Suppose that the function $f:[0,1]\rightarrow\mathbb{R}$ is integrable. Prove that $$\lim\limits_{n\rightarrow\infty}\frac{1}{n}\left[f\left(\frac{1}{n}\right)+f\left(\frac{2}{n}\right)+\cdots+f\left(\frac{n-1}{n}\right)+f(1)\right]=\int_{0}^{1}f$$
As $f$ is integrable, there is a partition sequence $\{P_n\}\in[0,1]$, for all $n$ satisfies $$\lim\limits_{n\rightarrow\infty}U(f,P_n)=\lim\limits_{n\rightarrow\infty}L(f,P_n)=\int_{0}^{1}f$$ For each index $n$, we have $$L(f,P_n)\leq \frac{1}{n}\left[f\left(\frac{1}{n}\right)+f\left(\frac{2}{n}\right)+\cdots+f\left(\frac{n-1}{n}\right)+f(1)\right]\leq U(f,P_n)\qquad(1)$$ Thus, $$\lim\limits_{n\rightarrow\infty}L(f,P_n)\leq \lim\limits_{n\rightarrow\infty}\frac{1}{n}\left[f\left(\frac{1}{n}\right)+f\left(\frac{2}{n}\right)+\cdots+f\left(\frac{n-1}{n}\right)+f(1)\right]\leq\lim\limits_{n\rightarrow\infty}U(f,P_n)$$ This gives the result as desire.
Can someone check this solution? I am not sure this is right or not. I have a feeling that inequality $(1)$ isn't right. Thanks.
The expression $$\lim\limits_{n\to\infty}\dfrac1n \sum\limits_{i=1}^n f\left(\dfrac in\right)$$ by the substitution $\Delta x = \dfrac 1n$ can be presented as the integral sum $$\lim\limits_{\Delta x\to0}\sum\limits_{i=1}^n f\left(\xi_i\right)\Delta x$$ for the function $f(x)$ on the interval (0,1) for splitting it into equal intervals $\Delta x_i=\Delta x$ as the terms $\xi_i=i\Delta x$ of their respective sections on the interval (0,1).
By definition, an integral is a limit of a sequence of integral sums when $\Delta x_i \to 0$, if it exists, regardless of the partition and select the points $\xi$ at each of the sections.
So $$\lim\limits_{n\to\infty}\dfrac1n \sum\limits_{i=1}^n f\left(\dfrac in\right) = \int\limits_0^1f(x)dx$$ by definition