If $f$ is continuous on $[0,1]$ where $\sigma_n = \{0,\frac{1}{n}, \frac{2}{n}, . . . . . \frac{n}{n}\}$ and where $x^*_k$ is any point on the interval $[\frac{(k-1)}{n}, \frac{k}{n}]$ (where $k = 1, . . . ., n$)
I'm trying to show that $L[f, \sigma_n] \leq \frac{1}{n}\sum^n_\mathrm{k=1}f(x^*_k) \leq U[f, \sigma_n]$.
Then I'm trying to show that $lim_\mathrm{n\rightarrow\infty}\frac{1}{n}\sum^n_\mathrm{k=1} f(x^*_k) = \int^1_0 f$
Here is my approach:
Since $f$ is continuous on the interval $I_k$, we know it obtains a maximum and a minimum on $I_k$.
Say that for the maximum, there exists $x_\mathrm{max} \in I_k$ such that $M[f, I_k] = f(x_\mathrm{max})$. For the minimum, there exists $x_\mathrm{min} \in I_k$ such that $m[f, I_k] = f(x_\mathrm{min})$. Hence we know that for any $x^*_k \in I_k$, we know that $$m[f, I_k] \leq f(x^*_k) \leq M[f, I_k]$$
Then we just have to multiply the above equation by $|I_k|$ and sum over $k$ to get $$L[f, \sigma_n] \leq \frac{1}{n}\sum^n_\mathrm{k=1}f(x^*_k) \leq U[f, \sigma_n]$$
For the second part, I'd say that $U[f, \sigma_n] - L[f, \sigma_n] < \epsilon$ by uniform continuity of $f$. I'm trying to use $L[f, \sigma_n]$ to work toward the integral. Not entirely certain where to go from here.