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One of the standard definitions of Riemann Integral is as follows:

Let $f$ be bounded on $[a, b]$. For any partition $P = \{x_{0}, x_{1}, x_{2}, \ldots, x_{n}\}$ of $[a, b]$ and any choice of points $t_{k} \in [x_{k - 1}, x_{k}]$ the sum $$S(P, f) = \sum_{k = 1}^{n}f(t_{k})(x_{k} - x_{k - 1})$$ is called a Riemann sum for $f$ over $P$ and tags $t_{k}$. The norm of $P$ denoted by $||P||$ is defined as $||P|| = \max_{k = 1}^{n}(x_{k} - x_{k - 1})$. A number $I$ is said to be Riemann integral of $f$ over $[a, b]$ if for any arbitrary $\epsilon > 0$ there exists a $\delta > 0$ such that $$|S(P, f) - I| < \epsilon$$ for all Riemann sums $f$ over any partition $P$ with $||P|| < \delta$. When such a number $I$ exists we say that $f$ is Riemann integrable over $[a, b]$ and we write $$I = \int_{a}^{b}f(x)\,dx$$

Suppose that if $f$ is in $R[0, 1]$. Does it necessarily follow that $$\lim_{n \to \infty}\frac{1}{n}\sum_{k = 1}^{n}f\left(\frac{k}{n}\right) = \int_{0}^{1}f(x)\,dx?$$

My attempt : Since $f$ is in $R[0,1]$, for any $\epsilon>0$, exist $\pi_0$ s.t for every $\pi$ which is refinement of $\pi_0$, $U(f,\pi)-L(f,\pi)<\epsilon$. If there exist $k$ s.t $\pi' = \{0,\frac{1}{k}, ..., 1\}$ is refinement of $\pi_0$, then it is done. But when if $\pi_0$ contains irrational point, then any $\pi_0$ cannot be refinement of $\pi_0$, Is there any other method to prove this?

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YES.

Set $$ P_n=\left\{t_0=0,t_1=\frac{1}{n},t_2=\frac{2}{n},\ldots,t_n=1\right\}. $$ Then $$ \frac{1}{n}\sum_{k=1}^nf\left(\frac{k}{n}\right)=S(P_n,f), \quad \text{with $t_k=\frac{k}{n}$}. $$

If we choose an arbitrary $\varepsilon>0$, then, according to the definition in the OP, there exists a $\delta>0$, such that if $\|P\|<\delta$, then $|S(P,f)-I|<\varepsilon$.

Hence if $n_0\in\mathbb N$, such that $\delta>\frac{1}{n_0}$, then $$ \left|\frac{1}{n}\sum_{k=1}^nf\left(\frac{k}{n}\right)-I\right|<\varepsilon, $$ for all $n\ge n_0$.

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