Here is the question I am struggling with:

Assume $f \in R[0, 1]$ and consider the sequence $(y_n)$, where $$y_n =\frac{1}{n} \sum_{i=1}^n \; f\left(\frac{i}{n}\right) .$$ Show that $\lim y_n = \int_0^1 f$.

So I can show that $y_{n} = S(f:P)$ which is the Riemann sum, but I can't figure out what I should do next. I figure I have to use the definition of a limit and somehow morph it into the definition of a Riemann integral, but I can't be sure. Any tips?

The definition of Riemann integral I am using is; there is $L \in \mathbb R$ such that for every $\epsilon > 0$ there is $\delta >0$ such that if $P$ is any tagged partition of $I$ with $\|P\|< \delta$ then $|S(f:P)−L|< \epsilon$.

  • $\begingroup$ hint : what is the geometrical meaning of $y_n$ ? $\endgroup$ – Glougloubarbaki Dec 4 '11 at 23:43
  • $\begingroup$ I suppose you could say that $y_{n}$ is the area under the function, but I still am confused $\endgroup$ – rioneye Dec 4 '11 at 23:49
  • $\begingroup$ well not exactly, the area under the function is its integral. $y_n$ is what you obtain with the rectangular method of approximation, with a step of $1/n$. $\endgroup$ – Glougloubarbaki Dec 4 '11 at 23:56
  • 2
    $\begingroup$ There are several (equivalent) textbook definitions of a Riemann integral, but different definitions lead to different steps in solving the problem, and hence different answers to your question. How is Riemann integral defined in your class? $\endgroup$ – user1551 Dec 5 '11 at 0:01
  • 2
    $\begingroup$ I edited the post and added $n \to \infty$ under the limit symbol. Check that it is ok. // Please add the relevant information to the question, so that they are not buried under the comments. $\endgroup$ – Srivatsan Dec 5 '11 at 0:23

The problem statement says that $f$ is Riemann integrable, thus $S(f; P)\rightarrow0$ for when $\|P\|\rightarrow0$. So, all you have to do is to identify the partition $P$ (or strictly speaking, the sequence of partitions $P_n$) in your problem and show that $\|P\|\rightarrow0$.

| cite | improve this answer | |

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.