$$ $$= \sup_i (x_i - x_{i-1})$$ where we have used that the sum on the right is telescoping and equals $1 - 0$. To say that something is Riemann integrable, we just need to show that if the mesh tends to zero this difference tends to zero. Since the term on the right actually is the mesh, we're done.

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