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?
 A: 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 $\frac{1}{2}$.
