2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Why do you think that the inequality is not right? The proof seems OK to me! :) $\endgroup$ – H. R. Dec 22 '15 at 0:55
  • 1
    $\begingroup$ The problem is that you have taken $P_n$ to be any partition so there is no guarantee that (1) holds for finite $n$. Instead since $f$ is integrable you are free to pick the partition sequence: take it to be the equally spaced one and (1) holds. $\endgroup$ – Winther Dec 22 '15 at 0:57
  • $\begingroup$ @Winther equally spaced one sequence, do you regular partition sequence? $\endgroup$ – Simple Dec 22 '15 at 1:05
  • $\begingroup$ I mean the same as you have in the sum $\{0,1/n,2/n,3/n,\ldots,1\}$ $\endgroup$ – Winther Dec 22 '15 at 1:06
  • $\begingroup$ @Winther Thanks, that clear my confusion. $\endgroup$ – Simple Dec 22 '15 at 1:09
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.