Show that if $f$ is monotone on $[a,b]$, every Riemann sum of $f$ is a Lesbesgue sum.
Intuitively, I think it's correct but I don't know how to express it in mathematical language. Here is my idea, $f$ is monotonic implies that $y_{i}^*\leq y_{i+1}^*\leq y_{i+2}^*\leq\cdots$ for $x_i^*\leq x_{i+1}^*\leq x_{i+2}^*\leq\cdots.$ and also monotonic function implies that the domain is an interval. The Riemann sum is denoted as below. $R(f;P) = \sum_{i=1}^{n}f(x_i^*)(x_i-x_{i-1})$. Here $f(x_i^*)$ is actually $y_i^*$ and we try to group all intervals $x_j-x_{j-1}$ such that $y_i^*=y_j^*$. Thus
$$R(f;P) = \sum_{i=1}^{k}y_i^*\sum_{j\in B}(x_j-x_{j-1})$$ where $B=\{j\in\mathbb{N}|y_i^*=y_j^*\}$. Therefore, $$R(f;P) = \sum_{i=1}^{k}y_i^*\sum_{j\in B}(x_j-x_{j-1})=\sum_{i=1}^{k}y_i^*.m(\{x\in A|y_{i-1}\leq f(x)<y_i\})=L(g;P)$$ (Note that$\sum_{j\in B}(x_j-x_{j-1})=m(\{x\in A|y_{i-1}\leq f(x)<y_i\})$because $ y_{i-1}\leq y_i^*<y_i$ and $m(\{x\in A|y_{i-1}\leq f(x)<y_i\})$ is an interval since it is monotone.)
Here is another question. Is the converse true ? If Riemann sum of $f$ is a Lesbesgue sum, then is $f$ monotone? ( I somehow feel like $x^2$ is the counterexample. Am I correct?)