If $f$ is Riemann integrable on $[0,1]$ then $\lim\limits_{n\rightarrow\infty}\frac{1}{n}\sum f\left(\frac{k}{n}\right)=\int\limits_{0}^{1} f(x)dx$ [closed]

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

Thanks

Solution 1:

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.