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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.

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    $\begingroup$ I don't understand what you don't understand... The last inequality that you write (the one that looks false to you) is obviously true, since $m$ and $M$ are the minimum and, respectively, maximum of $f$ on $[a,b]$. $\endgroup$
    – Alex M.
    Dec 28, 2015 at 20:47
  • $\begingroup$ @AlexM. Yes, I thought the same, then I thought that since $m=f(x_j)$ for some $j \in I$, where $I$ is our indexing set, and $M=f(x_k)$ for some $k \in I$, then we can reduce the inequality to $0 \le S \le 0$, where $S=\sum \limits_{i=1}^n$ with two few terms (namely $f(x_j) \wedge f(x_k)$). Although my thinking is most likely wrong. $\endgroup$ Dec 28, 2015 at 20:55

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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.

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  • $\begingroup$ Thanks, but is the step where I remove $\lim_{n \to \infty} \frac{b-a}{n}$ from all sides valid? $\endgroup$ Dec 28, 2015 at 20:36
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    $\begingroup$ Yes, in general what you can do to prove a $\lim$ inequality is : remove the limit, prove the inequality for $n$, $\textbf{prove that the limit exists}$, reapply the limit. $\endgroup$
    – mrprottolo
    Dec 28, 2015 at 20:39
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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.)

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