I need to prove that if $f$ is Riemann integrable $[0,1]$ then $$\lim\limits_{n\rightarrow\infty} \frac{1}{n}\sum f\left(\frac{k}{n}\right)= \int\limits_{0}^{1} f(x)dx$$

My idea is to recognize the right side defining the limit as a Riemann sum and use the uniform continuity.

I am not sure what to do next



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  • $\begingroup$ Yes, your idea should work. Start by splitting [0,1] into $n$ segments of length $\frac{1}{n}$ each and use this as a partition. $\endgroup$ – rtybase Jan 28 '17 at 23:25
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    $\begingroup$ If you're assuming that $f$ is Riemann integrable, then you shouldn't need uniform continuity—just recognizing that the expression inside the limit on the left-hand side is a Riemann sum should be enough. $\endgroup$ – Greg Martin Jan 29 '17 at 0:24
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    $\begingroup$ What's your definition of Riemann integrability? $\endgroup$ – zhw. Jan 29 '17 at 0:26
  • $\begingroup$ @rtybase, could you please explain it more? I started to prove it with 1/n<ẟ. Then I try to express this : 1/n∑f(k/n)-∫fdx $\endgroup$ – Pol Jan 30 '17 at 23:14

Take the following partition $$x_0=0<\frac{1}{n}<\frac{2}{n}<\frac{3}{n}<...<\frac{n-1}{n}<x_n=1$$ or $x_k=\frac{k}{n}$.

Now, the left Riemann sum is $$\sum_{k=1}^{n}f(x_{k-1})(x_k-x_{k-1})=\sum_{k=1}^{n}f\left(\frac{k-1}{n}\right)\left(\frac{k}{n}-\frac{k-1}{n}\right)=\frac{1}{n}\sum_{k=1}^{n}f\left(\frac{k-1}{n}\right)$$ And the right Riemann sum is $$\sum_{k=1}^{n}f(x_{k})(x_k-x_{k-1})=\sum_{k=1}^{n}f\left(\frac{k}{n}\right)\left(\frac{k}{n}-\frac{k-1}{n}\right)=\frac{1}{n}\sum_{k=1}^{n}f\left(\frac{k}{n}\right)$$ Combining this with the fact that $f$ is Riemann integrable (this is given) the limits of both left and right sums exist and are equal to $\int_{0}^{1}f(x)dx$. To understand this properly, it is important to understand the definition of Riemann integrability, which operates with "For $\forall \varepsilon >0$, there $\exists\delta$ such that for any partition ...", which also includes the one constructed above.


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