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Consider the following exercise:

If $f$ is uniformly continuous in $[a,b]$ then show that $$\int_{a}^{b}f(x)dx=\lim_{n\to\infty}\frac{b-a}{n}\sum_{i=0}^{n-1}f\left(a+\frac{i(b-a)}{n}\right).$$

The definition of Riemann integrable I have is:

A function $f$ is Riemann integrable in $[a,b]$ if and only if there is a number $L\in\mathbb{R}$ such that for every $\varepsilon>0$ there is a partition $P_{\varepsilon}$ such that the inequality $$\left|S(,f,P;\xi_0,...,\xi_{n-1})-L\right|<\varepsilon$$ holds for every partition $P=\{a_0,...,a_n\}$ finer than $P_{\varepsilon}$ and every election of points $\xi_i\in\{a_i,...,a_{i+1}\}.$ The number $L$ is defined as $$L=\int_{a}^{b}f(x)dx.$$ And $S(f,P;\xi_0,...,\xi_{n-1})$ denotes the Riemann sum of $f$ with respect to a partition $P$.

Is it necessary to use that definition of integral to prove the exercise? If so, how can I do it? Any help is welcome.

user926356
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    The essential hint is here: https://math.stackexchange.com/a/314028/42969 – Martin R Sep 15 '22 at 09:13
  • @MartinR Thanks for the hint. However, I am still unable to see how to derive one definition from the other – user926356 Sep 15 '22 at 20:53
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    A continuous function on a compact interval [a,b] is uniformly continuous – FDP Sep 16 '22 at 06:33
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    Use your definition to establish the definition based on norm (mesh) of partition (the proof is difficult). Then the result in your question uses uniform partition (subintervals are of equal length) and as number of intervals tends to infinity the norm tends to $0$. The result holds for any Riemann integrable function and continuity is not necessary. – Paramanand Singh Sep 16 '22 at 08:53

1 Answers1

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Further hint:

With $x_i = a+\frac{i(b-a)}{n}$, we have $x_{i+1}- x_i = \frac{b-a}{n}$ and

$$\left|\int_a^b f(x) \, dx - \frac{b-a}{n}\sum_{i=0}^{n-1}f\left(a+\frac{i(b-a)}{n}\right)\right|= \left|\sum_{i=0}^{n-1}\int_{x_i}^{x_{i+1}} f(x) \, dx - \frac{b-a}{n}\sum_{i=0}^{n-1}f\left(x_i\right)\right|\\= \left|\sum_{i=0}^{n-1}\int_{x_i}^{x_{i+1}} [f(x)-f(x_i)] \, dx \right| \leqslant \ldots $$

RRL
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  • How did you get the left hand side of the second absolute value? I mean everything that is before the minus sign. I understood everything but that. Thanks! – user926356 Sep 16 '22 at 02:33
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    @user926356: Are you asking why $\int_a^b f(x) , dx = \sum_{i=0}^{n-1}\int_{x_i}^{x_{i+1}} f(x) , dx$? – RRL Sep 16 '22 at 02:51
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    You know that if $a < c <b$ then $\int_a^b f(x) , dx = \int_a^c f(x) ,dx + \int_c^b f(x) , dx$ -- correct? Use that repeatedly for $a = x_0 < x_1 < \ldots < x_{n-1} < x_n = b$. – RRL Sep 16 '22 at 02:52
  • Thanks! That was precisely what I was asking. I understand now – user926356 Sep 16 '22 at 03:37
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    @user926356: You’re welcome. I assume you know the next steps: apply triangle inequality and bring in uniform continuity of $f$ to show LHS < $\epsilon$ when $n$ is large. – RRL Sep 16 '22 at 03:42