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I want to show that if $f$ is integrable on the interval $[0,1]$ then $\lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^n f\left(\frac{k}{n}\right)=\int_0^1f$.

I am using the definition of integrability in the sense of Riemann-Stieltjes:

$$\forall\epsilon>0:\exists P:\forall Q\geq P:\left|S(Q_n,f,I)-\int_0^1f\right|<\epsilon$$

Where $P$ and $Q$ are partitions of $[0,1]$.

I thought about using the definition for the partitions:

$$Q_n=\left(0,\frac1n,\frac2n,\cdots,\frac{n-1}n,1\right)$$ $$c_k=\frac kn$$

Then we have:

$$\left|S(Q_n,f,I)-\int_0^1f\right|=\left|\frac1n\sum_{k=1}^nf\left(\frac kn\right)-\int_0^1f\right|$$

The problem is that I don't know how I can show that this tends to zero since a $P$ appears in the definition of integral and I don't know what to do with it.

Pentachito
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  • The sum does not make sense when $k>n$. The upper limit in sum should be $n$ instead of $\infty$ and then the result follows by definition of Riemann integral. – Paramanand Singh Dec 09 '20 at 07:48
  • You're right, I was wrong when I wrote the limits.

    I fixed it.

     – Pentachito Dec 09 '20 at 14:07
  • The duplicate does not really resolve this question. The answer there just restates that $\lim\limits_{n\rightarrow\infty}\frac{1}{n}\sum f\left(\frac{k}{n}\right)=\int\limits_{0}^{1} f(x)dx$ because $f$ is Riemann integrable. The question here has to do with the equivalence of two definitions of Riemann integrability. OP works with the definition based on the existence for any $\epsilon$ of a partition $P_\epsilon$ such that all Riemann sums $S(Q,f)$ are within $\epsilon$ of the integral when $Q$ is a "refinement" of $P_\epsilon$ ... – RRL Dec 09 '20 at 18:30
  • The other definition states that $S(P,f) \to \int_0^1 f $ as $|P| \to 0$, i.e., the partition norm tends to zero. In effect what is being asked is why are these definitions equivalent. That must be proved and the proof is not entirely trivial. – RRL Dec 09 '20 at 18:33
  • @RRL the question starts with "I want to show that if $f$ is integrable on the interval $[0,1]$ then ...". Integrability is given. So, it applies to all partitions satisfying the definition(s). $Q_n$ can be constructed to satisfy the definition(s), thus it becomes a particular case. – rtybase Dec 09 '20 at 19:02
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    Integrability with the definition given as $\forall\epsilon>0:\exists P:\forall Q\geq P:\left|S(Q_n,f,I)-\int_0^1f\right|<\epsilon$ and $Q\geq P$ means $Q$ is a refinement of $P$. With that as the definition you don't have immediately that a sum converges to the integral just because the norm converges to zero. It is intuitively obvious and it is true but it is not just a trivial consequence of the first definition, $Q$ refines $P$ means that every point in $P$ is also in $Q$. For uniform partitions $P_m$ is not necessarily a refinement of $P_n$ when $m > n$. – RRL Dec 09 '20 at 19:11
  • See here for example. – RRL Dec 09 '20 at 19:12
  • That is exactly my problem. I have to use the definition of integral with partitions and refinements. – Pentachito Dec 09 '20 at 19:22
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    @RRL, anyway, showing the equivalence of the two definitions sounds too complicated for this exercise, imho. – rtybase Dec 09 '20 at 19:23
  • My problem is that $P$ could be any kind of partition and so I don't have $Q_n$ to be a refinement. – Pentachito Dec 09 '20 at 19:25
  • @Pentachito, two definitions are equivalent (see the wiki link and the proof there). I'd choose the easiest one for the problem. – rtybase Dec 09 '20 at 19:27
  • Actually the only thing I want I have to show is: – Pentachito Dec 09 '20 at 19:27
  • $\forall\epsilon>0:\exists P:\forall Q\geq P:|S(Q, f, I)-\int_{[a,b]}f|<\epsilon$ implies $\forall\epsilon>0:\exists N\in\mathbb{N}:\forall n\in\mathbb{N}\geq N:|S(Q_n, f, I)-\int_{[a,b]}f|<\epsilon$ – Pentachito Dec 09 '20 at 19:30
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    @rtybase: I agree -- if you are allowed to use either definition then this becomes a simple, almost trivial, problem where $|P_n| = 1/n \to 0$ implies $S(P_n,f) \to \int_0^1 f$ by definition. – RRL Dec 09 '20 at 19:32
  • Duplicate https://math.stackexchange.com/questions/3940980/f-riemann-integrable-on-0-1/3941014#3941014 ? – zkutch Dec 10 '20 at 04:23

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