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

Question

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?

Answer 1

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 $\frac{1}{2}$.

Answer 2

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.