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Definition of the Riemann integral in terms of Riemann sums : $$\int_a^b f(x) \,dx = \lim_{x\to \infty} \frac{b-a}{n}\sum_{j=1}^n f(c_{i})$$

where for each n

$$x_{0} = a, x_{1} = a + \frac{b − a} {n} , . . . , x_{i} = a + i \frac{b − a} {n} , . . . , x_{n} = b, $$ is independent of the choice $$c_{i} ∈ [x_{i−1}, x_{i} ].$$ Prove this result in the case when $f(x)$ is uniformly continuous.

A function is uniformly continuous on the interval $[a, b]$ if for $\varepsilon > 0$, there exists $\delta > 0$ so that,

if $|x − y| < \delta$ then $|f(x) − f(y)| < \varepsilon.$

I believe this is how it was given.

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    The things you prove are statements. The 'Riemann integral' is a type of operator, not a statement, so it is not the type of thing you can prove. Can you be more precise about what you want to know? – 2017-02-10
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    I believe I've made myself more clear in my question now. – 2017-02-10
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    I think what you are asking is to know how to show that the sequence of partial sums converges when $f$ is continuous? – 2017-02-10
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    @James You can't prove that the left-hand and right hand sums are the same for any function (if that's what you mean). The function would need to be Riemann integrable, which is essentially an assumption of some slightly stronger facts than the left-hand and right hand sum converge to the same thing. As Bernard said, you could however show it with different assumptions, like $f$ being continuous on $[a,b]$ – 2017-02-10
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    @James much better – 2017-02-10
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    My apologies, very new to this idea of proofs. – 2017-02-10
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    @James Not a prob. One more minor thing: you don't need a $\Delta x$ in your sum. The $(b-a)/n$ out front *is* $\Delta x$ – 2017-02-10
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    Duly noted. Are you aware of the solution to this problem? – 2017-02-10
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    There is a one line answer. Since $f$ is uniformly continuous on $[a, b] $ it is Riemann integrable on $[a, b] $ and therefore by definition of Riemann integral the desired equation in question holds for any choice on points $c_i$. – 2017-02-10
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    @James Reply with @ if you want me to see the reply. I am and am writing it up. – 2017-02-10
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    It appears that you also need to get familiar with proper definition of Riemann integral. Just have a look at the beginning of this answer http://math.stackexchange.com/a/1834341/72031 When a function is Riemann integrable then the desired equation holds irrespective of the choice of points $c_i$ as well as points $x_i$. – 2017-02-10

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If the limit exists then a proof that it is independent of which point you choose in the intervals is as follows:

Let $A^1_n$ be the sum for one choice and $A^2_n$ be the sum for the other (don't confuse the superscripts with exponents... nothing's being squared). Let's say the limit exists in one of the cases, say $A^1_n\to A.$ If we can show that $|A^1_n-A^2_n| \to 0$ then it follows that $A^2_n\to A$ as well. We can write prove this by writing $$ |A^2_n-A| = |A^2_n-A^1_n + (A^1_n-A)| \le |A_n^2-A_n^1| + |A^1_n-A| \to 0 + 0 = 0$$ where we used the triangle inequality.

So we want to prove $|A_n^2-A_n^1|\to 0.$ Now we will be more formal. By the definition of the limit, this means that for any $\epsilon>0$ we must be able to find an integer $N$ such that $|A_n^2-A_n^1| <\epsilon$ for all $n> N.$

So let $\epsilon$ be any positive real number. We will find an $N$ that satisfies the above definition.

We have $$ |A_n^1-A_n^2| = \left|\frac{b-a}{n}\sum_{j=1}^n(f(c_j^1)-f(c_j^2))\right| \le \frac{b-a}{n}\sum_{j=1}^n|f(c_j^1)-f(c_j^2)| $$ where $c^1_j$ are the first choice in each interval $j$ and $c^2_j$ are the second choice . We used the triangle inequality. We want to bound this expression.

We can use the uniform continuity property to bound that difference in the sum. Pick a $\delta$ so that $|f(x)-f(y)| < \frac{\epsilon}{b-a}$ whenever $|x-y|<\delta.$ Now observe that $c_j^1$ and $c_j^2$ are in the same partition interval and thus always within $(b-a)/n$ of each other, i.e. $$|c_j^1-c_j^2| < \frac{b-a}{n}.$$ So we can always make them closer together by picking $n$ large. In fact, if we pick $n >\frac{b-a}{\delta},$ then we have $|c_j^1-c_j^2| < \delta,$ which implies $|f(c_j^1)-f(c_j^2)| < \frac{\epsilon}{b-a}.$

So we let $N$ be any integer greater than $\frac{b-a}{\delta}.$ Then for $n>N,$ we have $$ |A_n^1-A_n^2| \le \frac{b-a}{n}\sum_{j=1}^n|f(c_j^1)-f(c_j^2)|\le \frac{b-a}{n} \sum_{j=1}^n \frac{\epsilon}{b-a} = \epsilon.$$

In words, the outline of the above proof is: "When the partition gets small, the values we pick to evaluate $f$ at get arbitrarily close together, which by uniform continuity means the function values over the entire partition get arbitrarily close together, which means the Riemann sums get arbitrarily close together."

Proving the limit exists requires further work.

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    The fact that the limit in question in independent of choice of $c_i$ is in fact part of definition of Riemann integral. I don't know why do we need to prove that. If $f$ is Riemann integrable then all Riemann sums for $f$ converge to the same value. – 2017-02-10
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    @ParamanandSingh Yes, but the exercise appears to be: "Using the definition of uniform continuity, prove that for a uniformly continuous function, in the case of evenly spaced partitions that the Riemann sum is independent of the tagged value." – 2017-02-10
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    I wonder what kind of textbooks give such pointless exercises. – 2017-02-10
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    @ParamanandSingh Thus the first paragraph of op ending with "is independent of the choice..." and the second starting with "prove this in the case that f is uniformly continuous" – 2017-02-10
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    @ParamanandSingh I agree it would be pointless if the answer were uniformly continuous -> riemann integrable -> special case. It's probably just to get the epsilons and deltas flowing... thus my very explicit and slow proof – 2017-02-10
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    Ok I think I got your point. The exercise makes sense if we are asked to show that for a uniformly continuous function the limit of that particular sum does not depend on choice of $c_i$. And this is what you have shown in the answer. But linking that sum with integral makes it pointless. – 2017-02-10
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    Here goes my +1 – 2017-02-10
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    @ParamanandSingh ahh I understand what you're saying about linking it with the integral. The question definitely still isn't perfectly posed... OP is new to proofs and is probably shaky on the definitions. Thanks. – 2017-02-10