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I want to Prove the following statement, I will be appreciate if some one help me to do that.

Let $f:[a,b]\to R$ and $f$ is bounded, show that if $f \in R$ ( Riemann integrable) and $\int_a^b fdx=A$ $\iff$ (if and only if) $$A= \lim_{d(p)\to 0} \sum_{k=1}^n f(x_k^*)(x_k-x_{k-1})$$
where $d(p)$ is $\text{max}_{1\le k\le n}|x_k-x_{k-1}|$, ( the diameter of $p$, $p$ is a partition of $[a,b]$ and $\{x_k^*\}_{k=1}^n$ are sample points.

Note: I just know that we should show that for any partition and for any sample point corresponding to partition there exist such a limit and this limit is between upper Riemann sum and lower Riemann sum !

quid
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mahsa
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    You need to tell us what your definition of $\int_a^b f,dx$ is. Saying that it is a "Riemann integral" is not enough: what is your definition of "Riemann integral"? – Jesse Madnick Dec 07 '15 at 04:13
  • What have you tried? Indeed, if you'd like have any hints; you should show others your attempts. :-) – Mikasa Dec 07 '15 at 04:14
  • $f$ is bounded and we want to show if there is such a limit, the value of limit is equal to the integral of $f$ in $[a,b]$ .And if $f$ is integrable the value of integral is equal to the value of limit. – mahsa Dec 07 '15 at 04:22
  • Actually I make the question better but I just know that we should show that for any partition and for any sample point corresponding to partition there exist such a limit and this limit is between upper Riemann sum and lower Riemann sum ! – mahsa Dec 07 '15 at 04:36
  • @mahsaghanbarpour: If you meant Actually I make the question better. But what I just know is to show for any partition and for any sample point corresponding to partition, there exists such a limit and this limit is between upper Riemann sum and lower Riemann sum. so add it below your question's body. – Mikasa Dec 07 '15 at 04:40
  • There are two definitions for the notion of Riemann Integral of a bounded function on a closed interval. I think you want to show the equivalence of these definitions. The proof is not obvious, but at the same time not too difficult either. Please see my answer http://math.stackexchange.com/a/1255344/72031 – Paramanand Singh Dec 07 '15 at 05:42

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