Deriving an inequality for Riemann sums

Question

I am reading a proof of the MVT for Integrals and the following inequality seems a bit non-intuitive and I'm not certain as to how it is derived (although the writer has seemingly indicated its derivation is from the fact that $m\le f(x_i)\le M, \forall x_i \in [a,b]$ and $m,M \in [a,b]$):

since $m \le f(x_i ^*) \le M$ for all $x_i ^* \in [a, b]$, we have

$$\lim \limits _{n \to \infty} \frac {b-a} n \sum \limits _{i=1} ^n m \le \lim \limits _{n \to \infty} \frac {b-a} n \sum \limits _{i=1} ^n f(x_i ^*) \le \lim \limits _{n \to \infty} \frac {b-a} n \sum \limits _{i=1} ^n M$$

The furthest I got in attempting to derive this inequality was by cancelling out $\lim \limits _{n \to \infty} \frac{b-a}{n}$ (I'm not sure whether this is valid or not) then trying to prove $\sum \limits _{i=1}^n m \le \sum \limits _{i=1}^n f(x_i) \le \sum \limits _{i=1}^n M$, which itself does not look true.

Answer 1

If $m\le f(x_1)\le M$ and $m\le f(x_2)\le M$, then if you sum those two you get $$2m\le f(x_1)+f(x_2)\le 2M.$$ By induction you should be able to prove $$\sum_{i=1}^n m \le \sum_{i=1}^n f(x_i) \le \sum_{i=1}^nM.$$

You can in the end multiply all by $\frac{a-b}{n}$ and take the limit, that is allowed, after you have proved that the limit actually exists.

Answer 2

This follows from the subsequent fact: Suppose $(a_n)_n, (b_n)_n$ are real sequences converging to $a,b$ respectively and $a_n \leq b_n$. Then $a\leq b$. So you have to understand why all the limits in the above calculation exist (e.g. the left most limit exists because the $n$ actually cancels with the sum.)